Find the Duplicate Number

The problem requires us to find the single duplicated number in an array without modifying the array and using only constant extra space.

Approach: Floyd’s Tortoise and Hare (Cycle Detection)

This approach uses the Floyd’s Tortoise and Hare (Cycle Detection) algorithm to identify the duplicate element.

  1. Find Intersection Point: Treat the array as a function f(x) = nums[x] that maps values to the index they point to. Using two pointers, one moving twice as fast as the other, find their intersection point inside the loop.

  2. Find the Entry Point: Use two pointers, one starting at the beginning of the array and the other at the intersection point. Move both pointers one step at a time. The point where they meet is the entry point of the loop, which corresponds to the duplicate number.

Code

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class Solution:
    def findDuplicate(self, nums: List[int]) -> int:
        # Step 1: Find the intersection point
        tortoise = nums[0]
        hare = nums[0]
        while True:
            tortoise = nums[tortoise]
            hare = nums[nums[hare]]
            if tortoise == hare:
                break

        # Step 2: Find the entry point of the loop
        ptr1 = nums[0]
        ptr2 = tortoise
        while ptr1 != ptr2:
            ptr1 = nums[ptr1]
            ptr2 = nums[ptr2]

        return ptr1

Explanation

  • The problem’s constraints (each integer being in the range [1, n] and nums.length == n + 1) guarantee that a cycle exists.
  • By finding the entry point of the loop, you can identify the duplicate number.

Key Takeaways

  • This approach does not modify the array and only uses constant extra space.
  • The time complexity is (O(n)), where (n) is the length of the array.
  • The space complexity is (O(1)) since we use only constant extra space.

Identifying Problem Isomorphism

“Find the Duplicate Number” can be mapped approximately to “Linked List Cycle II”.

The main connection between these two problems is that they both involve finding a point of repetition in a sequence - in one case a repeating integer in an array, and in the other a node in a linked list where a cycle begins.

The isomorphism is approximate because the data structures and problem specifics differ: one involves an array of integers and the other a linked list. However, the concept of detecting a cycle or a duplicate is a common theme, which can be addressed by similar approaches, such as Floyd’s cycle-finding algorithm.

Therefore, though the problems are not identical in nature, they do share a conceptual isomorphism that can be beneficial in deriving solutions for one another.

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def findDuplicate(self, nums: List[int]) -> int:
        N = len(nums) - 1
        seen = [0] * (N+1)
        for num in nums:
            if seen[num]:
                return num
            seen[num] = 1

Language Agnostic Coding Drills

  1. Understanding basic data types: integers, lists.
  2. Understanding how to define a function, function parameters, and return values.
  3. Understanding the length of a list.
  4. Understanding list index-based access and assignment.
  5. Understanding basic arithmetic operations like addition and subtraction.
  6. Creating and initializing a list with default values.
  7. Looping over a list using a for loop.
  8. Understanding conditional statements (if).
  9. Understanding the usage of return in a function.

Here are the units arranged in increasing level of difficulty:

  1. Understanding basic data types: integers, lists.
  2. Understanding basic arithmetic operations like addition and subtraction.
  3. Understanding how to define a function, function parameters, and return values.
  4. Understanding the usage of return in a function.
  5. Understanding the length of a list.
  6. Creating and initializing a list with default values.
  7. Understanding list index-based access and assignment.
  8. Looping over a list using a for loop.
  9. Understanding conditional statements (if).

This code is used to find the first duplicate number in the list of numbers passed into the function findDuplicate. The function loops over the list nums, for each number it checks if the number has been seen before (checked by seeing if seen[num] is true). If it has been seen before, it returns that number because it is the first duplicate. If it has not been seen before, it sets seen[num] to 1 to indicate that this number has been seen. This is a basic usage of a boolean list to track which numbers have been encountered before in the list.

Targeted Drills in Python

These drills are meant to reinforce the understanding of each concept individually.

  1. Understanding basic data types: integers, lists.
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# Define some integers and lists. Try some basic operations with them.
a = 5
b = 10
int_list = [1, 2, 3, 4, 5]
print(a + b)
print(int_list[2])
  1. Understanding basic arithmetic operations like addition and subtraction.
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# Perform some basic arithmetic operations
a = 10
b = 5
print(a + b)
print(a - b)
  1. Understanding how to define a function, function parameters, and return values.
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# Define a function that adds two numbers and returns the result
def add_numbers(a, b):
    return a + b

print(add_numbers(5, 10))
  1. Understanding the usage of return in a function.
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# Write a function that checks if a number is even or odd and returns a message
def check_even_odd(num):
    if num % 2 == 0:
        return "Even"
    else:
        return "Odd"

print(check_even_odd(3))
  1. Understanding the length of a list.
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# Define a list and find its length
numbers = [1, 2, 3, 4, 5]
print(len(numbers))
  1. Creating and initializing a list with default values.
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# Create a list of length 10, initialize with 0
zeroes = [0] * 10
print(zeroes)
  1. Understanding list index-based access and assignment.
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# Define a list and perform some index-based access and assignment
numbers = [1, 2, 3, 4, 5]
print(numbers[2])
numbers[2] = 100
print(numbers)
  1. Looping over a list using a for loop.
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# Define a list and loop over it, printing each element
numbers = [1, 2, 3, 4, 5]
for num in numbers:
    print(num)
  1. Understanding conditional statements (if).
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# Write a condition that checks if a number is greater than another
a = 10
b = 5
if a > b:
    print("a is greater than b")
else:
    print("a is not greater than b")
  1. Problem specific: Using a boolean list to track elements.
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# Define a list of numbers and another list to track if a number has been seen
nums = [1, 3, 4, 2, 2]
seen = [0] * 5
for num in nums:
    if seen[num]:
        print(f"Duplicate: {num}")
        break
    seen[num] = 1

10 Prerequisite LeetCode Problems

For 287. Find the Duplicate Number, the following are good preparation:

  1. 136. Single Number

    • Relevant because: It teaches bit manipulation and dealing with duplicates. However, it’s simpler because there’s no space constraint.

  2. 26. Remove Duplicates from Sorted Array

    • Relevant because: It deals with removing duplicates in an array, though you can modify the array in this case.

  3. 217. Contains Duplicate

    • Relevant because: It’s another problem focused on identifying duplicates but without the space constraints.

  4. 268. Missing Number

    • Relevant because: Similar to finding a duplicate, but in this case, you are finding a missing number, which is the reverse problem.

  5. 448. Find All Numbers Disappeared in an Array

    • Relevant because: It helps you understand how to identify discrepancies in an array without modifying it.

