Contains Duplicate

Here’s a simple implementation to check if the given array contains any duplicates:

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from typing import List

class Solution:
    def containsDuplicate(self, nums: List[int]) -> bool:
        return len(nums) != len(set(nums))

This code leverages the set data structure to remove duplicates and compares the length of the original list with the length of the set. If the lengths are not equal, that means there were some duplicates, and the function returns true. Otherwise, it returns false.

“Contains Duplicate” introduces the concept of using hash maps (or dictionaries in some languages) to track the occurrence of elements in an array. Here are a few simpler problems:

1. “344. Reverse String”: This problem deals with in-place reversal of a string or an array, which is a basic operation in many other problems.
2. “905. Sort Array By Parity”: Here, you sort an array based on a simple condition. This can help you get comfortable with array manipulation.
3. “1. Two Sum”: This problem is more difficult than the first two, but easier than “Contains Duplicate”. It requires you to find a pair of numbers that add up to a target, introducing you to the concept of using a hash map to speed up search operations.
4. “121. Best Time to Buy and Sell Stock”: This problem is a bit more complex, involving a common pattern in dynamic programming problems. However, it can be solved with a simpler greedy algorithm, making it a good stepping stone.

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# @param {Integer[]} nums
# @return {Boolean}
def contains_duplicate(nums)
    duplicate = false
    set = Set.new

    nums.each do |n|
        if set.include?(n)
            return true
        else
           set.add(n) 
        end
    end
    return false
end

Clarification Questions

What are the clarification questions we can ask about this problem?

Identifying Problem Isomorphism

“Contains Duplicate” can be mapped to “Single Number”.

Both problems deal with the occurrence of numbers in an array. “Contains Duplicate” asks if any value appears at least twice in the array, while “Single Number” is looking for a number that appears exactly once in the array.

In both cases, you could use a hash map to keep track of the occurrence of numbers. For “Contains Duplicate”, you check for any number that appears more than once. In “Single Number”, you look for a number that appears exactly once.

“Contains Duplicate” is simpler as it only asks to identify whether a duplicate exists or not, while “Single Number” asks to identify the specific number that doesn’t have a duplicate.

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# Time complexity: O(n)
# Space complexity: O(n)
class Solution(object):
    def containsDuplicate(self, nums):
        hset = set()
        for idx in nums:
            if idx in hset:
                return True
            else:
                hset.add(idx)

Practice Phase Template

Here is a language-agnostic Practice Phase Template for the given solution:

1. Concepts

   • Concept 1: Understanding the structure of arrays.
   • Concept 2: Understanding of sets and their properties.
   • Concept 3: Understanding how to iterate through an array.
   • Concept 4: Understanding how to add elements to a set.
   • Concept 5: Understanding the lookup operation in a set.

2. Coding Drills

   • Drill 1: Write a code snippet to create and manipulate an array.
   • Drill 2: Write a code snippet to create and manipulate a set.
   • Drill 3: Write a code snippet to iterate through an array.
   • Drill 4: Write a code snippet to add elements to a set.
   • Drill 5: Write a code snippet to perform a lookup operation in a set.

3. Skills

   • Skill 1: Ability to create and manipulate arrays.
   • Skill 2: Ability to create and manipulate sets.
   • Skill 3: Ability to iterate through arrays.
   • Skill 4: Ability to add elements to a set.
   • Skill 5: Ability to perform a lookup operation in a set.

4. Practice Objectives

   • Objective 1: Write a function that checks if an array contains duplicate values using set operations.

Remember that it’s essential to understand each concept thoroughly before moving onto the next and continually practicing coding drills to develop your skills further. Practice objectives give you a goal to strive for and can help gauge your progress as you continue to learn and develop your skills.

Targeted Drills

1. Coding Drill 1: Write a code snippet to create and manipulate an array.

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# Create an array
nums = [1, 2, 3, 4, 5]

# Access elements in the array
print(nums[0])  # Prints: 1

# Modify an element of the array
nums[0] = 10
print(nums)  # Prints: [10, 2, 3, 4, 5]

2. Coding Drill 2: Write a code snippet to create and manipulate a set.

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# Create a set
hset = {1, 2, 3, 4, 5}

# Add an element to the set
hset.add(6)
print(hset)  # Prints: {1, 2, 3, 4, 5, 6}

# Remove an element from the set
hset.remove(1)
print(hset)  # Prints: {2, 3, 4, 5, 6}

3. Coding Drill 3: Write a code snippet to iterate through an array.

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# Iterate through the array and print each element
nums = [1, 2, 3, 4, 5]
for num in nums:
    print(num)

4. Coding Drill 4: Write a code snippet to add elements to a set.

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# Create an empty set
hset = set()

# Add elements to the set
hset.add(1)
hset.add(2)
hset.add(3)

print(hset)  # Prints: {1, 2, 3}

5. Coding Drill 5: Write a code snippet to perform a lookup operation in a set.

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# Create a set
hset = {1, 2, 3, 4, 5}

# Perform a lookup operation
if 1 in hset:
    print("Element found.")
else:
    print("Element not found.")

Note: Set lookup in Python is a fast operation, which takes constant time i.e., O(1). It’s one of the reasons why sets are used in problems requiring elimination of duplicates or membership tests.

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class Solution(object):
    def containsDuplicate(self, nums):
        hset = set()
        for idx in nums:
            if idx in hset:
                return True
            else:
                hset.add(idx)

Problem Classification

Domain Classification: This problem belongs to the domain of array processing and mathematical algorithms in computer science.

Problem Analysis:

1. Input: The input to this problem is an array of integers nums. The length of this array can be from 1 to 105. Each integer in the array will be in the range from -109 to 109.

2. Output: The output is a Boolean value - true if any value appears at least twice in the array, and false if every element is distinct.

3. Task: Given the array nums, the task is to determine if there are any duplicate values in it.

Components:

1. Array: An array of integers is the primary data structure in this problem.
2. Frequency Count: To identify duplicates, one approach would be to count the frequency of each number in the array.
3. Boolean Return: The problem requires a true or false result, indicating the presence or absence of duplicates.

Classification: This problem can be classified as a frequency count problem. It requires determining the frequency of each number in the array, and then checking if any number appears more than once. It can also be viewed as a problem of determining the uniqueness of elements in an array.

Problem Simplification and Explanation

This problem is asking us to figure out whether there are any repeated numbers in a list of numbers (the array). An array in programming is like a line of storage boxes, where each box holds a number (the elements of the array).

If we take this to a real-life scenario, think of it like a hat check at a party. When a guest arrives, they hand over their hat to the attendant who stores it in a box and gives them a ticket with the box number. Now, imagine that instead of a unique hat for each guest, they hand in a number. Your task is to find out if any two guests handed in the same number.

To do this, you could keep a record (like a frequency count) of each number given to you (you would write down each number and a tick next to it each time it was handed in). If any number gets more than one tick, then you know that it has been repeated. That’s essentially what we’re doing in this problem.

Here’s how this translates to the problem:

1. Array: This is your list of numbers, or the “hats” handed to you by guests.
2. Frequency Count: This is your record of how many times you’ve seen each number.
3. Boolean Return: If you see a number more than once (a number has more than one tick next to it), you return ’true’ (meaning yes, there’s a repeat). If not, you return ‘false’.

The key concept here is counting the frequency of each number (or element) in the list (or array). By keeping track of how many times we see each number, we can easily tell if any number has been repeated. This frequency count concept is a common one in programming and is used in many different problems.

Constraints

This problem statement has a few characteristics and constraints that can be leveraged to make solving it more efficient:

1. Distinct values: The problem is asking us to return true if any value appears at least twice in the array. This means we are interested in the uniqueness of the numbers in the array, which immediately brings to mind using a data structure that can efficiently handle unique elements, like a Set or a Dictionary (Hashmap).

2. Size of the array: The constraint “1 <= nums.length <= 105” tells us that the array can be quite large, so we need to find a solution with a good time complexity to avoid a timeout error. Ideally, we should aim for a time complexity of O(n) or better.

3. Value of the numbers: The constraint “-109 <= nums[i] <= 109” indicates the range of the values in the array is quite large. However, since we’re only concerned with whether or not values are duplicated, the actual value of the numbers doesn’t affect our approach - we’re not performing any numeric operations on them.

4. Patterns: We are looking for duplicate numbers. If we iterate through the array and keep a record of the numbers we’ve seen, any number we encounter that is already in our records is a duplicate. This points towards using a data structure that allows fast lookup like a Set or a Dictionary (Hashmap).

In summary, an efficient approach could be to iterate through the array and add each element to a Set or a Hashmap. If we encounter a number that is already in our Set or Hashmap, we know it’s a duplicate and return true. If we finish iterating through the array without finding any duplicates, we return false. This would give us a time complexity of O(n), which is quite efficient.

Thought Process

Looking at the problem statement, we understand that the task is to determine whether any number appears more than once in a given list of numbers.

This suggests that we need to somehow keep track of the numbers we’ve already encountered as we traverse the list. There are a few data structures we could use to do this, such as a list, a set, or a dictionary. However, a set would be the most efficient because it has O(1) time complexity for insertion and lookup.

The problem doesn’t ask for the duplicate values, it just asks whether there is a duplicate or not. This means as soon as we encounter a duplicate, we can return True without having to go through the rest of the list. If we go through the entire list without finding a duplicate, we return False.

With these insights, the problem can be solved using the following Python code:

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def containsDuplicate(nums):
    # Create an empty set to keep track of numbers we've already seen
    seen = set()
    # Iterate through the list
    for num in nums:
        # If the number is already in the set, return True
        if num in seen:
            return True
        # Otherwise, add the number to the set
        seen.add(num)
    # If we've gone through the entire list and haven't found a duplicate, return False
    return False

The above solution is efficient with a time complexity of O(n) where n is the length of the list. The space complexity is also O(n) for the set used to track the numbers already seen. This solution exploits the properties of set data structure for efficient lookup and uniqueness to solve the problem.

Solution Approach and Analysis

Approaching this problem involves a simple strategy. If you can imagine a big bag of marbles, each with a number on it, the problem is asking us if there are any two marbles in the bag with the same number.

Here’s how I would approach this:

1. Create a ‘record’: Imagine creating a separate, smaller bag where we store every unique marble that we’ve seen so far. This can be thought of as a record or a tracking system. In computer science terms, we call this a Hash Set or Hash Table.

2. Go through the marbles: Now, we start pulling marbles out of the big bag one by one, and for each marble, we check if it’s already in our smaller bag (i.e., the Hash Set).

3. Check for duplicates: If a marble we pull out is already in our smaller bag, that means we have found a duplicate, and we return True. If the marble is not in the smaller bag, we put it in there and move on to the next marble.

4. No duplicates: If we manage to pull out all the marbles without finding any duplicates, we return False.

Now, in terms of how specific changes to the problem’s parameters would affect the solution, consider these scenarios:

• If the array is empty or contains only one element: These are edge cases. An empty array or an array with just one element can never contain a duplicate, so we should return False right away.

• If the array contains all the same elements: In this case, as soon as we check the second element, we will find a duplicate and return True.

Let’s run through an example:

Suppose we have an array of [3, 1, 4, 1, 5]. We create an empty Hash Set. Starting from the first element, we add 3 to the set. Then, we move to the next element 1, which is also not in the set, so we add it. Next, we move to 4, which is not in the set, so we add it. Next, we move to the second 1. We check our set and see that 1 is already there, which means we’ve found a duplicate! Therefore, we return True.

This approach is fairly straightforward and efficient, with a time complexity of O(n) and a space complexity of O(n), where n is the length of the array.

Identification of Applicable Theoretical Concepts

Here are some key mathematical or algorithmic concepts that can help us simplify the problem:

1. Set Theory: The problem essentially asks us to identify if there are any repeating elements in a set of numbers. This is a fundamental concept in Set Theory, where a set is a collection of distinct objects.

2. Hashing: A hash function maps data of arbitrary size to a fixed size. In the context of this problem, we can use a Hash Table (or a Hashmap in some languages) to store our array elements. The beauty of a Hash Table is that it allows for efficient search, insert, and delete operations, all of which can be done in average constant time O(1).

3. Linear Time Complexity: The most efficient solution to this problem will have a time complexity of O(n), where n is the length of the array. This concept comes from the study of algorithms, particularly in the field of Big O notation, which is used in Computer Science to describe the performance or complexity of an algorithm.

4. Space Complexity: While trying to solve this problem efficiently, we also need to consider the space we’re using. Using a Hash Table for finding duplicates improves time complexity but also increases the space complexity. In this case, since the constraint on the size of the array is quite high (up to 10^5), the space complexity would also be O(n). It’s a trade-off we are making for a more time-efficient solution.

These concepts form the crux of the problem and using them will help solve it in a simpler and more efficient manner.

Coding Constructs

1. High-level problem-solving strategies: This code is using the technique of “Hashing” to solve the problem. The “Set” data structure, which internally uses a hash table, is used for storing the elements in the list. It takes advantage of the property of set that it does not allow duplicate values and it has constant time complexity for insertion and lookup.

2. Explanation to a non-programmer: This code is like a gatekeeper checking tickets at a concert. The gatekeeper keeps track of all tickets that have been seen. When a new person comes in, if their ticket has been seen before, the gatekeeper knows there’s a problem. In this case, our code is the gatekeeper, the people are the elements in our list, and the tickets are the values of these elements.

3. Logical constructs: The logical constructs in this code include a loop and conditional statements. The loop is used to iterate through all the elements in the list. The conditional statement (the if statement) is used to check if an element is already in the set.

