Missing Element in Sorted Array

10 Prerequisite LeetCode Problems

For “1060. Missing Element in Sorted Array”, the following are a good preparation:

  1. “278. First Bad Version” - This problem introduces the concept of binary search which can be useful in the follow-up part of the question that asks for a solution with logarithmic time complexity.

  2. “33. Search in Rotated Sorted Array” - Another problem that employs binary search which is useful in solving this problem in O(log n) time.

  3. “35. Search Insert Position” - The problem can help with handling sorted array and finding the position to insert a new element.

  4. “349. Intersection of Two Arrays” - This problem can help to understand the concepts of handling unique elements in an array which is a feature in the given problem.

  5. “268. Missing Number” - This problem involves finding a missing number in an array. Understanding this problem will be directly beneficial for the given problem.

  6. “448. Find All Numbers Disappeared in an Array” - Understanding this problem will help in dealing with missing elements in an array.

  7. “704. Binary Search” - This problem provides a practice of binary search in a one-dimensional array which can be helpful in finding the missing element.

  8. “153. Find Minimum in Rotated Sorted Array” - The way of dealing with sorted arrays in this problem can provide useful insights.

  9. “162. Find Peak Element” - This problem introduces a way of finding an element in a sorted array which is not in its usual place.

  10. “441. Arranging Coins” - This problem deals with finding a particular count in a sorted structure, which can be beneficial for the concept of finding the kth missing number.

These provide a solid understanding of handling sorted arrays, applying binary search, and dealing with unique elements, which are crucial in solving “1060. Missing Element in Sorted Array”.

Problem Classification

This problem can be classified under the “Arrays”. Specifically, it can be considered a “missing elements in a sorted array” problem.

‘What’ Components:

  1. An integer array ’nums’, sorted in ascending order and containing unique elements.
  2. An integer ‘k’, representing the position of the missing number we are interested in.
  3. The task to find the ‘kth’ missing number in the sequence starting from the leftmost element of the array.

The problem is a Search problem, specifically a Binary Search problem, given the sorted nature of the array and the requirement for a logarithmic time complexity solution.

We are asked to identify a specific element (the kth missing number) from an ordered list of elements (sorted array). This involves a process of elimination which is the basic concept behind search algorithms. Furthermore, given the sorted nature of the array and the requirement to have a solution with logarithmic time complexity, a Binary Search algorithm becomes the most suitable approach to this problem.

Clarification Questions

  1. Can the elements in the ’nums’ array be negative?
  2. What should we return if there are less than ‘k’ missing numbers?
  3. Is the ‘kth’ missing number always larger than the maximum number in the array or can it be in between the elements of the array?
  4. If the ’nums’ array is empty, what should be the output?
  5. If ‘k’ is greater than the total possible number of integers between the minimum and maximum values of ’nums’, how should the function behave?
  6. Can ‘k’ be zero or negative?
  7. What is the expected behavior if the ’nums’ array contains only one element?

Problem Analysis and Key Insights

The key insights from the problem statement are:

  1. The array is sorted in ascending order. This suggests that we might be able to apply binary search or similar techniques to solve the problem in logarithmic time.

  2. All elements in the array are unique, which simplifies the problem because we don’t have to handle duplicates.

  3. We are asked to find the kth missing number. This means we need to keep track of how many numbers are missing as we traverse the array.

  4. The difference between the current index and the number at that index gives us the number of missing elements before the current position.

  5. As the problem also asks for a logarithmic time complexity solution in the follow-up, it hints that a binary search approach might be needed. Instead of a linear scan of the array, we need to figure out how to divide and conquer.

Problem Boundary

The scope of this problem is primarily about understanding and implementing an algorithm to find the kth missing number in a unique and sorted integer array.

The problem requires the following:

  1. Understanding the relationship between the array’s indices, the values at those indices, and the missing numbers.

  2. Implementing a solution that first correctly identifies the missing numbers in the array.

  3. Ensuring that the solution is not just correct, but also efficient. The follow-up question suggests that an optimal solution should have a time complexity of O(log n), hinting at a binary search or similar approach.

  4. Testing the solution against various edge cases to confirm its correctness and efficiency.

  5. The constraints in the problem also indicate that the solution should be able to handle large inputs, as the length of nums can be up to 5 * 10^4 and k can be up to 10^8.

These elements together define the scope of the problem.

Establishing the boundaries of a problem is about understanding the constraints and the limitations that are stated explicitly or implicitly in the problem statement. For this problem, we can establish the boundaries as follows:

  1. Input Limitations:

    • The array, nums, has a length between 1 and 5 * 10^4. It is sorted in ascending order, and all elements are unique.
    • The variable, k, ranges from 1 to 10^8.
    • The elements of nums are positive integers, with values up to 10^7.

  2. Output Expectations:

    • The output is the kth missing number in the array. Given that the array and k can have quite large values, the output could also be a large integer.

  3. Computational Limitations:

    • The problem implies that an efficient solution is expected, suggesting that an optimal solution should have a logarithmic time complexity (O(log n)).

  4. Problem Constraints:

    • The problem doesn’t state anything about what to do if there are no missing numbers or if the number of missing numbers is less than k. This is a boundary that may need clarification.

  5. Functional Boundaries:

    • The problem focuses on finding missing numbers, meaning the function or algorithm doesn’t need to modify the array or perform any other tasks beyond this.

By understanding these boundaries, we can ensure our solution fits within the problem constraints and correctly solves the problem.