  6. 121. Best Time to Buy and Sell Stock

    • Relevant because: It teaches how to keep track of variables as you iterate through the array, which is a useful technique.

  7. 53. Maximum Subarray

    • Relevant because: It teaches Kadane’s algorithm, which is useful for solving array problems that require optimization.

  8. 283. Move Zeroes

    • Relevant because: You have to rearrange numbers within an array, a skill useful for optimizing space.

  9. 169. Majority Element

    • Relevant because: It also deals with finding a specific kind of number (majority element) within an array.

  10. 122. Best Time to Buy and Sell Stock II

    • Relevant because: This is another problem that teaches optimization and dealing with arrays.

These cover bit manipulation, in-place array modification, and element search, that will prepare you for 287. Find the Duplicate Number.

Problem Classification

The problem falls under the “Arrays” and “Mathematics” domain in computer science, specifically algorithmic problem-solving. It’s a classic example of finding a duplicate element under strict constraints.

‘What’

  1. Input: An array nums containing n + 1 integers where each integer is in the range [1, n] inclusive.

  2. Output: A single integer, which is the only duplicate in the array.

  3. Constraints:

    • The array cannot be modified.
    • Only constant extra space is allowed.
    • The integer range and array length constraints (1 <= n <= 105 and nums.length == n + 1).
    • All integers in the array are unique except for one.

  4. Type: This is a “Search” problem, more specifically a “Duplicate Search” under constrained conditions.

  5. Difficulty: Given the constraints of constant space and not modifying the array, the problem adds a layer of complexity, making it a medium-level problem.

  6. Optimization Criteria: The problem asks for a solution that adheres to space and modification constraints, meaning you’re expected to optimize for both time and space complexity.

The problem is a classic search problem but with constraints that make it challenging. It’s in the domain of algorithmic problem-solving, specifically focusing on arrays and mathematical properties to find a duplicate number.

You have well-defined input, output, and constraints. You’re not asked to sort the array or transform it in any way; you’re simply required to find the single duplicate integer.

The problem becomes more complex due to the constraints of not modifying the array and using only constant extra space. This rules out many straightforward but space-consuming approaches, pushing you toward optimizing the solution for both time and space complexity.

The constraints make this a medium-level problem ideal for testing problem-solving skills, especially in making trade-offs between time and space complexity while adhering to specified conditions.

Clarification Questions

  1. Can the array have multiple duplicates or is it guaranteed to have only one duplicate number?
  2. Are negative numbers allowed in the array?
  3. Is it guaranteed that the array will always contain n + 1 integers?
  4. What should be returned if there’s no duplicate found? Is that scenario possible based on the problem constraints?
  5. Is the input array guaranteed to be non-empty?
  6. What is the maximum size for n? (Although this is mentioned in the problem as 1 <= n <= 10^5)
  7. Can the array contain zeros?
  8. Is the array sorted or unsorted? Does sorting the array break any constraints?
  9. What is the expected time complexity for solving this problem?
  10. Are we looking for the first duplicate found while iterating, or any duplicate will suffice for the solution?

These can provide insights into edge cases or any limitations that can influence your approach to solving the problem.

Problem Analysis and Key Insights

Analyzing the problem statement for 287. Find the Duplicate Number yields the following key insights:

  1. Constant Space: The problem explicitly asks for a solution that uses constant extra space. This rules out methods that involve creating a new data structure to keep track of elements.

  2. Unmodified Array: You’re not allowed to modify the input array, so sorting or rearranging elements is not an option.

  3. Boundaries of Numbers: Each integer is in the range [1, n] inclusive. This is a very crucial hint as it gives information about the data you’re dealing with.

  4. Single Duplicate: There is only one number that appears more than once, simplifying the problem to some extent.

  5. Length of Array: The length of the array is n + 1, which means it’s guaranteed that there’s at least one duplicate.

  6. Constraints: There’s a constraint on n, namely 1 <= n <= 10^5. This gives an upper limit, which can be useful to consider when thinking about time complexity.

  7. Integer Values: All the integers in the array appear only once except for precisely one integer. This simplifies the problem as you only have to find one integer that violates this rule.

Based on these insights, this seems to be a problem that tests your understanding of array manipulation and algorithmic complexity, particularly in scenarios where memory use is restricted.

Problem Boundary

The scope of the problem is array manipulation under specific constraints:

  1. Data Structure: The problem is confined to an array of integers.

  2. Algorithm Complexity: The problem explicitly restricts the solution space to algorithms that use constant extra space and do not modify the original array. This sets a specific scope on the type of algorithmic approaches that are applicable.

  3. Element Range: The integers in the array have a defined range between 1 and n, inclusive. This restricts the domain of the problem to positive integers that have a particular upper and lower bound.

  4. Duplicates: The problem is not about identifying all duplicates or their frequency. The scope is narrowed down to identifying just one repeated number.

  5. Constraints: With constraints such as 1 <= n <= 10^5 and nums.length == n + 1, the scope of the problem is further limited. It implies you have to be conscious of time complexity due to the possible size of the array.

  6. Output: The expected output is a single integer, the number that is duplicated.

The scope is around finding an efficient algorithm to identify a single duplicate integer within a given array, without modifying the array and while using only constant extra space.

To establish the boundary of the problem “287. Find the Duplicate Number,” you’ll need to consider the following aspects:

  1. Input Constraints:

    • Array length is n + 1.
    • Each integer is between 1 and n.

  2. Algorithm Constraints:

    • Constant extra space.
    • Do not modify the array.

  3. Output:

    • A single integer that is the duplicate.

  4. Test Cases:

    • Consider the minimum and maximum values for n (1 <= n <= 10^5).
    • Consider arrays where the duplicate number is either at the beginning, middle, or end.

By clarifying these aspects, you’re defining what is inside and what is outside the problem boundary. For example:

  • Inside the boundary: Finding the duplicate using a read-only operation on the array.
  • Outside the boundary: Sorting the array or using additional data structures like sets or hash maps.

By establishing these boundaries, you ensure that you’re solving the problem within the given constraints and are not venturing into solution spaces that, while possibly correct, don’t meet the problem’s requirements.

Distilling the Problem to Its Core Elements

Fundamental Concept

The fundamental concept behind this problem is the cycle detection in a linked list. The array can be treated as a linked list where each element is a node and the value at each index is the pointer to the next node. The cycle in the “linked list” signifies the duplicate number.

Simple Description

Imagine you have a list of room numbers. You start at the first room, go to the room number indicated on a card in that room, then go to the room indicated in the new room’s card, and so on. If you find yourself back in a room you’ve already been to, that’s the repeated room number.