4. Algorithmic approach in plain English: Start with an empty set. Go through the list one by one. For each element, check if it’s already in the set. If it is, that means it’s a duplicate, so we can stop and say “Yes, there are duplicates”. If it’s not in the set, add it to the set and move on to the next element. If we go through the entire list and never find a duplicate, then we can say “No, there are no duplicates”.

5. Key steps or operations:

   • Iterate over the elements in the list.
   • For each element, check if it is already in the set.
   • If it is, return True, indicating that a duplicate has been found.
   • If it’s not, add it to the set.
   • If we finish checking all elements without finding any duplicates, return False.

6. Algorithmic patterns or strategies: The code uses the “Hashing” strategy along with the “Early Exit” pattern. Hashing is used to keep track of elements we’ve seen in constant time by storing them in a set. Early Exit is used to stop the execution as soon as we find a duplicate, without having to check the rest of the elements.

Language Agnostic Coding Drills

1. Concepts Identification:

   • Variables: The basic building block where we store our data.
   • Lists: A collection that allows storing multiple values in an ordered manner.
   • Set: A collection used to store unique elements.
   • Looping: Iterating over the list to access every element.
   • Conditional statements: Making decisions based on conditions.
   • Function definition: Creating a self-contained block of code to perform a specific task.

2. List of Concepts in order of increasing difficulty:

   • Variables: Basic construct in programming to store and manipulate data. It’s the simplest concept as it requires just the knowledge of data types.
   • Lists: A bit more complex as they are data structures holding multiple values and require understanding of indexing to access the data.
   • Looping: This requires understanding of flow control and how to repeat certain operations.
   • Conditional statements: Involves decision-making based on certain conditions. It requires a good understanding of operators and comparisons.
   • Set: An unordered collection of unique elements. It requires understanding of data structures and their properties. In this case, the understanding that a set only stores unique values and the operations associated with it like add and lookup.
   • Function definition: It involves encapsulating a block of code under a function to reuse it and make code more organized. It requires understanding of parameters, return statement and scope of variables.

3. Problem-solving Approach:

   • Start by defining a function that takes a list as input.
   • Within the function, create an empty set for storing the elements.
   • Iterate over the list. For each element in the list, check whether it is already present in the set.
   • If the element is present, that indicates a duplicate and we can return True from our function. This leverages the property of the set that it contains unique values.
   • If the element is not in the set, add the element to the set. This uses the add operation in the set.
   • If we finish iterating over the list without finding any duplicates, return False.
   • This entire approach leverages several basic programming concepts like variables, loops, conditionals, set data structure, and function definition. It also uses a problem-solving strategy called “Early Exit” where we stop the function as soon as we find a duplicate.

Targeted Drills in Python

1. Python-based Coding Drills:

   • Variables:

     1 2	x = 10 print(x)

   • Lists:

     1 2	myList = [1, 2, 3, 4, 5] print(myList[2]) # Accessing list element at index 2

   • Looping:

     1 2	for i in range(5): print(i)

   • Conditional statements:

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     x = 10 if x > 5: print("x is greater than 5") else: print("x is less than or equal to 5")

   • Set:

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     mySet = set() mySet.add(1) mySet.add(2) print(mySet) # Output: {1, 2}

   • Function definition:

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     def greet(name): print(f"Hello, {name}!") greet("John")

2. Problem-specific concept - checking if a value is in a set:

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   mySet = {1, 2, 3} if 2 in mySet: print("2 is in the set") else: print("2 is not in the set")

   This drill is crucial because the problem requires checking for duplicate values, which can be quickly done using a set.

3. Integrating all the concepts:

   We can integrate these concepts by defining a function that uses a set to track unique values. The function iterates over the input list and, for each element, checks if it is already in the set. If it is, the function returns True (indicating a duplicate). If it isn’t, the function adds the element to the set. If the function goes through the entire list without finding any duplicates, it returns False.

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   def containsDuplicate(nums): # Create an empty set numSet = set() # Iterate over the list for num in nums: # If the number is already in the set, we have a duplicate if num in numSet: return True # Add the number to the set numSet.add(num) # If no duplicates found, return False return False

   This solution integrates all the Python coding concepts we’ve drilled into a complete solution for the problem.

Problem Classification

1. Data Structures: The problem involves handling a collection of data items, which is typically managed using data structures. In this case, an array or list is given as input.

2. Set Operations: The problem asks to check if there are any duplicate elements in the list, a task that can be easily accomplished using a set.

3. Search Problem: The problem essentially involves searching for duplicate items in a list, a common computational problem.

Language Agnostic Coding Drills

1. Understanding the Problem Statement: The first step is to understand the problem. The problem is asking to determine if a list (array) contains any duplicates.

2. Concept of Data Structures: An understanding of what data structures are and how they can be used is crucial. In this problem, we will be using a set data structure.

   Drill: Try creating different types of data structures (like lists, sets, dictionaries) and perform basic operations on them.

3. Set Data Structure: In the context of this problem, it is important to understand how a set works. A set is a collection of distinct elements. They cannot contain duplicates, which makes it an ideal choice for this problem.

   Drill: Write code to create a set, add elements to the set, and check if a certain element is in the set.

4. Loops and Iterations: This problem also requires looping over the elements of the given list.

   Drill: Write code that loops over a list and performs any operation, such as printing the elements.

5. Conditionals: You need to know how to use if-else conditions to check whether a number is already in the set or not.

   Drill: Write code that uses if-else statements to check certain conditions.

6. Combining Concepts: Once all these individual parts are understood, it’s about putting them together. You would create an empty set, then iterate over the list. For each number, you would check if it is in the set or not. If it is, that means the list contains duplicates and you return True. If not, you add the number to the set and continue.

This process of breaking down the problem makes it easier to understand and solve. The individual drills for each of the key components not only help in solving this problem, but also sets the groundwork for tackling similar problems.

Targeted Drills in Python

1. Understanding the Problem Statement

   No coding required for this step.

2. Concept of Data Structures

   Drill: Create a list, a set, and a dictionary, then add some elements and print them.

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   # List lst = [1, 2, 3] print(lst) # Set st = {1, 2, 3} print(st) # Dictionary dct = {"a": 1, "b": 2, "c": 3} print(dct)

3. Set Data Structure

   Drill: Create a set, add elements, and check if a certain element is in the set.

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   st = set() st.add(1) st.add(2) st.add(3) print(st) if 2 in st: print("Element exists in the set")

4. Loops and Iterations

   Drill: Loop over a list and print each element.

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   lst = [1, 2, 3, 4, 5] for element in lst: print(element)

5. Conditionals

   Drill: Use if-else statements to check a condition.

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   a = 5 if a > 3: print("a is greater than 3") else: print("a is not greater than 3")

6. Combining Concepts

   Drill: Check for duplicate elements in a list.

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   lst = [1, 2, 3, 2, 1] st = set() for element in lst: if element in st: print("Duplicates found") break else: st.add(element) else: print("No duplicates found")

Remember, practice makes perfect. These drills are designed to reinforce your understanding of the fundamental concepts and make you more comfortable with Python coding. You can modify these drills or create your own to practice further.

Q&A

Can you solve it in N time without using extra memory? I.e. no hash set or dictionary and not a n^2 solution where you check every other element.

If the array is unsorted and we are not allowed to use any additional memory, it is impossible to find duplicate elements in linear time without modifying the array. The reason is that without keeping a record of the elements we have seen (which would require extra memory), we would have to check each element against all others (resulting in a quadratic runtime).

However, there is a solution with O(N log N) time complexity and O(1) space complexity, assuming that we are allowed to modify the array. The idea is to sort the array first and then look for duplicates. Here is the Python code for that:

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def containsDuplicate(nums):
    nums.sort()
    for i in range(1, len(nums)):
        if nums[i] == nums[i-1]:
            return True
    return False

The sorting operation is O(N log N), and the subsequent loop is O(N), so the total time complexity is O(N log N). However, Python’s built-in sort function list.sort() uses a sorting algorithm (TimSort) that requires a small amount of additional memory, so if you are not allowed to use any additional memory at all, this solution might not be allowed. In other languages, you might be able to use an in-place sorting algorithm like heapsort to keep the space complexity at O(1).

return !(nums.size()==set(nums.begin(),nums.end()).size());

This line of code is a concise way of checking for duplicates in a list or array nums. Here’s how it works:

1. set(nums.begin(),nums.end()): This creates a set from the elements in nums. A set is a data structure that doesn’t allow duplicates. So if nums has any duplicate elements, they’ll be removed in this new set.

2. nums.size()==set(nums.begin(),nums.end()).size(): This checks if the size of nums is equal to the size of the newly created set. If there were any duplicates in nums, the size of the set would be smaller, and this condition would be False. If there were no duplicates, the sizes would be equal, and this condition would be True.

3. !(nums.size()==set(nums.begin(),nums.end()).size()): The ! operator negates the result of the condition. So this expression is True if there were duplicates in nums (i.e., nums has a larger size than the set), and False otherwise.

This is a Python code equivalent to your one-liner:

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def contains_duplicate(nums):
    return len(nums) != len(set(nums))

This code also has a time complexity of O(N), where N is the size of the input list nums, and it uses O(N) extra space to create the set. The set() function and the len() function each take O(N) time, so the total time complexity is O(N) + O(N) = O(N).

return len(nums) != len(set(nums))

This line of Python code is checking if there are any duplicate elements in the input list nums. Here’s how it works:

1. set(nums): This part of the code creates a set from the list nums. A set in Python is a collection of unique elements, meaning it does not allow for duplicates. If there are any duplicate elements in nums, they would be removed in the new set.

2. len(nums) != len(set(nums)): This part of the code checks if the length of the original list nums is different from the length of the set. If they are different, it means there were duplicates in nums (because the set removed them and therefore is shorter). So, this part of the code will return True if there are duplicates and False otherwise.

So, the overall function return len(nums) != len(set(nums)) is a simple one-liner to check for duplicates in a list.

Please note that while this operation is efficient, it does not meet your constraint of not using extra memory because the creation of a set from the list nums requires additional space.

I can understand why Arrays.sort() is a better option in terms of space but how usingArrays.Sort() (3ms) is giving faster results than using HashSet (5ms) ?

The performance difference you’re observing between Arrays.sort() and HashSet in Java could be attributed to several factors:

1. Nature of the data: If the array has lots of duplicates, adding elements to a HashSet might lead to more collisions and consequently, more time complexity to resolve these collisions. This would result in slower performance compared to sorting.

2. CPU Cache Utilization: CPUs have a small amount of very fast cache memory. When data is accessed sequentially (like in sorting algorithms), the CPU can prefetch data into cache, making the access faster. This type of optimization is generally more beneficial for sorting (which is memory access pattern friendly due to its sequential nature) compared to hash based data structures (where memory accesses are scattered).

3. Overhead of Object Wrapping: In Java, HashSet handles primitive types (like int) as their wrapper object equivalents (like Integer). The overhead of object wrapping and unwrapping can add to the execution time.

4. Overhead of Hashing: Hashing each element to place it in the HashSet can also introduce overhead, particularly for larger data sizes.

It’s worth noting that while Arrays.sort() might be faster for some specific inputs and conditions, HashSet has a constant average time complexity which can be significantly faster for certain other types of input. Therefore, the optimal choice depends on the specific context and requirements of the problem.

Remember that micro-benchmarks like these may not always give an accurate reflection of real-world performance due to the factors mentioned above as well as others like Java’s Just-In-Time (JIT) compiler optimizations, garbage collection, etc. Therefore, always consider the trade-offs and choose the approach that best fits your use case.

How are hashset based solution slower than the sorting based solutions in cpp?

There are a few reasons why a HashSet-based solution might appear slower than a sorting-based solution in C++, particularly for small datasets:

1. Memory Access Pattern: Sorting algorithms often have predictable, sequential memory access patterns. This allows for efficient use of the CPU cache, reducing the time required to fetch data from memory. In contrast, hash tables (such as unordered_set in C++) have unpredictable, scattered memory access patterns, which lead to less efficient use of the CPU cache.

2. Hashing Overhead: Each insertion into a hash set requires computation of a hash function, which can take a significant amount of time, particularly if the hash function is complex.

3. Collision Handling: If the hash function results in a collision (i.e., two or more elements have the same hash), the hash set must use a resolution strategy to handle the collision, which adds additional computation time.

4. Allocation Overhead: Hash sets dynamically allocate memory as elements are inserted. Each allocation requires time, which can slow down the operation.

For larger datasets, the constant time complexity of hash set operations can often outweigh these costs, making the hash set-based solution faster than the sorting-based solution. However, for small datasets, the overhead of using a hash set can make it slower than simply sorting the array.

As always, it’s important to consider the characteristics of your specific problem and dataset when choosing an algorithm. In some cases, the increased speed of hash set operations for large datasets may be worth the extra memory usage. In other cases, the lower memory usage and potentially faster execution time for small datasets may make sorting a better choice.

My original solution was to use an ArrayList to see if there has been a duplicate. My solution passed all TestCases but I got a TIme Limit Exceeded. I looked at the solution and saw it used a Set so I changed my ArrayList to a Set and it worked. Why does a Set work but not an Array List?

The primary difference between a Set and an ArrayList (or a List in general) lies in the time complexity for search operations.

When you use an ArrayList, to check if an element already exists in the list, you essentially need to traverse the list until you find the element. In the worst-case scenario (where the element doesn’t exist or is at the end of the list), this operation takes linear time, i.e., O(n), where n is the number of elements in the list.

On the other hand, a Set (specifically a HashSet) is designed for efficient lookup operations. It uses a technique called hashing to store and retrieve elements in constant time, i.e., O(1), irrespective of the number of elements in the set.

So, in your problem where you need to check for duplicates, a Set significantly outperforms an ArrayList as the dataset grows large. That’s why your solution using Set worked within the time limits but the one using ArrayList didn’t.