PYTHON 3 - Time Limit Exceeded (can someone pls review my code)?

class Solution: def missingElement(self, nums: List[int], k: int) -> int: res= [] for n in range(min(nums), max(nums)+k): if n not in nums: res.append(n) if len(res) == k: return(res[k-1])

The time complexity of your current solution is high due to the use of the “in” operator to check whether an element is in the list “nums”. This operator traverses the list, leading to an O(n) operation. Combined with the for loop, the overall time complexity becomes O(n^2), which could lead to a Time Limit Exceeded error for large inputs.

Here is an optimized approach with a time complexity of O(log(n)), which uses binary search:

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class Solution:
    def missingElement(self, nums: List[int], k: int) -> int:
        # If the kth missing number is larger than the last element of the array
        if k > (nums[-1] - nums[0] + 1) - len(nums):
            return nums[-1] + k - ((nums[-1] - nums[0] + 1) - len(nums))
        
        left, right = 0, len(nums) - 1
        while left < right - 1:
            mid = (left + right) // 2
            # If the number of missing numbers in the left half is >= k
            if (nums[mid] - nums[left] - (mid - left)) >= k:
                right = mid
            else:
                k -= nums[mid] - nums[left] - (mid - left)
                left = mid

        return nums[left] + k

This solution works by first checking if the kth missing number is larger than the last element in the array. If it is, we directly return the result. If not, we perform a binary search to find the kth missing number. We keep updating the left and right pointers and decreasing k accordingly. If the number of missing numbers in the left half is more than or equal to k, we move the right pointer. Otherwise, we move the left pointer and update k. Finally, we return the kth missing number in the found range.

Distilling the Problem to Its Core Elements

  1. Fundamental Concept: This problem is based upon the principle of Binary Search. It’s an optimization problem that requires searching in a sorted array in a way that reduces the number of comparisons made, thus reducing the time complexity to O(log n).

  2. Simplest Description: Imagine you have a list of unique numbers arranged in increasing order, but some numbers are missing. You are tasked with finding the ‘kth’ missing number. For example, in the list [1, 3, 5], the first missing number is 2, the second missing number is 4.

  3. Core Problem: The core problem we are trying to solve is identifying the ‘kth’ missing number in a sorted array.

  4. Key Components: The key components of the problem include:

    • A sorted list of unique numbers
    • The value of ‘k’, indicating the sequence of the missing number we want to find
    • An understanding of the Binary Search algorithm

  5. Minimal Set of Operations:

    • Use Binary Search to find the range in which the ‘kth’ missing number lies.
    • Calculate the number of missing numbers in that range.
    • If the number of missing numbers is less than ‘k’, adjust ‘k’ and the range accordingly.
    • Repeat the steps until the ‘kth’ missing number is found.

Visual Model of the Problem

This problem can be visualized using the concept of an array and index-based manipulation.

Let’s consider an example: nums = [4,7,9,10], k = 3. If we imagine the numbers from 4 to 10 on the number line, it will look something like this:

Number Line: 4 _ 7 _ _ 9 10 Index Line: 1 2 3 4 5 6 7

Here, the “_” represents the missing numbers. As we can see, there are gaps between the numbers in the ’nums’ array. The problem wants us to find the ‘kth’ missing number. In this case, ‘k’ is 3, so we’re looking for the third missing number.

Using the number line, we can visually see that the first missing number is 5 (the second number on the index line), the second missing number is 6 (the third number on the index line), and the third missing number is 8 (the fifth number on the index line).

So, in this example, the output should be 8.

This visualization can help us understand the problem statement and also guide us towards the solution. We need to develop an algorithm that can efficiently perform this calculation even for large arrays and large ‘k’ values. This is where the concept of binary search can come in handy.

Problem Restatement

Absolutely! Here’s my interpretation of the problem:

The problem provides us with a list of unique integers that are sorted in ascending order. However, there are some integers missing from this sequence. Our task is to identify the ‘kth’ missing number in this sequence.

To make it clear, let’s consider an example where we have the list nums = [4,7,9,10] and k = 3. Here, the missing numbers in sequence are [5,6,8], and the third missing number is ‘8’, which would be our output.

The problem further specifies that the length of the list ’nums’ will be at least 1 and at most 50,000. Each number in the list will be a unique integer between 1 and 10^7. The value of ‘k’ will be an integer between 1 and 10^8.

The challenge here is to find a solution that has a logarithmic time complexity, which suggests that we should aim for a solution based on binary search or a similar strategy.

Abstract Representation of the Problem

Here is an abstract formulation of the problem:

We have a monotonically increasing sequence of unique integers with one or more gaps within its range. Our objective is to find the kth missing element from the sequence.

A gap is defined as a missing element that would be present if the sequence was complete with no missing elements. The sequence can start from any integer and is not restricted to starting from 1.

The problem can be formulated as a function f(sequence, k) that returns the kth missing number in the sequence.

This abstract representation retains the key elements of the problem: the increasing sequence of integers, the concept of a gap in the sequence, and the task of finding the kth gap.

It’s worth noting that this problem, in its essence, is about understanding and manipulating ordered sequences.

Terminology

Yes, there are a few specialized terms and technical concepts crucial to understanding this problem:

  1. Monotonically Increasing Sequence: This is a sequence where every element is greater than or equal to the preceding one. In this problem, nums is a monotonically increasing sequence.

  2. Unique integers: These are integers where no two elements are the same. The problem specifies that the elements of the sequence are unique.

  3. kth Missing Number: This refers to the kth number that is not present in the sequence, but would be if the sequence was a perfect arithmetic progression (i.e., had a constant difference between all consecutive elements). Our task in this problem is to identify this number.

  4. Arithmetic Progression: This is a sequence of numbers in which the difference between any two consecutive numbers is constant. The sequence in this problem has gaps, and we need to find the kth number that would make it a perfect arithmetic progression.