Core Problem

The core problem is to find the single duplicate number in an array of n+1 integers where every integer is between 1 and n, without modifying the array and using constant space.

Key Components

  1. Array of integers: Stores n+1 integers between 1 and n.
  2. Duplicate Number: A single number that appears more than once in the array.
  3. Space Constraint: Use only constant extra space.
  4. Read-only Array: No modification to the original array is allowed.

Minimal Operations

  1. Iterate through the array to simulate a “linked list traversal.”
  2. Use two pointers (Floyd’s Tortoise and Hare technique) to detect the cycle.
  3. Once a cycle is detected, find the start of the cycle which will be the duplicate number.

These operations encapsulate the essence of solving this problem within the given constraints.

Visual Model of the Problem

Visualizing this problem can be very helpful for understanding its intricacies. Here’s how to do it:

Linked List View

  1. Imagine each element of the array as a node in a linked list.
  2. The value of each node is an index that points to another node in the list.
  3. For instance, with nums = [1,3,4,2,2], start at index 0, which has value 1. Then go to index 1, which has value 3, and so on.
  4. You’ll notice that you end up in a cycle: 2 → 4 → 2 → 4 → 2 → …
  5. This cycle signifies the duplicate element.

Graphical View

  1. Draw circles for each index in the array.
  2. Connect them using arrows according to the values in the array.
  3. For example, if nums[0] = 1, draw an arrow from the circle labeled ‘0’ to the circle labeled ‘1’.
  4. You’ll notice a cycle in this graphical representation.

Array Index View

  1. Write down the array: [1,3,4,2,2].
  2. Write down the steps you would take to traverse it, based on the array’s values.
  3. It will look like: Start -> 1 -> 3 -> 2 -> 4 -> 2 -> 4 -> ...
  4. Mark the point where the loop starts. In this example, it’s the second appearance of ‘2’.

Visualizing the problem in these ways helps in understanding how to detect the duplicate number using the concept of cycles.

Problem Restatement

You have an array with integers ranging from 1 to n. The array has n+1 elements, meaning one number is repeated. Your task is to find the repeated number. You can’t change the array, and you’re limited to using only a constant amount of extra space. All numbers appear just once except for the repeated one, which can appear more than once.

Abstract Representation of the Problem

In an abstract sense, you’re dealing with a finite sequence S of n+1 elements where each element belongs to a set N of integers from 1 to n. One element in S is repeated. The task is to identify the repeated element while adhering to constraints on memory and manipulation of S. The sequence S has the property that each element appears exactly once except for the repeated element. Your goal is to identify the duplicate without altering S and using constant additional memory.

Terminology

  1. Array: A data structure that stores multiple items of the same type in contiguous memory locations. The problem revolves around an array of integers.

  2. Constant Space: The problem asks for a solution that uses only a fixed amount of additional memory regardless of the array size. This is crucial for the solution’s efficiency.

  3. In-place Algorithm: An algorithm that transforms input using no auxiliary data structure. Though the problem doesn’t explicitly demand an in-place solution, it does stipulate that you can’t modify the original array, making this concept relevant.

  4. Constraints: Boundaries defined for the problem, such as the array length and the range of numbers in it. These give clues about the algorithmic complexity suitable for solving the problem.

Understanding these terms is essential for grasping both the problem and the types of solutions that are viable.

Problem Simplification and Explanation

You have a list of numbers. Every number appears once except for one, which appears twice. Your task is to find that number.

Key Concepts:

  1. Array: Think of it as a list of seats in a movie theater.
  2. Duplicate Number: Imagine there are tickets assigned to each seat, but one seat has two tickets.
  3. Constraints: You can’t move people (numbers) out of their seats (array), and you have limited pocket space (constant space) for any tools you use to find the duplicate.

Interaction:

  • You must go through the list (array) and find out which number (seat) has been repeated without rearranging the seats or using extra storage space.

Analogy: Imagine you are in a classroom where each student has a unique ID card. All students are seated. The IDs match the seat numbers. Suddenly, you realize two students share the same ID. You can’t ask them to stand or leave the room. Also, you don’t have any additional space to write down or keep records. You must figure out which ID is duplicated with these restrictions.

In essence, you need to find a clever way to identify the duplicate ticket or ID, adhering to the set rules.

Constraints

Specific Characteristics to Exploit:

  1. Range of Numbers: All numbers are between 1 and n (inclusive). This allows us to leverage the array indices for comparisons, as they also fall within the same range (0 to n).

  2. Constant Space: Since we can’t use extra memory proportional to the size of the array, methods that don’t require additional arrays or data structures are ideal.

  3. Single Duplicate: Knowing that there is only one duplicate can eliminate the need for general duplicate-checking algorithms and allow more focused methods.

  4. Length of Array: The array length being n+1 with elements ranging from 1 to n ensures that at least one number will be repeated.

  5. Non-modification: Since the array can’t be modified, algorithms like sorting are off the table, which narrows down the kinds of techniques you can apply.

Patterns or Ranges to Use:

  • The close relationship between the numbers in the array and their indices can be crucial. This relationship allows for techniques like cycle detection.

By identifying these characteristics, we can zero in on algorithms that meet the constraints while efficiently solving the problem. For instance, you might consider using the Floyd’s Tortoise and Hare (cycle detection) technique to find the duplicate number in constant space and linear time.

Key Insights from Constraints:

  1. Array Length: The length of the array being n+1, with elements ranging between 1 to n, signals that there must be at least one duplicate due to the Pigeonhole Principle.

  2. Constant Space: The constant space requirement rules out solutions that would involve extra data structures like hash maps or sets to store visited numbers.

  3. No Modification: Being unable to modify the array means we can’t use sorting or rearrangement methods. This also rules out any approach that would require swapping elements to their “correct” positions.

  4. Single Duplicate: Knowing there’s only one duplicate simplifies the problem. We don’t have to account for scenarios where there may be multiple duplicates, which makes cycle detection methods viable.

  5. Element Range: The constraint that the elements are between 1 and n (inclusive) is significant. This is crucial for exploiting array indices as a means of checking for duplicates without extra space.

By analyzing these constraints, we understand not only the limitations but also the avenues available for solving the problem. For instance, cycle detection methods become attractive due to the constraints of constant space and the presence of a single duplicate.

Case Analysis

Certainly, let’s explore some examples that cover various conditions:

Categorizing Test Cases

Minimal Case

  • Input: [1, 1]
  • Output: 1
  • Reasoning: The smallest array with length n+1 = 2. It clearly has one duplicate, which is 1.