In general, if you need to check for the existence of elements frequently, a Set or a HashMap is usually a better choice than a List or an ArrayList.

dups={} for idx, num in enumerate(nums): if num in dups: return True

       dups[idx]=num

return False it works but fails on case like [3,3] should return True but returns False

The problem with your code lies in this line: dups[idx]=num. In your current code, you’re using the index (idx) as the key in your dictionary dups and the number (num) as the value. This means that when you’re checking if num in dups, you’re actually checking if num is one of the keys in the dictionary, which it is not since the keys are the indices.

You should actually be using the number itself (num) as the key and the index (idx) as the value (or you could just use any value if you don’t need the index). This way, you’re correctly checking if the number is already in the dictionary.

Here’s how you can fix your code:

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dups={}
for idx, num in enumerate(nums):
    if num in dups:
        return True
    dups[num] = idx  # num is the key, idx is the value
return False

Now the code will correctly return True for the input [3,3].

The constraints in a problem statement usually define the range and limitations of the input parameters. These constraints are meant to help us understand the bounds within which our solution should work, which in turn can help inform our choice of algorithm, data structures, and optimization techniques.

The constraint you mentioned, “integer”, typically implies a 32-bit signed integer in the context of coding problems. However, it’s important to note that the maximum number of elements in an array or list isn’t solely determined by the type of the elements. Rather, it’s usually a separate constraint that might be given in the problem statement (e.g., “1 <= nums.length <= 105”).

If no such constraint is given, it’s typically safe to assume that the input could be quite large. In a real-world setting, it’s always a good idea to clarify these kinds of details if they’re not specified. For a coding problem, the test cases should be designed to check whether your solution works within the expected bounds. If your solution is exceeding the time limit, it may be necessary to optimize it for larger inputs.

How to build a bitmap to solve this problem? Thanks!

Using a bitmap (also known as bitset) is a great way to solve this problem efficiently, especially if the input range is known and not excessively large. A bitmap uses a bit array where each boolean value is stored in a single bit. Here’s a general outline of how you could apply this to the problem of determining if an array contains any duplicates.

1. Initialize a bitmap with enough bits to cover the entire range of possible input values. For example, if you know your inputs are always between 0 and 10^9, you might create a bitmap with 10^9 bits (though in practice this would be impractically large).

2. Iterate over the array. For each number, check if its corresponding bit in the bitmap is set. If it is, then that number has appeared before and you can return True.

3. If the bit was not set, then this is the first time we’ve seen this number. Set the corresponding bit in the bitmap.

4. If you’ve gone through the entire array without finding any duplicates, return False.

Here’s some Python-like pseudocode illustrating this approach:

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def contains_duplicate(nums):
    bitmap = BitArray(10**9)  # A hypothetical BitArray class

    for num in nums:
        if bitmap[num]:
            return True
        else:
            bitmap[num] = True

    return False

Please note that this pseudocode uses a hypothetical BitArray class for the sake of clarity. Python doesn’t have a built-in bit array type, but you could use a library like bitarray or use an integer as a bit array if the range of inputs is small.

Remember that a bitmap would be impractical for this problem if the range of inputs is very large (like 10^9), as it would require a prohibitive amount of memory. This technique works best when the range of inputs is relatively small compared to the number of inputs.

Also, keep in mind that this solution assumes that all the input numbers are non-negative. If negative numbers are possible, you’d need to offset the numbers to make them non-negative before using them to index into the bitmap. For instance, if inputs are in the range -10^9 to 10^9, you can add 10^9 to each input to ensure they are all non-negative.

can anyone please provide the c code for this which is easier to understand as I’m just a begineer.

Here is a simple implementation in C, using sorting to check for duplicates:

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#include <stdio.h>
#include <stdlib.h>

int compare (const void * a, const void * b) {
    return (*(int*)a - *(int*)b);
}

int containsDuplicate(int* nums, int numsSize) {
    qsort(nums, numsSize, sizeof(int), compare);
    for(int i = 1; i < numsSize; i++) {
        if(nums[i] == nums[i-1]) {
            return 1;
        }
    }
    return 0;
}

int main() {
    int nums[] = {1,2,3,4,1};
    int numsSize = sizeof(nums)/sizeof(nums[0]);
    if (containsDuplicate(nums, numsSize)) {
        printf("Contains duplicate\n");
    } else {
        printf("No duplicates\n");
    }
    return 0;
}

This program first sorts the array using the qsort function. Then, it iterates through the array and checks if the current number is the same as the previous number. If it is, then it returns 1 (true), indicating that a duplicate was found. If it makes it through the entire array without finding a duplicate, it returns 0 (false).

You can run this program with different inputs to see how it works. Just change the nums array in the main function.

Note: C does not support dynamic arrays as in Python or other higher-level languages, so you’ll need to know the size of the array upfront.

Guys, want your opinion on this solution, how can i improve it? The time complexity is 82ms

var containsDuplicate = function(nums) { const n = nums.length; let hasDuplicates = false; for(i=0;i<n;i++) { let j = i + 1; if( j >= n){ j = 0; } if(nums[i] === nums[j]) { hasDuplicates = true; break; } if( j === 0){ break }

	 }
   	
	return hasDuplicates 

};

Your code currently checks if an element is the same as the next element in the array, which won’t correctly determine if the array contains duplicates unless they’re adjacent in the array.

Moreover, your solution runs in O(n^2) time because for every element in the array, it compares it to every other element in the array. For large inputs, this could be very inefficient.

Here is an improved version of your function using a JavaScript Set. A Set in JavaScript is similar to a Set in mathematics. It is a collection of unique elements. The advantage of using a Set over an Array is that searching for an element in a Set is O(1), whereas, in an Array, it is O(n).

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var containsDuplicate = function(nums) {
    const uniqueNums = new Set();

    for(let i = 0; i < nums.length; i++) {
        if (uniqueNums.has(nums[i])) {
            return true;
        }
        uniqueNums.add(nums[i]);
    }

    return false;
};

In this function, we create a Set uniqueNums. For each number in nums, we check if it’s already in uniqueNums. If it is, we immediately return true because we’ve found a duplicate. If it’s not, we add it to uniqueNums. If we make it through the entire array without finding a duplicate, we return false.

This function runs in O(n) time, which is more efficient than the original version. The reason it’s more efficient is that both inserting an element into a Set and checking if a Set contains an element are O(1) operations. So we perform 2n O(1) operations, which makes the total time complexity O(n).

I used pre computation, but I am getting one test case wrong when I run the same code in GDB compiler its giving the correct ans

const int N = 1e5 + 10; int hsh[N]; class Solution { public: bool containsDuplicate(vector& nums) { for(auto num:nums){ hsh[num]=hsh[num]+1; } for(long long int i=0;i<N;i++){ if(hsh[i]>1){ return 1; }} return 0;} };.

Is it because the global array is not initialized by zero in this environment or something

In C++, global and static variables are initialized to zero by default. However, the problem with your code lies in the line where you access the array hsh[num]. Here, num can be a negative integer because the problem statement does not constrain the numbers to be positive. Since arrays in C++ are zero-indexed and do not support negative indices, you are likely experiencing out-of-bounds access, leading to undefined behavior.

Another thing is, you’re assuming that nums array won’t contain a number greater than 1e5 + 10, while the constraints say that elements in nums can go up to 1e9, which could potentially cause an out-of-bounds error.

Here is a modified version of your code that avoids these issues:

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class Solution {
public:
    bool containsDuplicate(vector<int>& nums) {
        unordered_map<int, int> hsh;
        for(auto num : nums){
            hsh[num]++;
            if (hsh[num] > 1) {
                return true;
            }
        }
        return false;
    }
};

In this code, I used std::unordered_map instead of an array for hsh. An std::unordered_map in C++ is a hash table that allows for fast O(1) access to individual elements directly by their values. This way, we can handle negative numbers and avoid potential out-of-bounds errors, which might be the reason for the incorrect results you were getting with your original code.

I’m using C and my code doesn’t work in TestCase 2. I’m trying to use a Hashset to complete this.

int hash[numsSize]; for(int i = 0; i < numsSize; i++){ hash[i] = 0; } for(int i = 0; i < numsSize; i++){ for(int j = 0; j < numsSize; j++){ if(nums[i] != hash[j]) { hash[i] = nums[i]; } else { return true; } } } return false; }

There’s a logical mistake in your code which might cause an incorrect output.

You’re trying to use the hash array like a hashtable, where you store the value of the nums array in the corresponding index of the hash array. But the way you have structured your loop leads to errors.

You’re going through the hash array for every element in the nums array. Then you’re comparing the nums array element with each element in the hash array. If it’s not found, you’re adding it to the hash array in the current index from the outer loop. This might lead to overwriting values already present in the hash array if there are duplicate values in the nums array which are not at the same index in the hash array.

Unfortunately, C doesn’t have built-in support for hashset or hashtable data structures. However, we can simulate a similar structure using the arrays, but you’d need to handle hash collisions and the array size might become an issue for larger integers.

For simplicity, the most straightforward way to solve this in C would be to sort the array first, then scan for duplicate adjacent elements. This does, however, change the time complexity from linear to O(n log n) due to the sort operation. If it’s required to have a solution in linear time, you might need to use more advanced data structures available in other libraries or implemented by yourself.

Here’s the sorting approach in C:

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int compare (const void * a, const void * b) {
  return ( *(int*)a - *(int*)b );
}

bool containsDuplicate(int* nums, int numsSize){
    qsort(nums, numsSize, sizeof(int), compare);
    for(int i = 0; i < numsSize-1; i++){
        if(nums[i] == nums[i+1]) {
            return true;
        }
    }
    return false;
}

In this code, the qsort function is used to sort the nums array. Then, the sorted array is iterated over, checking for equality between each pair of adjacent elements. If any pair is found to be equal, the function immediately returns true. If no equal pair is found after checking all pairs, the function returns false.

why does (len(nums) > len(set(nums))) have a better runtime than len(set(nums))!=len(nums) ?

Actually, both expressions len(nums) > len(set(nums)) and len(set(nums)) != len(nums) should have a similar runtime, as they are fundamentally doing the same operation.

The process involved in both expressions includes converting the list nums to a set (which eliminates duplicates) and then comparing the lengths of the original list and the set.

The primary operation in these expressions which takes computational time is the conversion of the list to a set. The comparison operation (> or !=) is a constant-time operation, and the time it takes does not change with the size of the input, so it should not have a significant impact on the runtime.

There might be minor fluctuations in runtime when you submit your solution due to various factors like server load or other factors, but fundamentally, both these expressions should have similar runtimes.

In practical terms, you can use either expression depending on your preference, as they both effectively check whether any duplicates exist in the list.

I don’t think there is a possible solution using only C programming (no sets, no vectors, just C). If somebody find a solution, please let me know.

While it’s true that C doesn’t have built-in data structures like sets or vectors that make it easy to solve this problem, it’s still possible to find a solution using just C. However, it may not be as efficient as the solutions you could implement in languages that do have these features. Here’s a way you can do it:

1. Sorting the array: Since C doesn’t have a built-in hash set data structure to check for duplicates efficiently, one alternative approach is to first sort the array. C has a built-in function for this, qsort(). After sorting, any duplicate elements will be adjacent to each other.

2. Checking for duplicates: Iterate over the sorted array and check if the current element is equal to the next one. If you find any pair of equal adjacent elements, you can return true, indicating that the array does contain duplicates.

Here’s an outline of the code:

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#include <stdlib.h>

int compare(const void * a, const void * b) {
   return (*(int*)a - *(int*)b);
}

bool containsDuplicate(int* nums, int numsSize){
    qsort(nums, numsSize, sizeof(int), compare);
    for(int i = 0; i < numsSize - 1; i++) {
        if(nums[i] == nums[i+1]) {
            return true;
        }
    }
    return false;
}

Please note that sorting the array takes O(n log n) time and checking for duplicates takes O(n) time. Thus, this solution is less efficient than the one that uses a hash set or dictionary, which could solve the problem in O(n) time. But in pure C, without additional data structures, this is a straightforward and feasible approach.

class Solution { public boolean containsDuplicate(int[] nums) { List listy = Arrays.asList(nums); List listx = new ArrayList(); for (Object i: listy) { if (!listx.contains(i)) { listx.add(i); } } return !(listy.size() == listx.size());

} }

Can i ask why this doesnt work

In your code, you’re using the Arrays.asList function to convert an array to a list. However, this method does not work as expected with an array of primitives like int[].

Java treats each primitive array as a single object, so Arrays.asList(nums) doesn’t create a list of integers but a list containing a single object, which is the array itself. So, listy.size() is always 1, not the length of the nums array.

To fix this, you need to change your array of primitives (int[]) to an array of objects (Integer[]). Then, Arrays.asList() will work as you expect:

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class Solution {
    public boolean containsDuplicate(Integer[] nums) {
        List<Integer> listy = Arrays.asList(nums);
        Set<Integer> setx = new HashSet<>(listy);
        
        return !(listy.size() == setx.size());
    }
}

Also, I changed listx to a Set, because set.add(i) and set.contains(i) operations are faster (both O(1)) than list.add(i) and list.contains(i) operations (O(n) in the worst case for ArrayList). If you’re checking for duplicates, using a Set is typically more efficient.

this is an easy one and there are many ways to do it, I did it by just writing a single line if to return a boolean by comparing length of list and and its set. Interestingly I found online that len function’s time complexity is just O(1) because underneath it doesn’t call a function and perform some operations. The interpreter maintains length stored for everything and len just fetches it. I am thinking this is the best and efficient possible, although I may not seem to understand what happens by using the set conversion. Anyone knows what process happens underneath set conversion and its time, space complexities, please make it known. and any more efficient way in case mine isn’t.