  5. Logarithmic Time Complexity: In computer science, logarithmic time complexity (O(log n)) is one of the most efficient algorithms for large-scale problems. It means that as the size of the input increases, the number of steps it takes to complete an operation does not increase linearly but logarithmically. The problem asks for a solution with this time complexity, indicating that the solution needs to be efficient for large inputs.

These terms and concepts are key to formulating an approach to the problem and to understanding and implementing an efficient solution.

Problem Simplification and Explanation

Sure, let’s simplify the problem.

Think of this problem as looking for a specific person in a line-up. The people are standing in increasing order of their heights, and all are unique. You are told the height of the shortest and the tallest person. However, some people in this line-up decided not to stand according to their exact height order. Your task is to find the kth person whose height order is missing from the line-up.

The line-up of people corresponds to the given sorted array nums. Each person’s height corresponds to the number in the array. The kth person whose height is missing corresponds to the kth missing number in the array.

The array nums is sorted in ascending order, and all elements are unique. The kth missing number is not the kth element in the array, but the kth number that would appear if the array was a perfect arithmetic progression, i.e., there was a constant difference between all consecutive elements.

The main challenge here is that we don’t know the exact “step” (difference between consecutive elements) in the perfect arithmetic progression. We need to infer it from the given elements in nums and the value of k.

A key observation is that if there were no missing numbers, the kth number in nums would have a specific value. If there are missing numbers before the kth number, the actual kth number in nums will be larger. The difference tells us how many numbers are missing.

However, if we simply start from the beginning and move towards the kth number, the process can be quite slow for large arrays or large values of k. The problem suggests that we should aim for a logarithmic time complexity solution, which usually hints towards a binary search algorithm.

By carefully choosing the middle point in our binary search, we can check how many numbers are missing before this point and decide whether the kth missing number is before or after it, effectively “halving” the search space with each step.

To summarize, this problem is like looking for a missing person in a line-up where not everyone is standing according to their height order, and we have an efficient strategy to speed up the search.

Constraints

There are a few characteristics in this problem that can be leveraged for an efficient solution:

  1. Sorted array: The input array nums is sorted in ascending order. This property is very important because it allows us to perform binary search, which is a highly efficient algorithm with a logarithmic time complexity (O(log n)).

  2. Unique elements: All the elements in the array are unique. This means that we don’t have to deal with duplicate values, simplifying our search for the missing number.

  3. Arithmetic progression: If there were no missing elements, nums would be an arithmetic progression, meaning the difference between any two consecutive elements would be constant. This allows us to easily calculate what the value of any element should be if there were no missing elements, and compare it to the actual value to determine how many elements are missing.

  4. Numerical ranges: The constraints that 1 <= nums[i] <= 10^7 and 1 <= k <= 10^8 give us some bounds on the possible values. However, they are quite large, so we cannot simply create an array of all possible values and count missing elements, as this would require too much memory. The large ranges of possible values indicate that we need an efficient algorithm that doesn’t rely on storing all possible numbers.

By identifying these characteristics and understanding how they can be exploited, we can develop an efficient approach to solve this problem.

Analyzing the constraints of the problem can provide some key insights that can guide the development of an efficient solution:

  1. Size of the array: The constraint 1 <= nums.length <= 5 * 10^4 indicates that the input array could be quite large. This suggests that we should aim for a solution that avoids iterating over the entire array multiple times. It also suggests that a solution with time complexity greater than O(n) could be too slow.

  2. Value of elements: The constraint 1 <= nums[i] <= 10^7 means that the values in the array could also be quite large, but they’re still within the range that can be handled by standard integer types in most programming languages.

  3. Value of k: The constraint 1 <= k <= 10^8 tells us that the kth missing number could be quite far into the sequence of missing numbers. This suggests that a solution which iteratively finds and counts missing numbers could be too slow, especially if k is near its maximum value.

  4. Array is sorted and unique: These characteristics of the array allow us to use binary search, which is a very efficient algorithm with a time complexity of O(log n).

These insights suggest that an efficient solution to this problem will likely involve using binary search to find the kth missing number in the array, taking advantage of the fact that the array is sorted and contains unique elements.

Case Analysis

Let’s consider additional examples to cover a wider range of scenarios:

  1. Single Missing Number (minimum input size):

    • Input: nums = [1, 3], k = 1
    • Output: 2
    • In this case, the input list has the smallest possible size (2), and there’s only one number missing (2). Since k = 1, we return the first missing number.

  2. No Missing Numbers Before K:

    • Input: nums = [1, 2, 3, 4, 5], k = 1
    • Output: 6
    • Here, there are no missing numbers in the array, so the kth missing number is simply the last number in the array plus k.

  3. Multiple Missing Numbers:

    • Input: nums = [2, 7, 11, 16], k = 3
    • Output: 5
    • There are several numbers missing from the array, and the third missing number is 5.

  4. Large K Value:

    • Input: nums = [1, 2, 4, 5], k = 100
    • Output: 102
    • Even though there are missing numbers in the array, k is large enough that the kth missing number is outside of the array.

  5. Missing Numbers at the Start:

    • Input: nums = [5, 6, 7, 8], k = 3
    • Output: 3
    • The missing numbers are at the start of the array, and the third missing number is 3.

  6. Large Input Size:

    • Input: nums = [i for i in range(1, 50001)], k = 1
    • Output: 50001
    • The array is the maximum size allowed by the constraints, and there are no missing numbers in the array. The kth missing number is simply the last number in the array plus k.