Single Element Range

  • Input: [2, 2, 1]
  • Output: 2
  • Reasoning: All elements except one form a range from 1 to 2. The duplicate is 2.

Multiple Element Range

  • Input: [3, 2, 4, 4, 1]
  • Output: 4
  • Reasoning: The elements form a range from 1 to 4. The duplicate is 4.

Middle Duplicate

  • Input: [1, 3, 2, 4, 3]
  • Output: 3
  • Reasoning: The duplicate is not at the edges but in the middle of the sorted array.

All Elements Repeated Once Except the Duplicate

  • Input: [2, 1, 3, 3, 4, 5]
  • Output: 3
  • Reasoning: All numbers from 1 to 5 are there, with 3 being the duplicate.

Sorted Array with Duplicate at End

  • Input: [1, 2, 3, 4, 5, 5]
  • Output: 5
  • Reasoning: The array is sorted, and the duplicate is at the end.

Sorted Array with Duplicate at Start

  • Input: [1, 1, 2, 3, 4, 5]
  • Output: 1
  • Reasoning: The array is sorted, and the duplicate is at the start.

Unsorted Array

  • Input: [4, 1, 3, 2, 4]
  • Output: 4
  • Reasoning: The elements are unsorted, testing the algorithm’s ability to find the duplicate without sorting.

Edge Cases

  1. Minimum Array Length: Already covered under ‘Minimal Case’.
  2. Duplicate at Start or End: Covered under ‘Sorted Array with Duplicate at Start’ and ‘Sorted Array with Duplicate at End’.
  3. All Elements are Unique Except One: Covered under ‘All Elements Repeated Once Except the Duplicate’.

By analyzing these test cases, you can uncover key aspects of the problem like handling of sorted/unsorted arrays, duplicates at start/end/middle, and the minimum case scenario. Make sure your solution can handle these different cases to ensure it is robust.

Visualizing these cases can often help in understanding the problem better. Here are some ways to visualize the various categories:

1. Minimal Case

Array: [1, 1] Visualization:

1 - 1

2. Single Element Range

Array: [2, 2, 1] Visualization:

1 - 2 - 2

3. Multiple Element Range

Array: [3, 2, 4, 4, 1] Visualization:

1 - 2 - 3 - 4 - 4

4. Middle Duplicate

Array: [1, 3, 2, 4, 3] Visualization:

1 - 2 - 3 - 3 - 4

5. All Elements Repeated Once Except the Duplicate

Array: [2, 1, 3, 3, 4, 5] Visualization:

1 - 2 - 3 - 3 - 4 - 5

6. Sorted Array with Duplicate at End

Array: [1, 2, 3, 4, 5, 5] Visualization:

1 - 2 - 3 - 4 - 5 - 5

7. Sorted Array with Duplicate at Start

Array: [1, 1, 2, 3, 4, 5] Visualization:

1 - 1 - 2 - 3 - 4 - 5

8. Unsorted Array

Array: [4, 1, 3, 2, 4] Visualization:

4 - 1 - 3 - 2 - 4

In each visualization, the dash - separates the elements. Duplicates are visually next to each other, making it easier to spot them. Visualization helps in understanding the structure of the input array and the position of duplicates, which is crucial for problem-solving.

From analyzing the different cases, the following key insights emerge:

  1. Array Length: Since the array has n+1 integers with all integers being between 1 and n, the array length effectively sets the range for possible numbers.

  2. Uniqueness Constraint: All integers appear only once except for precisely one integer, which simplifies the task of finding the duplicate.

  3. Position Irrelevance: The duplicate number can be at any position in the array. It doesn’t have to be next to its counterpart. This insight hints that sorting might not be required.

  4. Constant Space: The problem requires solving without modifying the array and using only constant extra space. This constraint rules out methods like sorting the array or using a hash table for counting.

  5. Numerical Range: All integers are positive and within a specific range. This can be exploited to find an efficient algorithm.

  6. Edge Cases: The duplicate integer can be either the smallest or the largest integer in the array. Hence, any algorithm should not be biased towards the beginning or end of the array.

These insights help in formulating a strategy for solving the problem and understanding what to watch out for when testing the solution.

Identification of Applicable Theoretical Concepts

Several mathematical and algorithmic concepts can be leveraged to solve this problem efficiently:

  1. Cycle Detection: Since the array contains n+1 numbers but only n are unique, it forms a cycle. Floyd’s Tortoise and Hare (Cycle Detection) can be applied here to find the duplicate.

  2. Mathematical Properties: The sum of the first n natural numbers can be calculated using the formula n * (n + 1) / 2. Knowing the sum of unique numbers and the actual sum of the array can sometimes provide clues about the duplicate, although this may not adhere to the constant space constraint.

  3. Binary Search: In a variant approach, we can use Binary Search on the number range, not the array, to find the duplicate. This fits within the constraints of not modifying the array and constant space.

  4. Pigeonhole Principle: Since there are n+1 numbers but only n are unique, at least one number must be duplicated. This mathematical principle reassures us that a duplicate must exist.

  5. In-Place Swap: Though the problem restricts modifying the array, in-place swap is often used for similar problems to rearrange numbers to their correct positions. Understanding why this can’t be used here reinforces the requirement for another approach.

  6. Hashing: Normally, one could use a Hash Table to keep track of elements seen so far, but the constant space constraint rules this out. Yet, understanding why we can’t use hashing can guide us toward a suitable algorithm.

  7. Pointers: Two-pointer techniques can sometimes simplify array and linked list problems. While you can’t modify the array here, understanding two-pointer methods can help.

By identifying these mathematical and algorithmic properties, you can select an approach that both fits within the constraints and solves the problem efficiently.

Simple Explanation

Imagine you have a bag of marbles. Each marble has a number written on it. The numbers start from 1 and go up to a certain number, let’s say 5. Normally, you should have one marble for each number: one with 1, one with 2, and so on up to 5. But in this bag, there’s one extra marble that makes it different. One number is repeated.

Your task is to find out which number is on that extra marble without dumping them all out or rearranging them in any way. Also, you can only look inside the bag a limited number of times, so you can’t just pick each one and keep it aside.

In other words, you have to find the number that appears more than once with the least amount of digging into the bag.

Metaphor: It’s like playing “Guess the Odd One Out” where you have a set of items, and you know only one item appears more than once. You have to figure out what that one item is without changing the order of items or taking them out.

Problem Breakdown and Solution Methodology

Let’s dive into how to tackle this problem. The problem statement can be likened to having a loop in a linked list; you want to find the point where the loop starts. You have a sequence, and at some point, the sequence starts to repeat. Here’s how to solve it:

Steps to Solve the Problem:

  1. Initialize Two Pointers: Start with two pointers, a slow one and a fast one, both pointing at the first element in the array.