You’re correct that the len function in Python has a time complexity of O(1) because Python internally keeps track of the length of its collections (like lists, tuples, sets, etc.).

In your solution, you’re using the set function to create a set from a list. A set in Python is an unordered collection of unique elements. When you pass a list to the set function, it iterates over the list and adds each element to the set. If an element is already in the set, it is not added again, ensuring that the set contains only unique elements.

The time complexity of creating a set from a list is O(n), where n is the number of elements in the list. This is because each element in the list needs to be visited at least once.

In terms of space complexity, creating a set from a list will take O(n) additional space, because in the worst-case scenario (when all elements are unique), a new set with the same number of elements as the list will be created.

Your solution, which creates a set and then compares the lengths of the list and the set, is quite efficient for this problem. It has a time complexity of O(n) and a space complexity of O(n), which is reasonable for a problem of this nature. The only overhead is the additional space used to store the set.

Hii all this is code with time complexity O(1) and space complexity O(n)

class Solution { public: bool containsDuplicate(vector& nums) { int size1= nums.size(); set s(nums.begin(),nums.end()); int size2= s.size(); if(size1==size2){ return false; } return true; } };

This is indeed a valid solution to the problem, but let’s clarify the time complexity.

While your code’s space complexity is O(n), which means it requires a linear amount of additional memory proportional to the input size, the time complexity is not O(1). O(1) means constant time, regardless of the size of the input, which is not the case here.

When creating the set from the vector, each element of the vector is processed. The set operation in C++ involves adding elements to the set, which, in the worst-case scenario, could involve rehashing the set if the load factor gets too high. This operation on average takes O(log n) time, where n is the number of elements in the set. Since you’re adding each element of the vector to the set, the total time complexity of creating the set is O(n log n).

Therefore, the total time complexity of this function is O(n log n) rather than O(1). The space complexity is O(n), as you correctly mentioned.

I use 2 methods but I can say it is quite similar

1,

Sort Array in Ascending order Loop from 0 to nums length - 1 Compare array a[i] and a[1+1] If they are equal then return true When finish the loop, return False 2,

Using HashSet Unique Loop i from begining to the end If Unique containt a[i] then return True Otherwise, Unique.add a[i] Finish Loop then return False Method 1: Memory Optimized Method 2: Runtime Optimized

How can I balance both?

Balancing runtime and memory optimization often depends on the specific needs of the program or the system it’s running on. However, here are some general ways you might find a balance between the two methods you mentioned for this problem:

1. Runtime and Memory Tradeoff: As you have noticed, the first method (sorting the array and checking for duplicates) has a time complexity of O(n log n) and a space complexity of O(1), while the second method (using a hashset to check for duplicates) has a time complexity of O(n) and a space complexity of O(n). If you want to find a balance, you might need to be okay with a bit higher time complexity to reduce the space complexity, or vice versa.

2. Customized Data Structures: In some cases, you might be able to create a customized data structure that is more space efficient than a hashset but still gives you faster lookup times than sorting the array. This could be complex and might not be worth it for a problem like this, but it’s an option in some cases.

3. Probabilistic Data Structures: If an approximate answer is acceptable, you could consider using a probabilistic data structure like a Bloom filter. A Bloom filter can tell you if an element is definitely not in a set or may be in the set. This could potentially reduce your space usage while still giving you fast lookup times.

Remember, the right balance between runtime and memory optimization often depends on the specific constraints of your problem and system. What is most important is to understand the tradeoffs you’re making and choose the best strategy for your needs.

how one can get dynamic array without using any third variable

If you want to resize an array dynamically in Python, the preferred way is to use a list. Lists in Python are dynamic arrays and you can add and remove items without using a third variable. Here is an example:

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# Initialize the list (dynamic array)
nums = [1, 2, 3, 4, 5]

# Add an item to the end
nums.append(6)
print(nums)  # Outputs: [1, 2, 3, 4, 5, 6]

# Remove the last item
nums.pop()
print(nums)  # Outputs: [1, 2, 3, 4, 5]

As you can see, the list nums is resized without the need for a third variable. The append method adds an item to the end, and the pop method removes the last item.

However, if you’re working with a language that doesn’t have built-in support for dynamic arrays (like C or C++), resizing an array without a temporary variable could be more challenging. You would generally need to use a function provided by the language’s standard library (like realloc in C) to resize the array, and this can involve creating a new block of memory, copying the elements over, and then freeing the old block of memory.

Best time complexity solution is not correct but accepted !!!!

                    input    [ 5 , 35 , 5 ] 

int hash(int key, int inputSize){ return abs(key) % inputSize; }

bool containsDuplicate(int* nums, int numsSize){ int hashTableSize = 2 * numsSize; int *hashTable = (int *) malloc(hashTableSize * sizeof( int ));

for (int i = 0; i < numsSize; i++) { if (hashTable[hash(nums[i], hashTableSize)] == nums[i]) { return true; } else { hashTable[hash(nums[i], hashTableSize)] = nums[i]; } }

return false; }

The code you provided is implementing a Hash Table with an open addressing technique, but it lacks collision resolution. Here, we are using an array hashTable as the Hash Table and the hash() function to calculate the index for each element in nums.

But the problem is, the same index can be calculated for different numbers. For example, if numsSize is 10, both 5 and 15 will be mapped to the index 5. So if we have a situation where we first insert 15 into the hash table and then we try to insert 5, the code will erroneously think that there’s a duplicate even though 5 and 15 are different.

To fix this, you can use chaining (a linked list at each hash table index to handle collisions) or another collision resolution technique such as linear probing, quadratic probing, or double hashing.

However, note that implementing these strategies in C would require more code and it wouldn’t be a constant time solution anymore. It would be average case O(1) (constant time), but worst case could be O(n) (linear time) if all keys collide.

Moreover, because your hashTable isn’t initialized, it contains random values which can lead to false positives when checking for duplicates.

A simple and effective solution for this problem in Python would be to utilize a Set to check for duplicates, since Set operations are typically O(1). Here’s an example:

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def containsDuplicate(nums):
    return len(nums) != len(set(nums))

This code works because a Set only contains unique elements. If the size of the Set is different from the size of the input list, it means there are duplicates in the list.

can’t we do it to soemthing silimar to Moore’s Voting algorithm? we can have a counter variable to track if a number has occured twice….if it has simply return true? please help

Unfortunately, the Moore’s Voting algorithm won’t work for this problem. The Moore’s Voting algorithm is used to find the majority element in a list, i.e., an element that appears more than n/2 times where n is the size of the list.

The problem statement here asks you to identify if there is any element that appears more than once in the array, irrespective of the size of the array. This is a different problem from what the Moore’s Voting algorithm solves.

For example, consider the array [1,2,3,1]. Here, the number 1 appears twice, but it’s not the majority element of the array (as it doesn’t appear more than n/2 times). But we still have to return True, since there is a duplicate.

A counter variable alone wouldn’t be enough to solve this problem, because it would only keep track of how many elements we’ve seen, not which elements we’ve seen. We need to keep track of the specific elements that we’ve already encountered to identify duplicates. This is typically done using a data structure like a Set, which has O(1) average time complexity for inserts and lookups, making it very efficient for this kind of problem.

For example, in Python, a solution could be:

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def containsDuplicate(nums):
    num_set = set()
    for num in nums:
        if num in num_set:
            return True
        num_set.add(num)
    return False

This function works by iterating over the list of numbers and adding each number to a set. If it encounters a number that’s already in the set, it returns True (indicating that there is a duplicate). If it gets through the entire list without finding any duplicates, it returns False.

I don’t understand why this code does not work for the test case [1,2,3,1]: public: bool containsDuplicate(vector& nums) { map<int, int> mpp; for(int i=0; i< nums.size(); i++){ if(mpp[nums[i]] >= 2){ return true; } else{ mpp[nums[i]]++; } } return false; }

The problem in your code lies in the condition if(mpp[nums[i]] >= 2). This line is checking if the count of the number nums[i] in your map mpp is greater than or equal to 2. However, this check is performed before you increment the count in the line mpp[nums[i]]++, which means that the count in your map will only ever be 0 or 1 at the time of the check, and so it will never be 2 or greater.

This can be fixed by simply incrementing the count before performing the check. Here’s the corrected code:

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public:
bool containsDuplicate(vector<int>& nums) {
    map<int, int> mpp;
    for(int i=0; i< nums.size(); i++){
        mpp[nums[i]]++;
        if(mpp[nums[i]] >= 2){
            return true;
        }
    }
    return false;
}

In this corrected version of the code, we increment the count of each number as we encounter it. Then, we check if the count of the number is 2 or more. If it is, that means we’ve encountered a duplicate, and so we return true. If we go through the entire list without finding any duplicates, we return false.

can anyone suggest me why this code is not working?

class Solution { public: bool containsDuplicate(vector& nums) { int n=1e5+10; int b[n]; int flag; for(int i=0;i<sizeof(b);i++) { b[nums[i]]++; } for(int i=0;i<sizeof(b);i++) { if(b[i]>1) { flag=1; } } if(flag==1) { return true; } else { return false; } } };

Your code has several issues:

1. You are trying to initialize an array b of size 1e5+10 with automatic storage duration (a local variable), which might lead to stack overflow. Instead, you should use dynamic memory allocation (std::vector<int> in C++).

2. You do not initialize the array b. In C++, local variables are not initialized automatically to 0, unlike global variables. So, they can have any arbitrary value initially. Use std::vector<int> b(n, 0); to initialize a vector of size n with all values as 0.

3. You are using sizeof(b) to find the number of elements in b. In C++, sizeof(b) will give you the size of the array in bytes, not the number of elements. When you use it in a for loop, you are likely exceeding the bounds of the array. Use b.size() instead for vectors.

4. flag is not initialized. In C++, local variables are not initialized automatically. So, it can have any arbitrary value initially, and your check for flag==1 may give incorrect results. Initialize flag as int flag = 0;

5. nums[i] can be negative according to the problem constraints, and you’re using it as an index. In C++, arrays can’t have negative indices.

Here’s a corrected version of your code:

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class Solution {
public:
    bool containsDuplicate(vector<int>& nums) {
        int n = 1e5 + 10;
        vector<int> b(n, 0);
        int flag = 0;
        for(int i = 0; i < nums.size(); i++) {
            b[nums[i] + 1e9]++; // Since nums[i] can be negative, add a large number to use it as an index
        }
        for(int i = 0; i < b.size(); i++) {
            if(b[i] > 1) {
                flag = 1;
                break;
            }
        }
        return flag == 1;
    }
};

Remember, this solution only works when the input nums contains integers in the range [-1e9, 1e9], and the size of nums is less than 1e5. The size of b vector is 1e5+10 and may not be enough to cover all possible integer values in nums.

what wrong in this!!!

class Solution { public: bool containsDuplicate(vector& nums) { for(int i = 0; i <nums.size(); i++) for (int j=i+1; j< nums.size(); j++) if (nums[i]==nums[j]) return true;

return false; } };

There is nothing syntactically wrong with your code. It’s just that your approach has a time complexity of O(n^2), where n is the size of the input array. This is because you are using two nested for loops to compare each pair of numbers in the array. For large input arrays, this will result in a timeout error, meaning your solution is not efficient enough.

Your code performs a brute-force search to find any duplicates. While this will work for small inputs, an optimal solution would use a data structure like a Set or HashMap to track the numbers you’ve already seen, allowing you to check for duplicates in constant time.

Here’s an example of how you could do it:

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class Solution:
    def containsDuplicate(self, nums: List[int]) -> bool:
        num_set = set()
        for num in nums:
            if num in num_set:
                return True
            num_set.add(num)
        return False

In this code, we iterate over the array once, adding each number to a set. Before adding a number to the set, we check if it’s already in there. If it is, that means it’s a duplicate, so we return True. If we get through the whole array without finding any duplicates, we return False.

This solution is much faster than the original one, as it only has to iterate over the array once, giving it a time complexity of O(n), where n is the size of the array.

Ask yourself how to store each element and its appearance?

To store each element and its appearance in Python, you could use a dictionary. In Python, a dictionary (dict) is a built-in data type that allows you to store key-value pairs.

For this specific problem, where we want to store each element from a list (nums) and its appearance, the element itself can be the key, and its frequency (appearance) can be the value.

Here is an example:

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def element_appearance(nums):
    appearance_dict = {}
    for num in nums:
        if num in appearance_dict:
            appearance_dict[num] += 1
        else:
            appearance_dict[num] = 1
    return appearance_dict

In this example, the function element_appearance takes a list of integers (nums) as an input, and returns a dictionary where the keys are the unique elements from nums, and the values are the number of times each element appears in nums.

This is done by iterating over the list nums. For each number (num) in nums, if num is already a key in the dictionary (appearance_dict), we increment its associated value by 1. If num is not already a key in the dictionary, we add it to the dictionary with an associated value of 1. This way, we end up with a dictionary that accurately represents each element and its frequency of appearance.

class Solution { public boolean containsDuplicate(int[] nums) { Arrays.sort(nums); for (int i = 0; i < nums.length - 1; i++) { if (nums[i] == nums[i+1]) { return true; } } return false; } }

this code i have copy and pate to my intellije idle but it not work so how can i run ?

also package Array;

import java.util.Arrays;

public class DuplicateArray { public static void main(String[] args) { int[] nums = {1,2,3,1}; containsDuplicate(nums); } public static boolean containsDuplicate(int[] nums) { Arrays.sort(nums); for (int i = 0; i < nums.length - 1; i++) { if (nums[i] == nums[i+1]) { return true; } } return false; }

} also i have written this code but not shows any output it shows only below line Process finished with exit code 0

plzz help me

The reason your code doesn’t show any output is that while you’re calling the containsDuplicate(nums) method in your main method, you aren’t doing anything with the result. This method returns a boolean value (either true or false), and you should print this result to the console in order to see it.