By analyzing these test cases, we can confirm our understanding of the problem, ensure our solution handles all possible scenarios, and gain insights about how to take advantage of the problem’s constraints.

Let’s visualize each of the cases:

  1. Single Missing Number:

    Array: [1, _ , 3]

    Here, the number 2 is missing, and since we are looking for the first missing number (k = 1), the output is 2.

  2. No Missing Numbers Before K:

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

    The missing number lies outside the array as there are no missing numbers within the array. The first missing number (k = 1) is 6.

  3. Multiple Missing Numbers:

    Array: [2, _, _, _, _, 7, _, _, _, 11, _, _, _, _, 16]

    Here, the numbers 3, 4, 5 are missing and since k=3, the output is 5.

  4. Large K Value:

    Array: [1, 2, _, 4, 5, _, _, _, …, _, _, _, _]

    Here, the missing numbers start from 3 but since k = 100, the 100th missing number will be way outside the existing array and equals to the last element in the array plus k, i.e., 102.

  5. Missing Numbers at the Start:

    Array: [_, _, _, _, 5, 6, 7, 8]

    The missing numbers are at the beginning of the array, and since k=3, the third missing number is 3.

  6. Large Input Size:

    Array: [1, 2, 3, …, 50000, _, _, _, …, _, _, _, _]

    In this case, all numbers up to 50000 are included in the array, but we’re looking for the first missing number, which will be the last element of the array plus k, i.e., 50001. This case tests the efficiency of our solution.

Through these visualizations, you can see how the position and amount of missing numbers influence the solution and why it’s essential to consider the value of k and the specific elements in the array.

Analyzing the different cases brings up several key insights:

  1. Position of Missing Numbers: The position of the missing numbers within the array matters. If they are at the beginning or scattered throughout the array, we might need to traverse the array from the start to find the kth missing number.

  2. Number of Missing Elements: The total number of missing elements within the array influences the solution. If there are fewer missing elements than k within the array, the kth missing number is located outside the array.

  3. Value of k: The kth missing number could be inside or outside the array depending on the value of k and the distribution of the missing numbers. If k is greater than the count of missing numbers in the array, the kth missing number will be the last element in the array plus the remaining count.

  4. Efficiency Matters: In cases where the array size is large, an efficient solution is crucial to avoid time limit exceeded errors.

  5. Array is Sorted: The sorted nature of the array is a valuable property. This enables efficient searching strategies such as binary search.

By understanding these insights, we can form a better problem-solving strategy that accounts for these different scenarios.

Identification of Applicable Theoretical Concepts

Indeed, there are some fundamental mathematical and algorithmic concepts that we can leverage to make this problem more manageable:

  1. Arithmetic Progression: The problem is about an arithmetic sequence or progression where the difference between all consecutive terms is constant. This property can be leveraged to calculate the expected element at any index.

  2. Binary Search: Binary search is a powerful algorithm that can find an element in a sorted array in logarithmic time complexity (O(log n)). Given that the input array is sorted, binary search can be utilized to efficiently find the position where the missing element should be.

  3. Cumulative Sum: The cumulative sum or prefix sum is a sequence of partial sums of a given sequence. It can be applied here to find the number of missing numbers before each index.

By applying these concepts, we can devise a more efficient solution that reduces the time complexity from linear to logarithmic. For example, we can use binary search to find the boundary where the kth missing number resides, and then calculate the exact value.

Simple Explanation

Sure, imagine you have a book with numbered pages from 1 to 100. But, some pages are missing. Now, you are given a task to find out the 5th missing page number.

But there is a twist, you don’t have to flip through the pages one by one from start to end (which would take a long time). Instead, you can quickly flip to the middle of the book, then check if the middle page is missing or not. Based on that, you can decide which half of the book you should continue looking in. This way, you can find the missing page number much faster. This is what we are trying to do with the given problem.

In this problem, the pages of the book are the numbers in the given list, and the missing pages are the missing numbers from the list. The 5th missing page in our example corresponds to the ‘kth’ missing number in the problem. The method of quickly flipping to the middle of the book is similar to the “binary search” method we can use to solve this problem efficiently.

Problem Breakdown and Solution Methodology

Sure, let’s break down the approach to this problem:

Step 1: Initial Observations We are given a sorted list of unique numbers. Some numbers are missing, and we need to find the kth missing number. The sorted nature of the array is an important observation that allows us to use a binary search to find the solution efficiently.

Step 2: Binary Search Just like you might use a binary search to look for a word in a dictionary, we can use it here to find the kth missing number. Binary search works by repeatedly dividing the search space in half. If the missing number is in the first half, we ignore the second half and vice versa.

Step 3: Implementing Binary Search We start by initializing two pointers at the beginning and end of the array. We calculate the mid index, then we calculate the number of missing numbers before the mid index. We do this by subtracting the value at the mid index with the first number, and then subtracting the number of elements before the mid index.

If the number of missing numbers before the mid index is greater than or equal to k, we move our right pointer to mid. Otherwise, we move our left pointer to mid + 1.

We repeat this process until our search space is reduced to a single element.

Step 4: Finding the Exact Missing Number Once our search space is reduced to a single element, we have found the block of numbers where our kth missing number is. We can find the exact missing number by adding the remaining difference to the number at the left pointer.

Step 5: Dealing with Changes in Problem’s Parameters The beauty of this binary search approach is that it works efficiently even if the size of the array or the value of k increases. It will still find the kth missing number in logarithmic time, meaning it will take significantly less time compared to other approaches such as scanning through the array.

Now, let’s use an example to illustrate this approach. Suppose we have nums = [4,7,9,10], k = 3.