  2. Detect Loop: Move the slow pointer one step and the fast pointer two steps. Continue this until they meet. This confirms that a loop (or repetition) exists.

  3. Find the Start of the Loop: Reset one of the pointers to the beginning. Move both pointers one step at a time until they meet again. The point where they meet is the start of the loop, i.e., the repeated number.

  4. Return the Result: Once the repeated number is identified, return it as the output.

Metaphor:

Think of it like a racetrack where one runner is faster than the other. They start at the same point, and if the track forms a loop, they’ll eventually meet again. To find the starting point of that loop, one runner goes back to the start while the other stays put. They then start running at the same speed and meet at the loop’s starting line.

How Parameters Affect the Solution:

  • If the range of numbers [1, n] changes, it doesn’t impact the solution as the algorithm relies on the structure rather than the values.
  • If the constraint of constant space is removed, other techniques like Hashing can be used, but it’s not applicable here due to the constant space requirement.

Example:

Let’s say the array is [1, 3, 4, 2, 2].

  1. Initialize two pointers: slow = 1, fast = 1.
  2. Move slow one step and fast two steps: they become 3 and 4, respectively. Continue until they meet. They meet at 2.
  3. Reset slow to the start: slow = 1, fast = 2.
  4. Move them one step at a time: slow = 3, fast = 3; slow = 4, fast = 4; slow = 2, fast = 2.
  5. They meet at 2, which is our answer.

This approach efficiently solves the problem while respecting all given constraints.

Inference of Problem-Solving Approach from the Problem Statement

Understanding key terms and concepts is crucial for mapping out a solution approach. Here are the essential terms for this problem:

  1. Array: The problem revolves around an array of integers. The array serves as our primary data structure, dictating which algorithms can be applied.

  2. Integers: The array contains integers, which means arithmetic operations and comparisons are straightforward.

  3. Range [1, n]: This constraint simplifies the problem because we know the minimum and maximum values that can appear in the array.

  4. n + 1 Integers: The array length is n + 1, with integers from 1 to n. This condition signifies that there is at least one integer that must repeat.

  5. Repeated Number: This is what we need to find. Our algorithm must be designed to identify this specific element.

  6. Constant Extra Space: This constraint rules out data structures like hash tables that use additional space proportional to the array length. We must solve the problem in-place.

  7. No Modification: We can’t modify the array, so sorting or rearranging elements is not allowed.

Strategy or Method:

  1. Array & Integers: Since we are dealing with an array of integers, looping through the array and using indices are fundamental operations.

  2. Range [1, n]: Knowing the range can sometimes offer optimization, but in this case, it serves more as a confirmation that a repeat must exist.

  3. n + 1 Integers: This gives us the clue that a “cycle” or “loop” must exist in the array, which leads us to the “Floyd’s Tortoise and Hare” algorithm to detect cycles.

  4. Repeated Number: Our objective informs the algorithm choice. We’re not looking for the maximum, minimum, or any such value but a repeating number, confirming that we need an algorithm effective in identifying duplicates.

  5. Constant Extra Space: This constraint limits our choices of algorithms, further narrowing us down to “Floyd’s Tortoise and Hare” for cycle detection, as it doesn’t require extra space.

  6. No Modification: Being unable to modify the array removes sorting algorithms from our set of choices, making the cycle detection method even more appropriate.

By understanding each term and constraint, we are guided to use a cycle detection algorithm that works in constant space and without modifying the array.

Visual aids like tables and diagrams can be powerful tools for understanding problems. Here’s how to recognize the key properties using these tools:

  1. Array & Integers

    • Table: Create a table with two columns: one for the index and another for the integer value at that index. This can help you visualize the arrangement of numbers and explore patterns.

  2. Range [1, n]

    • Number Line: Draw a number line from 1 to n and mark each integer. This sets the boundary for possible integers and helps recognize that a repeat must occur within this range.

  3. n + 1 Integers

    • Table Extension: Extend your table to n+1 rows to capture all integers, highlighting that there must be one repeating integer.

  4. Repeated Number

    • Venn Diagram: Use a Venn diagram to show the universe of integers from 1 to n and identify the subset where duplication happens. It can help you focus on finding that overlap.

  5. Constant Extra Space

    • Legend: Use a separate column in your table or a sidebar in your diagram to note any temporary variables you might need. The goal is to ensure this sidebar stays minimal, complying with the constant extra space constraint.

  6. No Modification

    • Immutable Table: Make a visual note (maybe an asterisk or a different color) to indicate that the table values are not to be altered. This reinforces the constraint as you work on your solution.

  7. Cycle Detection

    • Graph: Use arrows to point from each number to the number at its index. This can help you visually track cycles, which would manifest as closed loops in the graph.

By employing these visual aids, you’ll have a more tangible grasp of the problem’s constraints and requirements, aiding in your solution design.

How did you infer from the problem statement that this problem can be solved using ?

Simple Explanation of the Proof

I’m having trouble understanding the proof of this algorithm. Could you explain it in a way that’s easy to understand?

Stepwise Refinement

Stepwise Refinement

  1. Understand the Problem

    • Read the problem statement carefully, noting the constraints.

  2. Visualize the Data Structure

    • Create tables, graphs, or diagrams to understand the distribution and relationships among numbers in the array.

  3. Identify Potential Approaches

    • Consider various algorithms and techniques like cycle detection or hash mapping.

  4. Select the Approach

    • Given the constraints, decide on the best-suited approach. In this case, the Floyd’s cycle-finding algorithm is a strong candidate.

  5. Plan

    • Outline how to implement your chosen approach. Detail the variables, loops, and conditions needed.

  6. Pseudocode

    • Write a rough pseudocode outlining your approach. This helps in visualizing the implementation steps clearly.

  7. Code

    • Translate the pseudocode into actual code. Test small parts as you go along to ensure they work as expected.

  8. Test

    • Run the code on multiple test cases, including edge cases to ensure it meets all constraints and solves the problem.

  9. Debug and Optimize

    • If the solution isn’t working as expected or could be optimized, refine the code.

  10. Review

    • Once the code works as expected, review the entire process to understand what worked and what didn’t.

Granular, Actionable Steps

  1. Initialize slow and fast pointers at the first element.
  2. Loop: Move the slow pointer one step and the fast pointer two steps until they meet.
  3. Reset: Move the slow pointer back to the start.
  4. Find Intersection: Move both pointers one step at a time until they meet again. This is your duplicate number.