You can modify your main method like this:

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public static void main(String[] args) {
    int[] nums = {1,2,3,1};
    boolean hasDuplicate = containsDuplicate(nums);
    System.out.println(hasDuplicate);
}

Here, I’ve created a boolean variable hasDuplicate to store the result of the containsDuplicate(nums) method. Then, I use System.out.println(hasDuplicate); to print the result to the console. If the array nums contains any duplicate elements, the program will print true; otherwise, it will print false.

Regarding running this code in IntelliJ IDEA, you should be able to do so by clicking the green arrow (Run button) on the toolbar, or by pressing Shift + F10. If you still encounter problems, make sure your project is correctly set up and that IntelliJ IDEA recognizes it as a Java project.

Hello everyone ! i 10 testcases are not passing could anyone pls help me. class Solution { public: bool containsDuplicate(vector& nums) { int flag=0; for(int i=0;i<nums.size();i++) { for(int j=i+1;j<nums.size();j++) { if(nums[i]==nums[j]) { flag=1; }

    }
}
if(flag==1)
{
    return true;
}
else
{
    return false;
}

}
};

Your code is attempting to solve the problem using a brute force method by comparing each element with every other element in the array. This is a valid approach but it is not efficient, which is why it might be failing for larger test cases due to exceeding the time limit. This approach has a time complexity of O(n^2) where n is the size of the input array.

Here’s an optimized approach using a data structure called a set, which stores unique elements. The time complexity of inserting an element in a set and checking if an element is in a set are both O(1) on average.

In Python, it would look something like this:

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class Solution:
    def containsDuplicate(self, nums):
        seen = set()
        for num in nums:
            if num in seen:
                return True
            seen.add(num)
        return False

The seen set stores the numbers we have seen so far in the loop. For each number in the array, we check if we have seen it before. If we have (i.e., if it’s in the seen set), then we return True because it’s a duplicate. If we haven’t seen it before, we add it to the seen set. If we finish the loop without finding any duplicates, we return False.

And here is the equivalent code in C++:

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class Solution {
public:
    bool containsDuplicate(vector<int>& nums) {
        unordered_set<int> seen;
        for (int num : nums) {
            if (seen.count(num)) {
                return true;
            }
            seen.insert(num);
        }
        return false;
    }
};

This C++ code works the same way as the Python code: it uses an unordered set (unordered_set in C++) to store the numbers we’ve seen, and checks for each number if it’s in the set. If it is, that means it’s a duplicate, and we return true. If it’s not in the set, we add it to the set. If we finish the loop without finding any duplicates, we return false.

why is this giving me the wrong answer for nums = [1, 2, 3, 4]. class Solution: def containsDuplicate(self, nums: List[int]) -> bool: unique = [] i = 0 while i < len(nums): if nums[i] not in unique: unique.append(nums[i]) i += 1 else: i += 1 return (’true’) if i == len(nums): return (‘false’)

The problem with your code is that you’re always returning ’true’ when you find a number in the unique list. But the way you’re updating your i index and your unique list doesn’t guarantee that the number was actually a duplicate when you return ’true’. You increment i even when a number was not in the unique list, which makes it so that the next time through the loop, you might return ’true’ even though the current number was not a duplicate.

Furthermore, you’re returning the string ’true’ and ‘false’, not the boolean values True and False.

A corrected version of your code might look like this:

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class Solution: 
    def containsDuplicate(self, nums: List[int]) -> bool: 
        unique = [] 
        for num in nums:
            if num not in unique: 
                unique.append(num) 
            else:
                return True
        return False

In this corrected version, we go through each number in nums. If it’s not in the unique list, we add it to unique. But if it is in unique, that means it’s a duplicate, and we immediately return True. If we go through all the numbers and don’t find any duplicates, we return False.

However, this solution has a time complexity of O(n^2) because for each number in nums, we’re checking if it’s in the unique list, and checking membership in a list takes O(n) time. We can improve this to O(n) time complexity by using a set instead of a list for unique, because checking membership in a set takes O(1) time:

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class Solution: 
    def containsDuplicate(self, nums: List[int]) -> bool: 
        unique = set()
        for num in nums:
            if num not in unique: 
                unique.add(num) 
            else:
                return True
        return False

for i in nums: return nums.count(i) > 1 why my solution not working ?

The issue with your solution is that the return statement is executed as soon as the first iteration of the loop happens. Hence, it doesn’t matter if there are any duplicates later in the list. The solution you’ve proposed will only check for multiple occurrences of the first element of nums.

Here’s a revised version that should work:

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def containsDuplicate(nums):
    for i in nums:
        if nums.count(i) > 1:
            return True
    return False

This version will loop through every element in nums, and if any element appears more than once, it returns True. If it completes the entire loop without finding any duplicates, it returns False.

However, this solution has a time complexity of O(n^2) because for each number in nums, we’re calling nums.count(i), which takes O(n) time. You can improve this to O(n) time complexity by using a set:

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def containsDuplicate(nums):
    return len(nums) != len(set(nums))

This version checks whether nums has any duplicates by comparing its length to the length of set(nums), which removes any duplicates from nums. If the lengths are different, that means there were duplicates in nums, so we return True. If the lengths are the same, that means there were no duplicates in nums, so we return False.

bool containsDuplicate(int* nums, int numsSize){ for(int i=0;i<numsSize-1;i++){ for(int j=i+1;j<numsSize;j++){ if(nums[i]==nums[j]){ return true; } }

} return false; } can anybody help I m not getting what’s wrong ? its not verifying few testcases

Your solution implements a simple comparison-based approach to finding duplicates. It does work for smaller arrays and has a time complexity of O(n^2), where n is the size of the input array. For each element, it checks all the other elements to see if there is a match. This is why it takes a long time for larger input arrays and may fail certain test cases on platforms with strict time constraints.

An alternative solution is to sort the array first and then check for any adjacent duplicate elements. The sorting operation will take O(n log n) time and the duplicate checking will take O(n) time, making the total time complexity O(n log n), which is faster for larger arrays.

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int compare(const void * a, const void * b) {
    return (*(int*)a - *(int*)b);
}

bool containsDuplicate(int* nums, int numsSize){
    qsort(nums, numsSize, sizeof(int), compare);
    for(int i = 0; i < numsSize - 1; i++){
        if(nums[i] == nums[i+1]){
            return true;
        }
    }
    return false;
}

This code sorts the array first using the qsort function and then iterates through the sorted array to find duplicates. If it finds any duplicate pairs, it returns true; if it finishes checking the entire array without finding any duplicates, it returns false. Note that this approach changes the order of the elements in the input array.

while i!=len(nums):

    if nums[i] == nums[i+1]:

why this condition is showing error

The error occurs because nums[i+1] may be out of range.

When i is the last index of the list, nums[i+1] will be trying to access an element that is out of the list’s range, hence it will raise an IndexError in Python.

To resolve this, you should stop the loop at the second last index. You can do this by changing your condition to while i != len(nums) - 1: or using a for loop instead:

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for i in range(len(nums) - 1):
    if nums[i] == nums[i+1]:

Remember to sort the list before checking for duplicates as equal elements will be adjacent in a sorted list.

Always make sure you’re not accessing list indices that could potentially be out of range. It’s a common source of errors in Python and many other languages.

I’m noticing I’m repeatedly getting Time Limit Exceeded on the last test, this is my solution:

impl Solution { pub fn contains_duplicate(nums: Vec) -> bool { let mut arr: Vec = vec![];

    for (i, num) in nums.iter().enumerate() {
        if arr.contains(&num) { return true; }
        arr.push(*num);
    }

    false
}

}

The time complexity of your solution is O(n^2) because for each element in nums, you’re checking if it is in the vector arr which in worst case takes O(n) time (when the element is not present in the vector). Therefore, for n elements, you’re doing n checks of O(n) time each, leading to O(n^2) time complexity overall. This is why you’re getting Time Limit Exceeded.

A more time-efficient solution would involve using a HashSet (which in Rust is provided by std::collections::HashSet) instead of a Vector. The advantage of a HashSet is that it can insert elements and check if elements are present in O(1) time on average, leading to an overall time complexity of O(n) for this problem. Here’s how you could modify your solution:

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use std::collections::HashSet;

impl Solution {
    pub fn contains_duplicate(nums: Vec<i32>) -> bool {
        let mut set = HashSet::new();

        for num in nums {
            if set.contains(&num) { 
                return true; 
            }
            set.insert(num);
        }

        false
    }
}

This solution works by creating an empty HashSet, then for each number in nums, checking if it is already in the set (which is a fast O(1) operation), and if so returning true, otherwise inserting the number into the set. If it gets through all numbers without finding a duplicate, it returns false. This avoids the time limit exceeded error by bringing the time complexity down from O(n^2) to O(n).

Hi, i just can’t understand what is the problem with my code. i’m sure it is supposed to work…

import java.util.*; class Solution { public boolean containsDuplicate(int[] nums) { int count=0; for(int i=0; i<=nums.length; i++){ for(int j=0; j<=nums.length;j++){ if(nums[i]==nums[j]){ count++; } } } if(count>2){ return true; } return false; } }

Your solution has a few issues. Let’s go through them:

1. Out-of-bounds Error: In your for loops, you are looping until i<=nums.length and j<=nums.length. However, in Java, arrays are 0-indexed, which means the index of the last element of the array is nums.length-1. So your loop should be i<nums.length and j<nums.length to avoid accessing out-of-bounds index.

2. Duplicate Counting: You are comparing every pair of elements in nums, including each element with itself. This is why you check if(count>2) at the end, but it’s not a correct solution, since it only works if the array has exactly one duplicated element and all others are unique. If the array has more than one set of duplicates, it will return true even if none of the duplicates are in the array more than twice.

3. Efficiency: The nested loops have a time complexity of O(n^2), which is not efficient for large input. You can solve this problem more efficiently using a HashSet in Java, which allows for O(1) time complexity for insertions and lookups.

Here is a corrected and more efficient version of your code:

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import java.util.*;

class Solution {
    public boolean containsDuplicate(int[] nums) {
        Set<Integer> set = new HashSet<>();
        
        for(int num : nums){
            if(set.contains(num)){
                return true;
            }
            set.add(num);
        }
        return false;
    }
}

This code creates a HashSet and then loops through nums. For each number, it checks if that number is already in the set. If it is, that means it’s a duplicate, and it immediately returns true. If not, it adds the number to the set. If it gets through all of nums without finding a duplicate, it returns false.

can somebody explain to me why this works

array = {}
for n in nums:
    if n in array:
        return 1
    array[n] = index
return 0

but if i change “index” to something else like “value” or “i” than it bricks ….

In your code snippet, array is a dictionary used to store numbers from nums as keys. The values stored in the dictionary are not actually used for anything; their only purpose is to allow checking whether a number is in the dictionary or not.

However, there seems to be a bit of confusion regarding the variable index in your code. If it is not defined elsewhere in your code, this would throw a NameError in Python because index is not defined.

If you replace index with value or i or any other variable name, it should work just fine as long as that variable is defined. If these variables are not defined, you will get a NameError.

Here’s the corrected version of your code:

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array = {}
for n in nums:
    if n in array:
        return 1
    array[n] = 1  # you can replace 1 with any value here
return 0

In this code, n represents each number in the nums list. If n is already in the array dictionary, that means it’s a duplicate, and the function immediately returns 1. If n is not in array, it adds n to the dictionary and continues to the next number. If it gets through all of nums without finding a duplicate, it returns 0. The actual value associated with each key in the dictionary is irrelevant in this case, so you can use any value.

Time Complexity -> O(nlogn)

`class Solution { public: bool containsDuplicate(vector& v) {

sort(v.begin(),v.end()); bool flag = false; for(int i=1;i<v.size();i++){ if(v[i]==v[i-1]){ return true; } }

return flag;

} };`

Your solution is correct. The code first sorts the vector, which takes O(nlogn) time complexity due to the sorting algorithm. Then it loops through the sorted array to check if any two adjacent elements are the same. If it finds a pair of identical elements, it returns true. If it doesn’t find any duplicate elements, it returns false.

The time complexity of the loop is O(n), since it scans through the vector once. However, the overall time complexity is still O(nlogn) because the sorting part is the most computationally expensive part of the algorithm. This is why we say the overall time complexity is O(nlogn).

Also, you don’t really need the flag variable in your code, as you’re directly returning true or false from within the function. It could be simplified like this:

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class Solution {
public:
bool containsDuplicate(vector<int>& v) {
    sort(v.begin(),v.end());
    for(int i = 1; i < v.size(); i++){
        if(v[i] == v[i - 1]){
            return true;
        }
    }
    return false;
}
};

This version of the code does exactly the same thing as yours but it’s a bit cleaner and easier to understand.

is there any one who can tell me time & space complexity in the following two cases for finding duplicates. return new Set(nums).size != nums.length; // case 1 nums.sort((a,b) => a-b); for(let i=0; i<nums.length; i++) { if(nums[i]===nums[i+1]) { return true; } else return false; } // case 2

Case 1: In this case, you’re creating a set from the array. The time complexity of creating a set from an array is O(n) because you need to iterate through every element in the array. The space complexity is also O(n) because in the worst-case scenario, every element is unique and needs to be stored in the set.