  1. We start with left = 0, right = 4.
  2. We find mid = 2, and nums[mid] = 9.
  3. We calculate the number of missing numbers before index mid: missing = nums[mid] - nums[left] - mid. In this case, missing = 9 - 4 - 2 = 3.
  4. Since missing >= k, we move our right pointer to mid: right = mid = 2.
  5. Repeat the process, we find mid = 1, and nums[mid] = 7.
  6. Now, missing = nums[mid] - nums[left] - mid = 7 - 4 - 1 = 2. Since missing < k, we move our left pointer to mid + 1: left = mid + 1 = 2.
  7. Our left pointer now points to the last number before our kth missing number. We calculate the remaining difference: diff = k - missing = 3 - 2 = 1. We add this difference to the number at the left pointer to get our kth missing number: 9 + diff = 10.

So, the 3rd missing number in the list [4,7,9,10] is 10.

Inference of Problem-Solving Approach from the Problem Statement

Sure, let’s identify the key terms or concepts that are crucial for solving this problem:

  1. Integer Array: This problem revolves around an array (or list) of unique integers. Knowing that we are dealing with an array informs us that we can use array-based solutions and methods, such as indexing to access individual elements directly.

  2. Sorted in Ascending Order: This information is critical. It informs us that we can take advantage of techniques suitable for sorted data, such as binary search, which can significantly improve the efficiency of our solution.

  3. Missing Numbers: The goal of this problem is to find missing numbers in the sequence. Understanding this guides us to look for a pattern or a way to predict which numbers are missing, rather than checking each possible number to see if it’s in the list.

  4. Kth Missing Number: The fact that we’re looking for the kth missing number (not just any missing number) implies that the missing numbers have an order and that we should be tracking the number of missing numbers we have found so far.

  5. Logarithmic Time Complexity: The follow-up question asks for an O(log(n)) solution. This means that the solution needs to be more efficient than just linearly checking each number in the array. This guides us towards binary search, which is a common logarithmic time complexity algorithm.

These key terms and concepts inform our approach by guiding us towards using a binary search algorithm to find the kth missing number efficiently in the sorted integer array.

You can create a table to help understand the properties of this problem. Let’s illustrate this using the first example from the problem statement: nums = [4,7,9,10], k = 1.

Create a table with three columns. The first column represents the index, the second column the number at that index in the array, and the third column the missing numbers from the previous number to the current number.

IndexNumberMissing Numbers
04
175, 6
298
310

From this table, you can clearly see the missing numbers between each number in the array, and you can count them to find the kth missing number.

For binary search, visualization is usually done by drawing the array as a line of numbers and showing how the search narrows down the search space by half at each step.

Let’s visualize the binary search process on the above example. For simplicity, let’s assume k=3 (so the target missing number is 8).

4   7   9   10
|   |   |   |

Initially, the whole array is the search space. The middle element is 7. The missing numbers before 7 are 5,6. So there are 2 missing numbers which is less than k (3). So the target missing number is not in the first half, we can remove it and only focus on the second half.

9   10
|   |

Now, we continue our search in the reduced space. The middle element is 9 and the missing numbers before 9 are 5,6,8. So there are 3 missing numbers which is equal to k, so the target missing number is 8.

This way, by repeatedly dividing the array in half and deciding which half to continue searching, we are performing a binary search. It’s a good way to visualize the binary search process for this problem.

The hint that the problem can be solved using binary search comes from the nature of the input and the constraints. The input is a sorted list of unique integers, and you’re asked to find a missing element in a specific position, which matches the typical use case for binary search.

Binary search is an efficient algorithm used to find an element in a sorted list. It works by repeatedly dividing the search space in half. If the target value is greater than the middle element, it searches the right half. If the target value is less than the middle element, it searches the left half.

In this problem, we can use a slightly modified version of binary search to find the kth missing number. Instead of directly searching for a value, we’ll be using binary search to find the range where the kth missing number lies.

Also, the problem statement specifically asks for a logarithmic time complexity solution in the follow-up question. Binary search is one of the most common algorithms with logarithmic time complexity, O(log n), suggesting it’s likely to be part of an efficient solution to this problem.

That’s how we can infer from the problem statement that binary search can be used to solve this problem.

Simple Explanation of the Proof

Sure, let’s break down the reasoning behind why a binary search approach would work for this problem.

The problem involves finding the kth missing number in a sorted array. The array is sorted in ascending order and all its elements are unique.

The binary search algorithm is used to find a specific value in a sorted array in logarithmic time. It does this by repeatedly halving the search interval. If the target value is greater than the middle element, the left half of the interval is discarded and the search continues on the right half. Conversely, if the target value is less than the middle element, the search continues on the left half.

In this problem, we’re not searching for a value that’s already in the array, but a missing value. This means we need to modify our binary search approach slightly. Instead of searching for a specific value, we’re trying to find the segment of the array where the kth missing value would be located if all the missing values were inserted back into the array.

Here’s how we do it:

  1. We start by calculating the number of missing numbers before the middle element of the array. We do this by subtracting the value of the first element and the index of the middle element from the value of the middle element.

  2. If the number of missing numbers before the middle element is less than k, it means that the kth missing number must be in the second half of the array, so we discard the first half and continue our search on the second half.

  3. If the number of missing numbers before the middle element is greater than or equal to k, it means that the kth missing number is in the first half of the array, so we discard the second half and continue our search on the first half.

  4. We repeat these steps until we’ve narrowed down our search to a single element. This element and the one after it are the bounds within which the kth missing number must lie. The kth missing number is then simply the first element plus k minus the number of missing numbers before the first element.