Independent Parts

  1. Finding the Meeting Point: The first loop to find the meeting point of slow and fast pointers can be considered an independent part.
  2. Finding the Duplicate: Once the meeting point is found, finding the duplicate is another independent step.

Repeatable Patterns

  1. Pointer Movement: The concept of moving slow and fast pointers is a repeatable pattern often seen in cycle detection problems.
  2. Loop Until Meet: Both in finding the meeting point and the duplicate, the loop-until-meet pattern repeats.

Breaking down the problem in this manner enables effective tackling of each part, and identifying patterns helps in recognizing similar problems in the future.

Solution Approach and Analysis

Step 1: Understand the Constraints

  • The first step is to understand that you can’t modify the array and you have to use constant extra space. This rules out approaches that sort the array or use additional data structures like sets or hash maps.

Step 2: Choose the Right Algorithm

  • Given the constraints, cycle detection algorithms are a good fit. Specifically, Floyd’s cycle-finding algorithm can be applied here. This algorithm is often used to detect cycles in linked lists but can be adapted for this array problem.

Step 3: Initialize Pointers

  • Initialize two pointers, slow and fast, at the first element of the array. Imagine these pointers as two runners on a racetrack, one slow and one fast.

Step 4: Detect Cycle

  • Move slow one step at a time and fast two steps at a time through the array. If there is a cycle (i.e., a duplicate), they will meet at some point.

Step 5: Find the Entry Point of the Cycle

  • Once slow and fast meet, move the slow pointer back to the beginning of the array. Now move both pointers one step at a time. The point where they meet again is the duplicate element.

Step 6: Return the Duplicate

  • The value at the meeting point is the duplicate number in the array.

Effect of Changes in Parameters

  • If the constraint of constant space were lifted, simpler methods like sorting or using hash maps could be employed.
  • If multiple duplicates existed, the cycle detection method would not work.

Example Case:

Input: [1, 3, 4, 2, 2]

  1. Step 3: slow = 1, fast = 1
  2. Step 4:
    • 1st iteration: slow = 3, fast = 4
    • 2nd iteration: slow = 4, fast = 3
    • 3rd iteration: slow = 2, fast = 2 (they meet)
  3. Step 5:
    • Reset slow = 1
    • 1st iteration: slow = 3, fast = 3
    • 2nd iteration: slow = 4, fast = 4
    • 3rd iteration: slow = 2, fast = 2 (they meet again)
  4. Step 6: The duplicate is 2.

By following this approach, you can find the duplicate element effectively within the given constraints.

Identify Invariant

The invariant in this problem is the existence of a single repeated number in the array, within the range [1, n]. This is a condition that holds true throughout the operation of the algorithm, irrespective of the various states that the program might go through. It helps ensure the correctness of the algorithm by providing a constant that remains unchanged during its execution.

This invariant helps guide the design of the algorithm. For example, the Floyd’s cycle-finding algorithm relies on the invariant of a cycle existing within the array to find the duplicate number. Because we know there is only one duplicate, we can be sure that the cycle detection will successfully find it.

Identify Loop Invariant

What is the loop invariant in this problem?

Is invariant and loop invariant the same for this problem?

Identify Recursion Invariant

Is there an invariant during recursion in this problem?

Is invariant and invariant during recursion the same for this problem?

Thought Process

To solve this problem, we can think of it in terms of a cycle detection problem. The constraints suggest that the array’s structure is like a linked list with a loop, since there’s one repeated number. This leads us to Floyd’s cycle-finding algorithm, also known as the “Tortoise and Hare” technique.

  1. Initialization: Start with two pointers, the “tortoise” and the “hare”, both at the first position.

  2. Find the Intersection Point: Move the tortoise one step and the hare two steps until they meet. They are guaranteed to meet because there’s a cycle (duplicate number).

  3. Find the Entry Point of the Cycle (Duplicate Number): Reset one pointer to the first position and keep the other at the meeting point. Move both one step at a time until they meet again. The point where they meet is the entry point of the cycle, which is the duplicate number.

Code

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def findDuplicate(nums):
    # Step 1: Initialization
    tortoise = nums[0]
    hare = nums[0]

    # Step 2: Find Intersection Point
    while True:
        tortoise = nums[tortoise]
        hare = nums[nums[hare]]
        if tortoise == hare:
            break

    # Step 3: Find Entry Point of Cycle
    tortoise = nums[0]
    while tortoise != hare:
        tortoise = nums[tortoise]
        hare = nums[hare]

    return hare

Cues from Problem Statement

  • The constraints specify that there’s exactly one duplicate, implying a cycle. This is a cue to use a cycle detection algorithm.
  • The constraints also mention that the solution should not modify the array and should use only constant space, ruling out sorting or hashing methods.

Key Insights

  • The problem can be visualized as a cycle detection problem in a linked list.
  • Given the constraints, Floyd’s cycle-finding algorithm stands out as the most applicable method for solving it.

This algorithm works in O(n) time complexity and O(1) space complexity, adhering to the problem’s constraints.

Establishing Preconditions and Postconditions

Parameters:

  1. Inputs to the method: The method takes a single input parameter nums.
  2. Types: nums is a list of integers.
  3. What they represent: This list represents the array of integers where each integer is in the range [1, n] inclusive, and there’s exactly one duplicate.

Preconditions:

  1. State of the program: No specific state of the program is necessary before the method is called.
  2. Constraints on input:
    • 1 <= n <= 10^5
    • The length of nums should be n + 1.
    • All integers in nums should appear only once except for precisely one integer, which appears two or more times.
  3. Specific state: None.

Method Functionality:

  1. Expected to do: The method is expected to return the integer that is duplicated in the array.
  2. Interaction with inputs: It reads the nums list and uses it to find the duplicate number.

Postconditions:

  1. State of the program: No change in the state of the program or the values of the parameters.
  2. Return value: The integer that is duplicated in the nums list.
  3. Side effects: None, as the array is not modified and constant space is used.

Error Handling:

  1. Response if preconditions are not met: The method assumes that the preconditions about the input are met. It does not include error handling for constraint violations.
  2. Exception or special value: No specific error handling is present. It is assumed that the input adheres to the problem constraints.

Problem Decomposition

Problem Understanding:

  1. In my own words: You’re given a list of integers that range from 1 to n. The list has a length of n+1, and all numbers in it are unique except for one that is repeated. The goal is to find that repeated number.
  2. Key components and requirements: The input list, the constraints on the list size and elements, and the requirement to find the duplicate element without modifying the array or using more than constant extra space.

Initial Breakdown:

  1. Major parts:
    • Understand the array structure.
    • Identify a method to find the duplicate without modifying the array.
    • Implement the method efficiently.