Case 2: The time complexity is dominated by the sorting operation. JavaScript’s built-in sort method uses a variant of quicksort in most engines, which has an average and best-case complexity of O(n log n) and worst-case complexity of O(n^2). However, JavaScript engines may use other sorting algorithms for smaller arrays (like insertion sort), which makes the actual time complexity in practice somewhat nuanced. After sorting, you’re looping through the sorted array once to check for duplicates, which is O(n). But because O(n log n) > O(n), we can say the overall time complexity is O(n log n).

The space complexity of JavaScript’s sort can vary between O(log n) and O(n), depending on the implementation. However, your sorting operation is in-place and doesn’t use any additional space proportional to the size of the input array. After sorting, your loop doesn’t use any extra space either, so the space complexity is O(1), constant space.

In conclusion:

Case 1: Time complexity is O(n), and space complexity is O(n). Case 2: Time complexity is O(n log n), and space complexity is O(1) or O(n) depending on the specific sorting algorithm’s implementation in the JavaScript engine.

class Solution { public: bool containsDuplicate(vector& nums) { int n= sizeof(nums); for(int i=0;i<n;i++){ for(int j=i+1;j<n;j++){ if(nums[i]==nums[j]){ return true; } } } return false; } };

This is my code i dont know why it is giving me a runtime error

In your code, you’re trying to get the size of the vector by using the sizeof() function, which is incorrect. In C++, the sizeof() function is generally used to find the size in bytes of a variable, type, or object. However, it won’t provide the correct size of a vector (i.e., the number of elements in the vector).

Instead, you should use the size() function to get the number of elements in the vector. Here’s how you should modify your code:

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class Solution {
public:
    bool containsDuplicate(vector<int>& nums) {
        int n= nums.size();  // use size() instead of sizeof()
        for(int i=0;i<n;i++){
            for(int j=i+1;j<n;j++){
                if(nums[i]==nums[j]){
                    return true;
                }
            }
        }
        return false;
    }
};

Please note that this solution has a time complexity of O(n^2) because of the two nested loops, which makes it inefficient for large inputs. If performance is a concern, you should consider using a different approach, such as sorting the array first or using a data structure like a set to keep track of the elements you’ve encountered.

class Solution { public: // approach: // sort and the check for duplicate

// merge function void merge(vector& nums, int left, int mid, int right) { int subArrayOne = mid - left + 1; int subArrayTwo = right - mid;

// Create temp arrays
int *leftArray = new int[subArrayOne],
     *rightArray = new int[subArrayTwo];

// Copy data to temp arrays leftArray[] and rightArray[]
for (int i = 0; i < subArrayOne; i++)
    leftArray[i] = nums[left + i];
for (int j = 0; j < subArrayTwo; j++)
    rightArray[j] = nums[mid + 1 + j];

int indexOfSubArrayOne
    = 0, // Initial index of first sub-array
    indexOfSubArrayTwo
    = 0; // Initial index of second sub-array
int indexOfMergedArray
    = left; // Initial index of merged array

// Merge the temp arrays back into array[left..right]
while (indexOfSubArrayOne < subArrayOne
       && indexOfSubArrayTwo < subArrayTwo) {
    if (leftArray[indexOfSubArrayOne]
        <= rightArray[indexOfSubArrayTwo]) {
        nums[indexOfMergedArray]
            = leftArray[indexOfSubArrayOne];
        indexOfSubArrayOne++;
    }
    else {
        nums[indexOfMergedArray]
            = rightArray[indexOfSubArrayTwo];
        indexOfSubArrayTwo++;
    }
    indexOfMergedArray++;
}
// Copy the remaining elements of
// left[], if there are any
while (indexOfSubArrayOne < subArrayOne) {
    nums[indexOfMergedArray]
        = leftArray[indexOfSubArrayOne];
    indexOfSubArrayOne++;
    indexOfMergedArray++;
}
// Copy the remaining elements of
// right[], if there are any
while (indexOfSubArrayTwo < subArrayTwo) {
    nums[indexOfMergedArray]
        = rightArray[indexOfSubArrayTwo];
    indexOfSubArrayTwo++;
    indexOfMergedArray++;
}
delete[] leftArray;
delete[] rightArray;

}

// mergesort function which includes merge function inside // begin is for left index and end is // right index of the sub-array // of arr to be sorted */ void mergeSort(vector& nums, int begin, int end) { if (begin >= end) return; // Returns recursively

int mid = begin + (end - begin) / 2;
mergeSort(nums, begin, mid);
mergeSort(nums, mid + 1, end);
merge(nums, begin, mid, end);

} // containsDuplicate will include mergesort function

bool containsDuplicate(vector<int>& nums) {
    int n = nums.size();

    // sort first
    int begin = 0;
    int end = 0;
    mergeSort(nums, begin, end);

    // check for duplicate
      for(int i=0; i<nums.size()-1; i++){
        if(nums[i] == nums[i+1] || nums[i] == nums[n-1]){
            return true;
        }
    }
    return false;
}

};

Hello. I need help understanding why my code works for all the tests when I click “run” but when submitting, it doesn’t work for the array that includes zeros which is [0,4,5,0,3,6]. Thanks in advance.

Intuition My first thoughts were to sort the list and using a sorting method which I’ve most recently been using (merge sort). I’ll return true if we find a duplicate and return false if we don’t

Approach First we need to sort the list before checking for a duplicate. I’m not using STL (besides the vector)- because I choose not to. I use merge sort for sorting. I call the merge sort function within the containsDuplicate function after initializing variable n to be the size of the list.

After sorting, I check for the duplicate by traversing through the list using a for loop, and returning true if the current number and next number are the same OR if the current number and last number of the list are the same.

Your code is failing for the given test case because you’re incorrectly setting the begin and end variables before calling mergeSort(nums, begin, end);

The variables begin and end are supposed to represent the starting and ending index of the array section you’re sorting. In this case, you should be sorting the entire array, so begin should be 0 (the first index), and end should be n-1 (the last index). However, you’re setting both begin and end to 0, which means you’re not actually sorting anything.

You should update your code to correctly set the begin and end variables as follows:

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    bool containsDuplicate(vector<int>& nums) {
        int n = nums.size();

        // sort first
        int begin = 0;
        int end = n - 1; // Set end to the last index of the array
        mergeSort(nums, begin, end);

        // check for duplicate
        for(int i=0; i<nums.size()-1; i++){
            if(nums[i] == nums[i+1] || nums[i] == nums[n-1]){
                return true;
            }
        }
        return false;
    }

With this correction, your code will correctly sort the array before checking for duplicates, and should pass the test case.

Also, your current condition if(nums[i] == nums[i+1] || nums[i] == nums[n-1]) can be simplified to if(nums[i] == nums[i+1]) because after sorting, checking with the last element doesn’t make sense. It will only return the correct result for sorted arrays and is unnecessary in this context.

class Solution { public boolean containsDuplicate(int[] nums) { for(int i=0;i<nums.length;i++){ for(int j=i+1;j<nums.length;j++){ if(nums[i]==nums[j]){ return true; } } } return false; } } this is giving me “Time limit exceeded” Please help as from the logic of it I would think this works.

Your current solution works, but it has a time complexity of O(n^2) because you’re using a nested loop to check each pair of elements in the array. This is acceptable for small arrays, but when you’re working with large arrays (such as those in the test cases), this can cause your program to exceed the time limit.

A better approach is to use a data structure that allows faster lookups, such as a HashSet in Java. The HashSet data structure allows for O(1) lookup times on average, and you can use it to check whether an element has already been seen in the array.

Here is how you might modify your code to use a HashSet:

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import java.util.HashSet;

class Solution {
    public boolean containsDuplicate(int[] nums) {
        HashSet<Integer> seen = new HashSet<Integer>();
        for (int num : nums) {
            if (seen.contains(num)) {
                return true;
            }
            seen.add(num);
        }
        return false;
    }
}

This code iterates over the array nums just once, checking for each element whether it’s already in the HashSet seen. If it is, that means we’ve found a duplicate, and we return true. If not, we add the element to the HashSet. If we make it all the way through the array without finding a duplicate, we return false.

This solution has a time complexity of O(n), which should prevent the time limit exceeded error.

why does this one line python solution do so poorly, it’s O(n) right?

def containsDuplicate(self, nums: List[int]) -> bool: return len(set(nums)) < len(nums)

Your Python solution indeed has an O(n) time complexity, as it involves converting a list of n elements into a set. However, the space complexity is also O(n) because creating a set from the list requires additional space to store the set.

While this solution is simple and intuitive, it may not perform well with large inputs due to the additional space required to create the set. It’s important to note that space complexity can be a factor in performance considerations. Also, keep in mind that the process of creating a set from a list can be slower than expected due to hash computations for the elements of the list.

Moreover, the performance of Python, being an interpreted language, might not be as good as lower-level languages like C++ or Java for certain test cases, especially those with large inputs.

So, while the time complexity is O(n), other factors such as space complexity, the specific implementation of data structures in the language (Python’s set in this case), and the nature of the language itself can affect the performance of the solution.

xor method will not work.. {2,2,2,2}, take this example! Negation of index does not work because it modifies the array.

You’re correct. The XOR method won’t work in this case.

XOR has a property that a XOR a = 0 and a XOR 0 = a. So, if we XOR all the elements of the array, we would end up with zero if there are even occurrences of each element (including duplicates). But for the case where each duplicate appears an odd number of times, this method would fail.

For example, with the array {2,2,2,2}, XORing all elements would result in 0, which falsely suggests that there are no duplicates.

Moreover, using XOR will not be able to handle the situation where the array doesn’t contain any duplicates but includes 0, since a XOR 0 = a.

Thus, the XOR method is not a suitable solution for this problem. For handling such problems, hash maps or sorting can be used to track the occurrences of the numbers in the array.

The repeated number will always be at the junction between the straight path and the cycle because, the cycle exists only because two nodes reach to the same next as one value is repeated. We traverse the array and we go to the element at the index same as the current element value. Since there is a repeated value, we would be going to a index twice, hence there will be cycle if there is a repeated value. This method is only applicable to questions in which all elements are between 0 and array.length - 1 . Two pointer in a cycle, starting at the same point and moving at different speeds will always collide at some point. At any point when the slow and fast pointers collide the fast pointer must have covered twice the distance of the slow pointer because its moving two steps in every iteration while slow pointer moves only one. And where would the fast point move that extra distance? It could only move that in the cycle because that is only how it could come back the same point in the cycle, as the slow and fast are colliding at a point in the cycle, hence the distance moved by the slow pointer would be multiple of the length of the cycle. At last we would move both the slow and fast pointer by one step until they meet again and that would be our duplicate number.

The method you’re describing, often referred to as Floyd’s Tortoise and Hare (Cycle Detection) algorithm, is typically used to detect a cycle in a linked list. It can indeed be used to find duplicates in an array, but as you mentioned, it has the specific precondition that all elements are between 1 and n-1, where n is the size of the array.

Here is how the algorithm works:

1. Initialize two pointers: the tortoise (slow pointer) and the hare (fast pointer).

2. Move the tortoise by one step and the hare by two steps.

3. If there is a cycle (which represents a duplicate because two indices point to the same value), the hare and the tortoise will meet within the cycle.

4. Reset the tortoise to the start point, and keep the hare at the meeting point. Move both the tortoise and hare one step at a time.

5. The point where they meet again is the duplicate number.

Here is a Python implementation of the above logic:

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class Solution:
    def findDuplicate(self, nums: List[int]) -> int:
        tortoise = hare = nums[0]
        while True:
            tortoise = nums[tortoise]
            hare = nums[nums[hare]]
            if tortoise == hare:
                break
        
        # Find the "entrance" to the cycle.
        tortoise = nums[0]
        while tortoise != hare:
            tortoise = nums[tortoise]
            hare = nums[hare]
        
        return hare

This solution is predicated on the fact that, because of the duplicate, some index will eventually point to the same value and create a cycle. It is also based on the mathematical proof that the slow and fast pointers will meet at the duplicate number.

Remember, the array should not contain any zeros because zero as an index might lead to a situation where we are stuck at the first position, since we’d be trying to access the zeroth index again and again. In this case, the precondition is that all the numbers are positive and less than n.

hey thanks for explaining .i have a question. the moment we are reaching the junction btw straight line and cycle is the point of duplicate number . so why we need to traverse second time?

That’s a great question! The point where the hare (fast pointer) and tortoise (slow pointer) meet within the cycle isn’t necessarily the duplicate number - it’s just a point within the cycle.

The reason we traverse the array again is to find the starting point of the cycle, which is the duplicate number. We do this by resetting one pointer to the start of the array (but keeping the other pointer at the meeting point), and then moving both pointers one step at a time. When they meet again, that’s the starting point of the cycle, and thus the duplicate number.

Why does this work? Consider the distance each pointer travels:

• When the tortoise enters the cycle, the hare is somewhere within the cycle (since it moves faster). They will eventually meet inside the cycle, but not necessarily at the start of the cycle (the duplicate number).

• By the time they meet, both pointers have traveled a distance that is a multiple of the cycle’s length. The hare has traveled twice the distance of the tortoise.

• The distance from the start of the array to the start of the cycle (let’s call this distance “a”), plus the distance from the start of the cycle to the point where the two pointers meet within the cycle (let’s call this distance “b”), is equal to the total distance the tortoise has traveled.

• Therefore, the distance from the start of the array to the start of the cycle (“a”) is equal to the distance from the point where the two pointers meet within the cycle to the start of the cycle if you continue moving in the direction of the cycle.