This binary search approach works because the array is sorted and all elements are unique, allowing us to accurately calculate the number of missing numbers before any given element. The binary search enables us to locate the kth missing number in logarithmic time, which is more efficient than a linear search through all the elements of the array.

Stepwise Refinement

Sure, let’s refine the approach into more granular steps. Our overall plan is to use binary search to find the kth missing number. Here’s a step-by-step guide:

  1. Define the search space: Our search space is the array itself, we’re trying to find a missing number so the array elements act as bounds of the potential values the missing number can have.

  2. Start Binary Search: The binary search algorithm involves finding the middle element of our search space and deciding which half to continue searching in. Set the ’low’ pointer to the first index of the array and ‘high’ to the last.

  3. Calculate missing numbers: Calculate the number of missing numbers before the middle of the array. To do this, subtract the first value of the array from the middle value, then subtract the index of the middle value. This will give us the number of integers that are missing before the middle element in the array.

    missingBeforeMid = nums[mid] - nums[low] - (mid - low)

  4. Decide which half to search: If missingBeforeMid < k, it means the kth missing number is not in the first half, so we move our ’low’ pointer to mid + 1. However, we need to subtract the number of missing numbers before mid from k, because we are skipping those missing numbers.

    If missingBeforeMid >= k, the kth missing number is in the first half of the array. So, move the ‘high’ pointer to mid.

  5. Repeat Steps 3-4: We need to repeat this process until our search space is narrowed down to a single element. That is, until ’low’ is equal to ‘high’.

  6. Find the missing number: At this point, we have found two numbers in our array such that the missing number is between them. We can find the missing number by adding k to the lower bound.

The steps from 3 to 6 represent a repeating pattern in our solution as they are executed repeatedly in each iteration of the binary search algorithm.

This problem is a monolithic one, which means that the steps are dependent on each other and can’t be solved independently. The calculation of the missing number in step 6 is dependent on the binary search in steps 3-5, which is in turn dependent on defining the search space in step 1.

This refined plan takes our high-level binary search approach and breaks it down into granular, actionable steps that we can implement in code.

Solution Approach and Analysis

Of course, let’s break down the process:

  1. Understanding the problem: This problem is essentially about finding the kth missing number in an array. It’s as if we have a complete sorted set of unique numbers, and someone took out some numbers randomly. We are tasked to find out what would be the kth number that is missing from this set.

    • Analogies: You can think of this as a complete book series where each volume is a unique number. Now, some of these volumes are missing, and you are asked to find out which is the kth volume that is missing.

  2. Binary search: Since the array is sorted, we can make use of binary search to solve the problem. The idea here is to reduce our search space by half with each step. We take the mid-element and calculate the number of missing elements before it. If the number of missing elements before mid is less than k, it means the kth missing number lies in the second half. If not, it’s in the first half.

  3. Update and repeat: Based on the above comparison, we either update our low pointer to mid + 1 or the high pointer to mid, effectively halving our search space. We repeat this process until our search space is narrowed down to a single element.

  4. Find the missing number: At this point, we have found two numbers in our array such that the missing number is between them. We can find the missing number by adding k to the lower bound.

Let’s consider an example to illustrate this:

Input: nums = [4,7,9,10], k = 3
Output: 8

The array here should ideally be [4,5,6,7,8,9,10], but it’s missing 5, 6, and 8. We are to find the 3rd missing number.

We start with low=0 and high=3 (the indices). The mid index is (0+3)//2=1. The mid number nums[mid] = 7. The count of missing numbers before mid = nums[mid] - nums[low] - (mid - low) = 7 - 4 - 1 = 2.

Since the count of missing numbers before mid (2) is less than k (3), we know the 3rd missing number must be in the second half of the array. So, we update low to mid+1, and k to k - missingBeforeMid (k = k - 2 = 1).

Now, low=2, high=3. The mid index is (2+3)//2=2. The mid number nums[mid] = 9. The count of missing numbers before mid = nums[mid] - nums[low] - (mid - low) = 9 - 9 - 0 = 0.

Since the count of missing numbers before mid (0) is less than k (1), we again update low to mid+1, and k remains the same as there were no missing numbers before mid.

Finally, low=3, high=3. The count of missing numbers is 0, and k=1. Our 1st missing number will be nums[low] + k = 10 + 1 = 11, but since 11 exceeds our array, we should return nums[low - 1] + k = 9 + 1 = 10. This is incorrect, which suggests our mid calculation might need a small adjustment.

The issue here is with how we calculate mid, it should be (low + high + 1) // 2, to lean towards the right. This will give us the correct result of 8 for this example.

Changes in the problem’s parameters, like the value of k and the array length, will affect the runtime of our solution as binary search is dependent on the size of the search space. However, it wouldn’t affect the correctness of the solution as it’s designed to handle these variations.

The binary search approach provides an efficient way to handle large input arrays and large values of k due to its logarithmic time complexity. It exploits the fact that the array is sorted and uniquely valued, and cleverly reduces the search space by half with each step.

Identify Invariant

An invariant in the context of an algorithm is a condition that remains true throughout its execution.

In this problem, the invariant we maintain is that the kth missing number always exists in the range specified by the two pointers low and high that we use in our binary search approach.

This holds at the start since low points to the first element and high points to the last element, so the entire array is the range and the kth missing number must be in this range.

During each step, we calculate the number of missing elements before the mid index. If this count is less than k, we know the kth missing number must be in the second half, so we update low to mid + 1. Otherwise, it’s in the first half, so we update high to mid. In both cases, we’ve effectively halved our search space while ensuring the kth missing number is still in the range between low and high.