Subproblem Refinement:

  1. Understand the array structure:
    • Note that it’s 1-based, and elements are within the array’s index range.
  2. Identify a method to find the duplicate:
    • Choose an algorithm like Floyd’s Tortoise and Hare.
  3. Implement the method efficiently:
    • Make sure to only use constant extra space and don’t modify the array.

Task Identification:

  1. Array understanding: This is not a repeated task; it is unique to understanding this problem’s constraints.
  2. Algorithm choice and implementation: These could be considered repeated tasks in problems involving finding duplicates or cycles.

Task Abstraction:

  1. Array understanding: It’s a straightforward task—know what the array elements represent and their constraints.
  2. Algorithm choice and implementation: General enough to apply to other problems but specific to this problem’s constraints.

Method Naming:

  1. Array understanding: understandArrayConstraints
  2. Algorithm choice: chooseAlgorithm
  3. Implementation: findDuplicate

Subproblem Interactions:

  1. Order:
    • First, understand the array structure (understandArrayConstraints).
    • Then, choose an algorithm (chooseAlgorithm).
    • Finally, implement the chosen algorithm to find the duplicate (findDuplicate).
  2. Dependencies:
    • Understanding the array structure is essential before choosing an algorithm.
    • Algorithm choice directly influences implementation.

From Brute Force to Optimal Solution

Brute Force Solution

  1. Method: Nested Loops to Compare All Pairs
  2. Implementation in Python3:
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def findDuplicate(nums):
    for i in range(len(nums)):
        for j in range(i+1, len(nums)):
            if nums[i] == nums[j]:
                return nums[i]

  1. Inefficiencies:

    • Time Complexity: (O(n^2)) due to nested loops.
    • Space Complexity: (O(1)), but the time inefficiency is a major concern here.

Optimization Steps

  1. Idea: Sort the array first, then a simple linear scan can reveal the duplicate.

  2. Code Snippet:

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    def findDuplicate(nums):
    sorted_nums = sorted(nums)
    for i in range(1, len(sorted_nums)):
        if sorted_nums[i] == sorted_nums[i-1]:
            return sorted_nums[i]

  3. Impact:

    • Time Complexity: (O(n \log n)) for sorting.
    • Space Complexity: (O(n)) for sorted array.

Step 2: Utilize the Existing Array Order (Floyd’s Tortoise and Hare)

  1. Idea: The array contains (n + 1) integers where each integer is between 1 and (n), which means it’s like a linked list where nums[i] points to nums[nums[i]]. The duplicate number forms a cycle.

  2. Code Snippet:

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    def findDuplicate(nums):
    # Phase 1: Finding the intersection point
    slow = fast = nums[0]
    while True:
        slow = nums[slow]
        fast = nums[nums[fast]]
        if slow == fast:
            break

    # Phase 2: Finding the entrance to the cycle (duplicate)
    slow = nums[0]
    while slow != fast:
        slow = nums[slow]
        fast = nums[fast]

    return slow

  3. Impact:

    • Time Complexity: (O(n)) for the two while loops.
    • Space Complexity: (O(1)) extra space, no new arrays.

Summary

We started with a (O(n^2)) time complexity solution and optimized it down to (O(n)) time complexity and (O(1)) space complexity using Floyd’s Tortoise and Hare method. This meets the constraints of the problem and is the most efficient approach known for this specific problem setup.

Code Explanation and Design Decisions

Floyd’s Tortoise and Hare Method for Finding Duplicates

1. Initial Parameters

  • slow and fast: Pointers initialized to the first element (nums[0]). They traverse the array at different speeds to identify a cycle, which indicates a duplicate.

2. Primary Loop or Iteration

  • The while True loop (Phase 1) finds an intersection point within the cycle.
  • Each iteration advances the slow pointer one step and the fast pointer two steps.
  • The loop represents the cycle detection process. A cycle indicates a duplicate number.

3. Conditions or Branches within the Loop

  • if slow == fast: Checks for the intersection point in the cycle.
  • The condition signifies that both pointers meet at a common point within a cycle, thereby confirming a duplicate number.

4. Updates or Modifications to Parameters

  • slow = nums[slow] and fast = nums[nums[fast]]: These update the positions of slow and fast.
  • These changes advance the pointers and help to detect a cycle (duplicate number).

5. Invariant

  • The distance between slow and fast is always increasing by 1 at each step until they meet.
  • This invariant helps to ensure that if a cycle exists (i.e., a duplicate number), the pointers will eventually meet.

6. Significance of the Final Output

  • The value where the two pointers meet (slow) is the duplicate number.
  • This satisfies the problem’s requirement of finding the only duplicate in the array.

The method is designed to find the duplicate number using (O(n)) time complexity and (O(1)) space complexity, aligning with the problem’s constraints.

Coding Constructs

Floyd’s Tortoise and Hare Method for Finding Duplicates

1. High-Level Strategies or Techniques

  • Two-Pointer Technique: Uses a slow and a fast pointer to traverse the list.
  • Cycle Detection: Deploys Floyd’s Tortoise and Hare algorithm to find cycles, which in this context means finding a duplicate number.

2. Explaining to a Non-Programmer

This code is like having two people walk through a loop-shaped hiking trail to see if any section of the trail is walked twice (a duplicate). One walks faster than the other. If they meet at the same spot, that means some part of the trail must be a loop (a duplicate section).

3. Logical Elements or Constructs

  • Loop: For continuous traversal.
  • Conditionals: To check for an intersection point.
  • Pointers: To represent positions in the list.

4. Algorithmic Approach in Plain English

Start with two pointers at the beginning of the list. Move one pointer one step at a time, and the other two steps at a time. Keep doing this until they meet. Once they meet, start one pointer back at the beginning and move both one step at a time until they meet again. Where they meet the second time is the duplicate number.

5. Key Steps or Operations on Input Data

  • Initialize slow and fast pointers at the first element.
  • Move slow one step and fast two steps in a loop to find an intersection.
  • Reset slow to the start and move both pointers one step at a time to find the duplicate.

These steps are essential for identifying a cycle (duplicate number) in the array while adhering to the constraints of the problem.

6. Algorithmic Patterns or Strategies

  • Two-Pointer Technique: Simplifies the traversal of the list.
  • Cycle Detection: Specifically, Floyd’s Tortoise and Hare algorithm for finding cycles in a linked list, adapted here for an array.

This algorithmic pattern efficiently solves the problem in (O(n)) time and (O(1)) space.