That’s why, when you reset the tortoise to the start of the array and move both pointers one step at a time, they will meet at the start of the cycle (the duplicate number).

a more simpler approach to this question public int findDuplicate(int[] nums) {

    for (int num : nums) 
    {
        int idx = Math.abs(num);
        
        if (0 > nums[idx]) 
            return idx;
        
        nums[idx] = -nums[idx];
    }

    return -1;
 
}

optimsed version of second approach told by striver in this video ,in that we were using extra array to know that the value is visited or not here we are using given arrray tc = O(n), Sc = O(1)

This is indeed an optimized and simpler approach, but there’s one thing that needs to be considered. This method modifies the input array, which might not be acceptable in all situations. It depends on whether the problem statement allows input modification or not.

Here’s a brief explanation of the solution:

This solution works by iterating over the array, treating the value of each element as an index. For each value, we negate the element at the corresponding index in the array. Because the array values are all positive and within the range of valid indices, we can be sure that each value corresponds to a valid index.

If we encounter a negative value at an index, it means that the index has been visited before, thus we have found a duplicate.

This solution has a time complexity of O(n), and it doesn’t require extra space (space complexity is O(1)). However, it modifies the input array, which might be a problem in some cases. Also, this approach works because all numbers are positive and within the array’s index range. It won’t work with negative numbers or numbers greater than the size of the array.

Also, the value -1 being returned for no duplicate found could be an issue if negative numbers are allowed in the array. In such cases, the function would need to return a different value that can’t possibly be an element of the array, or throw an exception.

So it’s always good to understand the problem constraints fully before selecting an approach!

Intiution & proof:

1. First collision: (when fast meets slow in the cycle) (This will help to find whether cycle exists or not.) fast is just moving twice the speed as slow. slow is moving 1 unit , fast is moving at 2 units. so, fast = 2* slow. now take some distance n(consider it as a ring road :-p ) To cover distance n , slow needs n units fast needs 2n units right (fast =2*slow) This mean eventually they are meeting. I hope this makes sense.
2. Finding where collision occured (the entry point/duplicate number in this case): From (1)-> we already know there is some entry point (where cycle starts) & a meeting point (where fast meets slow). lets say distance from starting point to entrypoint = d1; distance from entrypoint to meetingpoint = d2; distance from meetingpoint to entrypoint (which makes cycle) = d3; slow covers d1 distance and d2 to reach meeting point, so slow = d1+d2; fast covers d1,d2 d3 and again d2 to reach meeting point, so fast = d1+d2+d2+d3 = d1+2d2+d3 from 1, we know fast = 2* slow d1+2d2+d3 = 2(d1+d2) which mean ( d1=d3 ) It concludes distance between d1 (starting & entry) equals d3(meetingpoint to entrypoint), thats the intuition behind starting fast pointer at head for the second time and moving both 1 unit, at equal distance theyll find the duplicate/entrypoint.

That’s correct, you have nicely explained the concept of Floyd’s Cycle Detection algorithm (also known as the Tortoise and the Hare algorithm), and how it applies to find duplicates in an array.

To summarize the intuition & proof:

1. First collision: This part of the algorithm helps identify if there’s a cycle in the sequence. The fast pointer moves two steps at a time, while the slow pointer moves one step at a time. Since the fast pointer is moving twice as fast as the slow pointer, if there is a cycle, they are guaranteed to meet at some point within the cycle.

2. Finding where the collision occurred: After detecting a cycle in the sequence, the next step is to find the starting point of the cycle which is also our duplicate number. Based on the distance equations you have described (d1 = d3), we restart one of the pointers from the start of the sequence and move both pointers one step at a time. They will meet again at the starting point of the cycle, which is the duplicate number.

This technique works well because it exploits the property of cycles and doesn’t require any additional space, hence the space complexity is O(1). However, it only works for problems where the numbers in the sequence are within the range from 1 to n (the length of the sequence) and there is exactly one duplicate. Otherwise, the algorithm would need modifications or a different approach altogether.

I have an alternate approach after finding the first collision point. We can fix the fast pointer and move the slow pointer only to find the length of the loop. Once we have the length, We can keep one pointer at the start and other at a distance of lenght. Move them one step each and find the meeting point. That gives the duplicate/starting point of loop.

Yes, you’re absolutely correct, and your approach is another variation of the Floyd’s Cycle Detection Algorithm. This is often referred to as Floyd’s “cycle length” method.

To summarize your approach:

1. Detect the cycle: Same as before, you move two pointers at different speeds (one at twice the speed of the other). They will eventually meet somewhere in the cycle.

2. Find the length of the cycle: Once a cycle is detected, you can keep one pointer stationary at the point of collision and move the other pointer one step at a time, counting the number of steps it takes to get back to the point of collision. This will give you the length of the cycle.

3. Find the start of the cycle (or the duplicate number): Now, you reset both pointers back to the start. Move one pointer ‘cycle length’ steps forward. Then, start moving both pointers at the same speed. When they collide again, this will be the start of the cycle (and also the duplicate number in this context).

This approach also takes advantage of the properties of cycles in the same way and is still an O(n) time complexity solution with O(1) space complexity. It can be useful when it’s important to know the length of the cycle or when it might affect the logic of the problem at hand.

This can be done using negation of indices in very easy manner. just check abs value in the condition so that to refrain from modification of array.

class Solution { public: int findDuplicate(vector& nums) { //map<int,int>ans; for(int i=0;i<nums.size();i++) { if(nums[abs(nums[i])-1] > 0) { nums[abs(nums[i])-1] = -nums[abs(nums[i])-1] ; } else return abs(nums[i]); }

    return 0;
}

};

Your approach to the problem is correct but it assumes that you can modify the input array. It works by essentially turning the nums array into a map where the indices map to whether or not that number has been seen.

Here’s the reasoning behind your code:

1. For every number num in the nums array, go to the position in the array corresponding to num - 1 (since indices are 0-based and num is 1-based) and negate the number there. This essentially marks that index as ‘seen’ or ‘visited’.

2. If ever you encounter a number num where the position in the array corresponding to num - 1 has already been negated (i.e., it’s less than zero), then num is the number that is repeated.

However, this approach does not work if the array cannot be modified, and it also does not handle the case where the array contains zeroes. Also, this approach might not work in some languages if the integers are unsigned, because you can’t store negative values in unsigned variables.

The time complexity of this approach is O(n) and the space complexity is O(1), if we are allowed to modify the input array.

I have one more approach which is also O(n) in time complexity and O(1) in space complexity. As many of us might find difficulty while solving question using linked list therefore what I am doing is that just traverse through an array and then whatever that value of the element is suppose a[0] = 4 , then we just reverse the sign of 3rd element in the array (assuming 0 based indexing) and we do that for whole . Once we completed traversing we only left only 1 positive element therefore our ans is index of that element + 1 ( because it’s 0 based index). Let’s understand this by an easy example suppose we have given an array a = [1, 3,2,2] and it’s also given that all the elements in an array is in the range 1 to n(inclusive) there we start traversing through an array. Initially a[0] = 1 therefore we invert the sign of 0th element(1-1) now our array becomes -1, 3,2,2 after that we move forward now we have a[I] = 3 therefore we invert the sign of a[2] which implies now our array is -1, 3,-2, 2 then we do that with rest of the elemts and one thing that you need to keep in mind is always take absolute value of the element because we are inverting therefore at last we have -1, 3,-2, 2 now the first positive value is 3 with index 1 implies our is 2. You might be thinking of that 2 is also positive but why it’s not the answer it’s because the index of 2 is 3 but n is 3 therefore it is not possible.

Yes, your approach is correct and quite efficient, as it solves the problem with O(n) time complexity and O(1) space complexity. However, like the previous approach that uses negation, this approach also modifies the input array, which might not be acceptable in some cases.

The approach iterates over each number in the array, treating the absolute value of each number as an index (with 1 subtracted to account for 0-based indexing) and then negates the value at that index in the array. This way, every index that corresponds to a number in the array will become negative.

The duplicate number will be the one that tries to negate an already negative value. Hence, when the first positive value is encountered in the array, the index of that number plus 1 (to account for 0-based indexing) will be the duplicate number.

Please note that, as you said, this approach assumes that all elements are in the range from 1 to n and there are no zeros in the array. It also doesn’t handle the case when there are multiple duplicate numbers.

Another point to be careful about is that this method changes the input array. In some problems, it might not be allowed to change the input array. You should always confirm whether you can modify the input array before choosing the method to solve the problem.

If we will watch all the videos then how can we develop problem-solving skills?(Assuming that we know basic DSA)I mean should we watch all the videos and practice the remaining problems or should we approach first and then for better solutions we should watch your video?

And if we are watching these videos first then what should we do for practice(remaining questions)?

Great question! Improving problem-solving skills, especially in the context of programming and algorithm design, involves a combination of learning, practice, and review. Watching educational videos can provide you with a strong theoretical foundation, but it is essential to apply that knowledge through practical problems to develop and hone your skills.

Here’s a suggested approach:

1. Learn the Basics: Before attempting to solve complex problems, you need a good understanding of the fundamentals. This includes basic programming principles, data structures, and algorithms. You should spend some time learning these concepts through books, online courses, and educational videos.

2. Practice: Once you’ve learned the basics, start solving problems on your own. Choose problems that are slightly above your current level to challenge yourself, but not so difficult that you feel overwhelmed.

3. Review Solutions: After attempting a problem, compare your solution with the ones provided in the videos or problem discussion forums. This helps you to learn different approaches and improve your solution design.

4. Iterative Learning: Even if you watched the video solutions, it’s helpful to revisit problems after some time and try to solve them without looking at the solutions. It helps to reinforce your understanding and see how much you’ve improved.

5. Participate in Contests: Participating in coding contests on platforms like Codeforces, HackerRank, LeetCode, etc., can help you deal with time pressure and expose you to a variety of problem types.

6. Understand Mistakes: When you encounter a problem you cannot solve, don’t just look at the solution and move on. Spend time understanding why you couldn’t solve it, and what you can learn from the experience.

Remember, problem-solving is a skill that takes time to develop, so don’t get discouraged if you struggle with some problems. Stay patient and persistent, and your abilities will improve over time.

I’m assuming n =9 and the input test array as nums={1,2,2,3,4,7,6,5,9,8}. So I believe the input array matches the expected constraints, can you please let me know if we will have to make any kind of check here? Or am i just assuming the nums array wrong?

Your assumption about the nums array seems to be correct according to the constraints of the problem typically associated with this type of array (such as LeetCode’s “Find the Duplicate Number”).

In this problem, the constraints are usually that:

1. Only one duplicate number exists in the array.
2. The array nums length is n + 1 (where n is the maximum value in the array).
3. All numbers in nums are in the range [1, n] inclusive.

Given these constraints, your array nums={1,2,2,3,4,7,6,5,9,8} does indeed meet these conditions. The length of the array is 10 (which is n + 1, where n is the maximum number in the array 9), and all numbers in the array are within the range [1, n] inclusive. The duplicate number in this case is 2.

These constraints are important because they allow for certain solution strategies, like the “Tortoise and Hare” cycle detection or index negation methods mentioned earlier, to be applied.

If there were more than one duplicate, or if the numbers in the array were not within the range [1, n], then those strategies would not necessarily work, and a different approach might be needed.

The below attached code also works. The problem with this code is this will modify the array(which according to the question is not to be done).

My argument here is the array can be easily restored in O(n) time complexity. So should we go with this kind of approach in the questions where it is asked not to modify the array?

for (int i = 0; i < nums.size(); ++i) {

        int index = abs(nums[i])-1;
        
        if (nums[index] < 0)
            return index+1;
        nums[index] = -nums[index];
    }

//This loop is to restore the original array. for(int i=0; i< nums.size(); i++){ nums[i] = abs(nums[i]); }

The answer to your question largely depends on the specifics of the problem statement and the interview context.

If the problem explicitly states that you should not modify the input array, then strictly speaking, you should comply with this requirement. Modifying the array, even if you restore it afterwards, technically violates this condition. In an interview situation, your solution might be rejected based on this, even though your approach could solve the problem correctly and efficiently.

However, if the problem does not clearly specify whether you can modify the input or not, then it might be acceptable to use this approach, especially if it provides significant performance benefits. It’s often a good idea to discuss your approach with the interviewer to see if they consider this to be a reasonable solution.

As a rule of thumb, it’s always a good idea to follow the problem’s instructions as closely as possible, and if you’re unsure about something, don’t hesitate to ask for clarification. It shows that you care about getting the details right, which is an important quality in software development.

To understand this solution, you just need to ask yourself this question. Assume the distance from head to the start of the loop is x1 the distance from the start of the loop to the point fast and slow meet is x2 the distance from the point fast and slow meet to the start of the loop is x3

What is the distance fast moved? What is the distance slow moved? And their relationship?

x1 + x2 + x3 + x2 x1 + x2 x1 + x2 + x3 + x2 = 2 (x1 + x2) Q: Why x1 + x2 + x3 + x2 = 2 (x1 + x2)?

Ans: Its just the relationship between the distance travelled by fast ptr and the distance travelled by the slow ptr. As fast ptr, runs twice the speed of slow ptr. Therefore, when they meet, distance travelled by fast ptr will be 2*(distance travelled by slow ptr).

Therefore, x1 = x3.

Hope this makes sense.

Yes, that’s correct. It’s essentially applying the logic behind the Floyd’s cycle detection algorithm (also known as the tortoise and hare algorithm) to find the duplicate in the array.

To explain a bit more:

• The slow pointer (tortoise) moves one step at a time, and the fast pointer (hare) moves two steps at a time.
• If there’s a cycle (duplicate number) in the array, the slow and fast pointers will eventually meet. This is because the fast pointer is catching up to the slow pointer from behind at a rate of one step per unit of time.
• Let’s say at this point, the slow pointer has moved x1 + x2 steps and the fast pointer has moved x1 + x2 + x3 + x2 steps. Because the fast pointer moves twice as fast as the slow pointer, we have 2 * (x1 + x2) = x1 + x2 + x3 + x2. Simplifying this equation gives us x1 = x3.
• This implies that the distance from the start of the array to the start of the cycle (the duplicate number) is the same as the distance from the meeting point of the pointers to the start of the cycle.