This invariant condition holds true for every iteration and helps guide our binary search approach to find the kth missing number in the array.

Identify Loop Invariant

The loop invariant here would be the same as the general invariant I just mentioned. In the context of the binary search loop:

  1. The kth missing number always exists within the range [low, high] in the sorted array.

At the beginning of each iteration, we calculate the missing numbers before the mid-index. If the count of missing numbers is less than k, we update the low pointer to mid + 1. If it’s more or equal, we update the high pointer to mid. By maintaining this loop invariant, we ensure that the kth missing number always falls within the bounds of low and high.

At the end of each iteration, the range [low, high] is halved while still containing the kth missing number. This loop invariant guides our binary search to efficiently find the kth missing number in the sorted array.

Yes, in this problem, the invariant and the loop invariant are the same. The invariant is a condition that is initially true and remains true after every iteration of a loop. The term “loop invariant” is used specifically to refer to invariants related to loops.

In the context of this problem, our loop invariant is: “The kth missing number always exists within the range [low, high] in the sorted array”. This condition is initially true before the loop starts, and it remains true after each iteration of the binary search loop.

Hence, the invariant and the loop invariant refer to the same condition in this case.

Thought Process

The basic thought process for solving this problem would be as follows:

  1. Understand the problem: We are given a sorted integer array where each element is unique. We are asked to find the kth missing number in this array. The key here is to note that the array is sorted, which opens up the possibility of binary search as an efficient way to solve this problem.

  2. Observe patterns: If there was no missing number, the nth number in this sequence should be nums[0] + n - 1 (considering 0-based index). The difference between the expected number and the actual number at index n will give us the count of missing numbers before index n.

  3. Formulate an approach: Our aim is to find the smallest n such that the count of missing numbers before n is at least k. Once we find such an n, we can easily determine the kth missing number.

  4. Binary search: We can use binary search to find this n. If the count of missing numbers at mid of the array is less than k, it means our n is in the second half of the array. Otherwise, it is in the first half.

  5. Calculate the kth missing number: Once we find the n, the n-1th index will have less than k missing numbers, so the kth missing number will be after nums[n-1]. We calculate it as nums[n-1] + k - missing_count(nums[n-1]).

Here is a Python code implementing the above approach:

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from typing import List

def missingElement(nums: List[int], k: int) -> int:
    def missing_count(idx):
        # Calculate number of missing elements up to idx
        return nums[idx] - nums[0] - idx

    n = len(nums)
    if k > missing_count(n - 1):  # If kth missing number is beyond the array
        return nums[-1] + k - missing_count(n - 1)

    # Binary search to find index where kth missing number lies
    left, right = 0, n - 1
    while left < right:
        mid = left + (right - left) // 2
        if missing_count(mid) < k:
            left = mid + 1
        else:
            right = mid

    # Calculate the kth missing number
    return nums[left - 1] + k - missing_count(left - 1)

This approach exploits the sorted nature of the input array and performs in O(log(n)) time complexity, where n is the length of the input array. It utilizes binary search to effectively narrow down the range in which the kth missing number lies.

Establishing Preconditions and Postconditions

  1. Parameters:

    • The method has two inputs: nums and k.
    • nums is a list of integers and k is an integer.
    • nums represents a sorted and unique list of integers, and k represents the index of the missing number we are interested in.

  2. Preconditions:

    • Before this method is called, it’s not necessary that the program must be in any specific state as this function does not rely on any global or external state.
    • Constraints on the input parameters are:
      • nums is an array of integers, 1 <= nums.length <= 5 * 10^4, 1 <= nums[i] <= 10^7, nums is sorted in ascending order, and all the elements are unique.
      • k is an integer, 1 <= k <= 10^8.

  3. Method Functionality:

    • This method is expected to find the kth missing number in the sorted array nums.
    • It operates solely based on the inputs provided and does not interact with any other state of the program.

  4. Postconditions:

    • After the method has been called, it doesn’t modify the input parameters or any global state of the program. It simply returns a result and leaves everything else unchanged.
    • The return value represents the kth missing number in the array.
    • The method does not have any side effects.

  5. Error Handling:

    • If the preconditions are not met, the Python interpreter will raise exceptions like TypeError if the inputs are not of expected types, or ValueError or IndexError during the computation inside the function.
    • The function itself does not have any specific error handling code. It’s the caller’s responsibility to handle any exceptions and ensure that the inputs meet the precondition.

Problem Decomposition

  1. Problem Understanding:

    • The problem is about finding the kth missing number in a sorted array. The array is unique, meaning there are no repeating numbers. The key requirement is to find the missing number that would have been at the kth position if the array was a complete series.

  2. Initial Breakdown:

    • The problem can be broken down into two broad stages:
      • Determine the gaps between elements in the array.
      • Find the kth missing number.

  3. Subproblem Refinement:

    • Determining the gaps between elements can be further broken down into comparing each element with its adjacent one.
    • Finding the kth missing number can be refined into iterating through the calculated gaps and reducing k until it reaches 0.

  4. Task Identification:

    • The tasks of comparing elements and reducing k are repeatedly done. These could be generalized into tasks of getting the difference between elements and iterating to find the missing number.

  5. Task Abstraction:

    • The task of getting the difference between elements is abstract enough as it only needs the current and next element.
    • The task of finding the missing number needs the gaps array and the value of k. It’s abstract as it operates on any array and k value.

  6. Method Naming:

    • The task of getting the difference between elements could be named calculate_gaps.
    • The task of finding the missing number could be named find_missing.

  7. Subproblem Interactions:

    • The tasks interact with each other in a sequential way. First, we calculate the gaps, then we use this gaps array to find the missing number.
    • They need to be performed in this order because the second task depends on the output of the first one. There is a clear dependency between these tasks.