Language Agnostic Coding Drills

1. Distinct Concepts in the Code

  1. Variable Initialization: Learning to declare and initialize variables.
  2. Array Traversal: Understanding how to move through an array.
  3. Pointers/Index Tracking: Keeping track of positions in an array.
  4. Loops: Using loops to automate repetitive tasks.
  5. Conditional Statements: Using if conditions to control logic.
  6. Function Calls: The ability to call a function and understand its return value.
  7. Two-Pointer Technique: Using more than one pointer to traverse an array.
  8. Cycle Detection: Understanding and identifying cycles in a structure.

2. Concepts by Increasing Difficulty

  1. Variable Initialization: Easy; Basics of declaring variables.
  2. Array Traversal: Easy; Next logical step after variables—iterating over them.
  3. Loops: Medium; Requires understanding of control structures.
  4. Conditional Statements: Medium; Branching logic can be complex but is fundamental.
  5. Pointers/Index Tracking: Medium; Requires a good grasp of arrays and variables.
  6. Function Calls: Medium; Understanding scope and return types.
  7. Two-Pointer Technique: Hard; Requires a strong understanding of arrays, pointers, and loops.
  8. Cycle Detection: Hard; A complex concept requiring an understanding of all the previous points and advanced logic.

3. Problem-Solving Approach

  1. Variable Initialization: Start by understanding the problem statement and initializing variables like slow and fast pointers, often set to the first element of the array.

  2. Array Traversal: The next task is to understand that the array needs to be traversed to find the duplicate number.

  3. Loops: Implement a loop that will allow both slow and fast pointers to move through the array until they meet.

  4. Conditional Statements: Within the loop, conditions will be required to check if the slow and fast pointers have met, which will break the loop.

  5. Pointers/Index Tracking: Master tracking the indices or pointers in an array as you traverse it, updating them based on specific rules (e.g., moving one step or two steps).

  6. Function Calls: If modularizing code, be ready to call functions that execute specific tasks, like resetting the slow pointer to the start of the array.

  7. Two-Pointer Technique: Implement the two-pointer technique, advancing one pointer by one step and the other by two steps, to find the cycle or duplicate number efficiently.

  8. Cycle Detection: The culmination of all steps is to find the duplicate number, which in this problem translates to finding a cycle. Use the pointers to detect this cycle and return the duplicate number.

Each drill contributes to the overall solution by breaking the complex task of finding a duplicate number in a constrained environment into smaller, more manageable tasks. By mastering each drill, you’re better equipped to implement the final, efficient solution.

Targeted Drills in Python

1. Python-Based Coding Drills for General Concepts

Variable Initialization

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# Initializing a variable named 'x' and assigning it a value of 10
x = 10

Array Traversal

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# Traversing through an array and printing each element
arr = [1, 2, 3, 4]
for element in arr:
    print(element)

Loops

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# Using a loop to print numbers from 1 to 5
for i in range(1, 6):
    print(i)

Conditional Statements

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# Using if condition to check if a number is even or odd
x = 4
if x % 2 == 0:
    print("Even")
else:
    print("Odd")

Pointers/Index Tracking

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# Moving through an array using index
arr = [1, 2, 3, 4]
index = 0
while index < len(arr):
    print(arr[index])
    index += 1

Function Calls

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# Defining and calling a function to print "Hello, World!"
def greet():
    print("Hello, World!")

greet()

Two-Pointer Technique

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# Using two pointers to find a pair that sums up to a given number in a sorted array
arr = [1, 2, 3, 5, 6]
target = 8
left = 0
right = len(arr) - 1

while left < right:
    current_sum = arr[left] + arr[right]
    if current_sum == target:
        print(f"Pair found: {arr[left]}, {arr[right]}")
        break
    elif current_sum < target:
        left += 1
    else:
        right -= 1

Cycle Detection

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# Detecting cycle in an array by moving at different speeds
arr = [1, 2, 3, 4, 2]
slow, fast = arr[0], arr[0]

while True:
    slow = arr[slow]
    fast = arr[arr[fast]]
    if slow == fast:
        print("Cycle detected")
        break

2. Problem-Specific Coding Drills

No problem-specific coding drills are necessary in this case, as the general drills cover all the coding concepts needed to solve this type of problem.

3. Integrating the Drills

  1. Start by initializing your variables and pointers. You might need a slow and fast pointer for cycle detection.

  2. Use a loop to traverse the array. Inside the loop, you can use the two-pointer technique and move them at different speeds to find a cycle.

  3. Incorporate conditional statements within the loop to decide when to break out of the loop or to update your pointers.

  4. Use index tracking or pointers to remember positions in the array as you move through it.

  5. If necessary, make function calls to modularize your code and keep it clean.

  6. Ultimately, use cycle detection logic to find the duplicate number, which would involve all the previous steps.

  7. Compile all these steps into a complete function, ensuring that each drill fits appropriately where it’s supposed to in the broader solution. Each drill feeds naturally into the next, finally culminating in a program that solves the problem effectively.

Q&A

Similar Problems

Here are 10 problems that use similar underlying concepts:

  1. Two Sum (Easy)

    • Similarity: Utilizes array traversal and index tracking to find pairs that add up to a target sum, much like the two-pointer technique we discussed.

  2. Contains Duplicate (Easy)

    • Similarity: Requires checking for duplicates in an array, which may employ an index or pointer method similar to our cycle detection example.

  3. Reverse Linked List (Easy)

    • Similarity: Involves traversal and pointer manipulation, similar to the techniques of index tracking or two-pointer methods.

  4. Find the Duplicate Number (Medium)

    • Similarity: Relies on cycle detection for finding the duplicate in an array, just like our cycle detection drill.

  5. Merge Two Sorted Lists (Easy)

    • Similarity: Uses pointers to iterate through two lists, much like the two-pointer technique.

  6. Remove Nth Node From End of List (Medium)

    • Similarity: Employs a two-pointer technique to find and remove a node, a strategy similar to finding a pair that sums up to a target.

  7. Valid Parentheses (Easy)

    • Similarity: Requires traversing through the string and keeping track of the elements using a stack. This is similar to our discussion on index tracking.

  8. Climbing Stairs (Easy)

    • Similarity: Involves breaking down a problem into smaller subproblems, akin to our subproblem refinement steps.

  9. Maximum Subarray (Easy)

    • Similarity: Makes use of variables to keep track of current and maximum sums as you traverse an array, similar to our use of variables and index tracking.

  10. Palindrome Linked List (Medium)

    • Similarity: Utilizes the two-pointer technique to find the middle of the list and then reverses the second half to check for palindromic structure.

Each of these problems involves one or more of the techniques we discussed, whether it’s array traversal, index tracking, or two-pointer techniques.