So, if we reset one of the pointers to the start of the array after they first meet and then move both pointers at the same speed, they will meet again at the start of the cycle, which is the duplicate number.

if input is not bound to meet then will it iterate infinitely?

The method of using slow and fast pointers (Floyd’s cycle detection algorithm) assumes that a cycle exists in the data structure being iterated over. In the context of finding duplicates in an array where all numbers are within the range from 1 to n (size of the array), a cycle will always exist if there’s a duplicate.

If the input doesn’t meet this assumption – for instance, if there’s no duplicate, or if there are numbers outside the 1 to n range – then the behavior of the algorithm is not defined. It might result in an infinite loop, or it might result in trying to access an element outside the bounds of the array, leading to an error.

So, you should always make sure that the assumptions of the algorithm are met by your input. If you are not guaranteed that there will be a duplicate, you need to modify the algorithm to handle this case or choose a different approach entirely.

how did you understand that the linked list cycle method will lead to the optimal solution? Moreover, How did you know, we need to use the tortoise method along with the linked list cycle method? Is there anything to see or understand from the question to get to know when to use this method?

The “Linked List Cycle” (Floyd’s Cycle Detection) method is a common technique to detect cycles in data structures, particularly linked lists. However, its usage is not limited to linked lists only. In many problems, it’s about seeing the hidden linked list or cycle in the structure.

In the problem of finding duplicate numbers in an array where the numbers are from 1 to n, we can intuitively map each number as an index in the array, thereby treating the array as a sort of linked list where each element points to another element based on its value.

If there is a duplicate number, this means there are at least two elements pointing to the same index, which creates a cycle in this “hidden” linked list. Thus, we can apply Floyd’s Cycle Detection algorithm to this problem.

Understanding when to use this method comes with practice and pattern recognition. The more problems you solve, the better you’ll become at recognizing these patterns.

In this particular case, some hints that this method might work include:

1. The numbers in the array are all within the range from 1 to n, which can naturally be mapped to array indices.
2. We’re looking for a duplicate, which in the “hidden linked list” interpretation would create a cycle.
3. The problem asks for constant extra space complexity, which rules out methods like hashing but allows for methods like Floyd’s Cycle Detection, which only needs two pointers.

Knowing these algorithm techniques and practicing a lot of problems is the key to developing the ability to recognize when a certain technique will be applicable.

I have an approach please correct. Let’s say I stand at index 0, now I see value at arr[0] and swap arr[0] with arr[arr[0]] , I repeat the same for the new arr[0] value , while swapping I check if both the swapping numbers are same, if yes I break the loop and print.

Your approach can potentially solve the problem, however, it will violate one of the key constraints of the problem which is “not to modify the array”. The swapping operation you mentioned would modify the array.

Also, there is another catch, while doing the swapping operation you might end up in an infinite loop for certain inputs. Consider an array [2, 1]. If you start at index 0, you will swap arr[0] with arr[arr[0]], which means you will swap arr[0] with arr[2]. This is already an out of bounds operation, assuming 0-based indexing, and it gets worse if you assume 1-based indexing and have a number in the array that is out of the bounds of the array size.

It’s crucial to always consider the constraints of the problem and verify your solution against various edge cases. Your approach might be useful in certain contexts but it’s not applicable here because it modifies the array and potentially leads to out of bounds exceptions or infinite loops.

cant we do like this :- first we calculate the sum of all element present in the array say it is arraySum , now we calculate the sum of n given natural number say it is sum then (arraySum - sum ) gives the duplicate element

Yes, your approach would work if there’s only one duplicate element and if all the elements from 1 to n are present in the array. However, if there are multiple duplicates or if some numbers are missing in the range 1 to n, then this approach won’t work. Also, if the numbers are very large, summing them could lead to integer overflow.

Additionally, the given problem also has a specific constraint: “You must not modify the array (assume the array is read only).” Using a summation approach does not violate this constraint and hence is valid in this context. But as I mentioned earlier, it won’t work in all cases.

Here’s how you could implement your approach in Python, assuming only one duplicate:

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def findDuplicate(nums):
    n = len(nums) - 1
    expected_sum = n * (n + 1) // 2
    actual_sum = sum(nums)
    return actual_sum - expected_sum

This will return the duplicate number. But remember, it only works under certain assumptions. The most reliable solutions would be the ones discussed earlier that use a set to detect duplicates, or Floyd’s Tortoise and Hare algorithm to find a cycle in the array indices.

We can also use count sort method right?

Count sort is not typically used for finding duplicate elements. Count sort is a sorting technique based on keys between a specific range. It works by counting the number of objects having distinct key values, then doing some arithmetic to calculate the position of each object in the output sequence.

For this specific problem, using a variation of counting (like counting frequency of each element) can indeed help to find duplicate elements, but it comes with increased space complexity. You’ll need to create a separate count array, so if the input array contains a large number of distinct elements, the space complexity could be high.

Here’s how you could implement this in Python:

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def findDuplicate(nums):
    count = [0] * (len(nums) - 1)
    for num in nums:
        if count[num - 1] != 0:
            return num
        count[num - 1] += 1

However, the problem statement specifies that the solution must not modify the input array and must use only constant extra space. A counting approach does not meet these constraints, so it’s not an optimal solution for this problem. The cycle detection approach (Floyd’s Tortoise and Hare algorithm) is the most efficient and suitable one given these constraints.

This is a variation of the repeating and missing numbers in n sized array problem.

Yes, you are correct. The problem of finding a duplicate in an array where every number is between 1 and n (inclusive) is indeed a variation of the problem of finding repeating and missing numbers in an array.

The “Find the Duplicate Number” problem is simpler in a way, because you only need to find the duplicate number, not a missing number. But in both problems, the constraint that all numbers are in the range 1 to n (inclusive) allows us to use the values as indices in a way that we can mark visited numbers or detect cycles.

However, these problems differ slightly in terms of constraints. In the “Find the Duplicate Number” problem, you are specifically asked not to modify the input array and to use only constant extra space. This means that some solutions that would work for the “repeating and missing numbers” problem (such as marking visited numbers by changing the sign of the corresponding element in the array) would not be applicable here.

how is the distance travelled by F pointer 2a?

In the problem of finding a duplicate number using the slow and fast pointer method (also known as the tortoise and hare method), we imagine a “cyclic” space created by the indices and values of the array.

The slow pointer (S) and the fast pointer (F) start at the same place. Each step for S is one unit, while each step for F is two units, effectively meaning F is travelling twice as fast as S.

When the fast pointer enters the cycle, it’s still moving twice as fast as the slow pointer. If the slow pointer moves ‘a’ units to meet the fast pointer inside the cycle, the fast pointer has moved ‘2a’ units because it is moving at twice the speed. In the time the slow pointer moves ‘a’ units, the fast pointer moves ‘2a’ units.

This is the reason why we say the fast pointer travels ‘2a’ distance during the time the slow pointer travels ‘a’ distance.

I didn’t really get why this approach should work part. Could someone elaborate on it a little more?

The idea behind this solution is to find a cycle in the graph represented by the array. Each array element can be considered as a pointer to another element in the array (using the value as an index). If there’s a duplicate value in the array, it means there are at least two elements point to the same next element, which forms a cycle.

The slow and fast pointers method, also known as Floyd’s Cycle Detection Algorithm or Tortoise and Hare method, is a way to detect cycles in a graph or linked list. Here is how it works:

1. You start with two pointers, a slow one that moves one step at a time, and a fast one that moves two steps at a time.
2. If there’s no cycle, the fast pointer will reach the end of the array first.
3. If there’s a cycle, at some point, the fast pointer and the slow pointer will meet inside the cycle, because the fast pointer is moving at twice the speed of the slow pointer. This point is not necessarily the duplicate number or the starting point of the cycle.
4. After the first meeting, reset the fast pointer back to the start point, and make it move at the same speed as the slow pointer. They will meet again at the start point of the cycle, which is the duplicate number.

The key point to understand here is that the distance from the start point to the cycle start point is the same as the distance from the first meeting point to the cycle start point.

The reason why it works is a bit mathematical. Assume the distance from the start to the cycle start is a, the distance from the cycle start to the first meeting point is b, and the distance from the first meeting point back to the cycle start is c. When the first meeting happens, the slow pointer has traveled a+b, the fast pointer has traveled a+b+c+b (which is the distance slow pointer traveled plus one loop in the cycle). Because the fast pointer travels at twice the speed of the slow pointer, a+b+c+b = 2*(a+b), so a = c. That’s why when we reset the fast pointer back to the start and move it at the same speed, it will meet the slow pointer at the cycle start point (the duplicate number).

var findDuplicate = function(arr) { for (let i = 0; i < arr.length; i++) { if (arr[Math.abs(arr[i]) - 1] < 0) return Math.abs(arr[i]); else arr[Math.abs(arr[i]) - 1] = -arr[Math.abs(arr[i]) - 1]; } return -1; };

Yes, this solution also works for the problem and it’s an optimized version of the approach where we use extra space to keep track of visited elements.

Here, you are basically using the array itself to keep track of visited numbers by making them negative. If you encounter a negative number, it means that you have seen this number before, and hence, you return the absolute value of this number.

However, one thing to note here is that this approach modifies the original array, which might not be desired in some scenarios. Also, this solution assumes that all numbers in the array are positive to start with, which is true for this problem as per the problem statement.

Here is how the code works:

1. Traverse the array.
2. For every element you encounter (take its absolute value to account for the numbers you’ve made negative), check the sign of the number at the index equal to the current element minus one (since array is 0-indexed and numbers start from 1).
3. If the number at that index is positive, it means you have not seen the current element before, so mark it as seen by negating the number at the corresponding index.
4. If the number at that index is already negative, it means you have seen this number before, so it’s a duplicate. Return the absolute value of the current element.

Remember to take the absolute value of arr[i] every time you use it, because you’ve been flipping signs in the array.

And finally, if you’ve gone through the entire array and haven’t found a duplicate, return -1 (or any indicator to show that no duplicate was found). However, according to the problem statement, we are guaranteed that there is exactly one duplicate, so this final return might not be necessary in this particular case.

Suppose there are 20elements in range (0-9). Then we mark index 0-19 for each element. So there will be a moment when all elements will be in cycle before we finsh visiting all index. What to do then?

The problem statement specifies that each integer is between 1 and n (inclusive) where n is the number of elements in the array, excluding the repeated number. In your case, the elements should be in the range (1-20) rather than (0-9), so there shouldn’t be a case where all elements will be in a cycle before we finish visiting all indices.

The problem also specifies that there’s only one duplicate number, but it could be repeated more than once. This ensures that there will be a cycle in the array.

In the scenario you described with 20 elements in the range (0-9), this problem’s assumptions do not hold. The methods discussed would not apply to such a scenario because the logic of those methods depends heavily on the specific constraints given in the problem.

If you have a problem with different constraints, such as multiple duplicates or a different range of possible values, you might need a different approach, such as using a frequency count or a hash map to track duplicates.

Keep in mind that when solving problems like this, the assumptions and constraints given in the problem are crucial to coming up with an optimal solution. If those change, the solution might need to change as well.

Suppose in the first index the element is “0” what should we do..?

According to the problem constraints as mentioned in a typical “find the duplicate number” problem, all the integers are within the range from 1 to n (inclusive) where n is the size of the array, so there shouldn’t be a “0” in the array.

However, if your problem has different constraints and does include “0”, you would need to modify the approach slightly. You cannot simply use the numbers as indices, since “0” would refer to the first position and cause a possible infinite loop if there’s a “0” in the array.

In such case, one workaround could be to add 1 to every number in the array before applying the methods we discussed. This would change the range from (0 to n-1) to (1 to n), which fits into the assumptions of the methods. But remember, this approach changes the input array, which might not be allowed in some situations.

Another approach is to use a HashMap or a Set to store the numbers that have been seen so far. If a number is seen again, then it’s a duplicate. This would work even if the numbers include “0” and won’t modify the array, but it does require O(n) extra space.

A suggestion, please try to think of the entire test case keeping constraints in check, and then you can put forward a question. The reason am saying this is this will help you grow, many others had the same doubt, but when they thought of the complete test case, they realised that the cycle was build in all cases. Keeping your case, can you give me the entire test case rather than hypothetically saying two numbers!

I appreciate your suggestion and it is indeed a great approach to learn and grow. It’s important to test our logic against different test cases, especially the edge cases, and always consider the problem constraints. This will not only help in detecting the errors in the logic but also in coming up with a more optimal and foolproof solution.

Taking your example, it’s mentioned that we have 20 elements within the range of 0 to 9. According to the problem constraints of the “find duplicate in an array” problem, all the elements are within the range 1 to n (where n is the size of the array), thus, your case would violate these constraints. If the range of the numbers is different, the problem becomes a different one and requires a different approach.

However, if you still want to solve it using a similar approach as discussed earlier, then it’s important to remember that there will be a cycle because of the duplicates, and not necessarily at the start of the traversal. Even if we start at some index, as we keep jumping according to the values, we’ll eventually enter a cycle (if one exists) because there are duplicate numbers. It’s also possible that all numbers form a cycle, in which case we can start at any index.

It’s also important to note that a “cycle” doesn’t always mean an infinite loop. It just means that the flow of traversing will eventually come back to a point it’s already visited before.

Could you please provide a more concrete test case, keeping in mind the constraints of the problem? That would be very helpful in providing a more specific answer.