From Brute Force to Optimal Solution

The given problem can be solved with a brute force approach, but it’s not efficient. Let’s start with the brute force solution and then move towards optimizing it.

Brute Force Solution

A naive, brute force approach is to iterate through the range of min(nums) to max(nums) + k, checking for each number if it is in the list. If it’s not, we append it to a results list. Once the length of the results list equals k, we return the k-1th element. Here is the code for that:

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class Solution:
    def missingElement(self, nums: List[int], k: int) -> int:
        res= []
        for n in range(min(nums), max(nums)+k):
            if n not in nums:
                res.append(n)
            if len(res) == k:
                return(res[k-1])

The inefficiency in this brute force approach is due to the fact that we are iterating over each number in the range and checking if it exists in the list. This checking operation if n not in nums: takes O(n) time for each number in the range, leading to a time complexity of O(n^2). This is highly inefficient when dealing with large inputs.

Optimizing the Solution

An optimized approach would be to take advantage of the fact that the list is sorted. We can simply iterate over the list, and for each pair of adjacent elements, we calculate the number of missing numbers between them (gap). If gap is less than or equal to k, we subtract gap from k. If gap is greater than k, it means our kth missing number lies between these two numbers. Hence, we return the number at the current index plus k. If we’ve gone through the list and haven’t found the missing number, it must be greater than the maximum number in the list. In this case, we return the max number plus the remaining k.

Here is the code for the optimized solution:

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class Solution:
    def missingElement(self, nums: List[int], k: int) -> int:
        for i in range(len(nums) - 1):
            gap = nums[i+1] - nums[i] - 1
            if gap >= k:
                return nums[i] + k
            else:
                k -= gap
        return nums[-1] + k

This optimized solution has a time complexity of O(n) and a space complexity of O(1), making it far more efficient than the brute force approach. This is a huge improvement especially when the size of the input list is large. The primary reasoning behind this optimization is the understanding that we can leverage the sorted nature of the list to more quickly locate the kth missing number.

Code Explanation and Design Decisions

  1. Initial Parameters: The initial parameters in this problem are nums and k. The nums parameter is a sorted list of distinct numbers, representing a range of integers with some numbers missing. The k parameter represents the ordinal number of the missing element we’re looking for. For instance, if k = 2, we’re interested in finding the second missing number in the list.

  2. Primary Loop: The main loop in this problem iterates over the nums list. Each iteration in the loop represents an inspection of a pair of adjacent numbers in the list. The purpose of each iteration is to find the kth missing number between the pair of numbers.

  3. Conditions: There’s a single if condition within the loop that checks if the gap between two adjacent numbers is greater than or equal to k. This is based on the understanding that if the gap (number of missing numbers) between two adjacent numbers is less than k, the kth missing number isn’t in that range.

  4. Updates: Inside the loop, we update the value of k by subtracting the gap from k. This is because we have “accounted for” the gap number of missing numbers, so we reduce our k by the gap to find the remaining missing numbers. If the gap is more than k, we’ve found our kth missing number and we return it.

  5. Invariant: The invariant in this problem is the value of k. At the start of each iteration, k represents the number of missing numbers we’re still looking for. During each iteration, k is decremented by the number of missing numbers found. This invariant helps us track our progress towards finding the kth missing number.

  6. Output: The final output of the code is the kth missing number. This output satisfies the requirement of the problem by identifying the kth missing number in the sorted list. If kth missing number is not in the list, then the number returned would be the maximum number in the list plus the remaining k, which also fulfills the requirements of the problem statement.

Coding Constructs

  1. High-Level Strategies: This code is using the strategy of iteration and condition checking to solve the problem. Specifically, it’s using the characteristics of sorted lists and mathematical operations to find the kth missing number.

  2. Explanation for Non-Programmer: Imagine you are given a sorted list of numbers where some numbers may be missing, and you’re tasked with finding the kth missing number. This code is essentially doing that by checking the difference between adjacent numbers. It’s like checking the difference between each pair of adjacent stepping stones to find the kth missing stone.

  3. Logical Constructs: The logical constructs used in this code are loops (to iterate through the list of numbers), conditionals (to check if the kth missing number is within the current pair of numbers), and arithmetic operations (to update the count of missing numbers and calculate the kth missing number when found).

  4. Algorithmic Approach in Plain English: The code starts by looking at each pair of numbers in the list. It checks if the difference between these numbers (minus one, since the numbers are distinct and sorted) is less than k. If it is, the code moves on to the next pair, subtracting this difference from k. If the difference is greater than or equal to k, it means the kth missing number is within this pair. The code then calculates and returns this missing number. If no missing number is found within the list, the code returns a number outside the list that would be the kth missing number.

  5. Key Steps/Operations: The key steps include iterating through the list, checking the difference between adjacent numbers, updating the count of missing numbers (k), and calculating the kth missing number when found.

  6. Algorithmic Patterns/Strategies: This code employs a common pattern of iteration over a list along with conditional checks. It utilizes properties of sorted lists (specifically, the fact that the difference between adjacent numbers tells us how many numbers are missing between them) and uses basic arithmetic operations for calculation and updates. This can be classified as a linear search strategy.

Language Agnostic Coding Drills

  1. Concept Identification:

Here are the distinct coding concepts contained in the problem’s solution:

  • Variable Initialization: Setting up variables to hold intermediate results and control the program flow.

  • Iteration: Going through each element in a data structure, like a list, typically using a for loop.

  • Conditional Checking: Implementing if-else statements to perform different operations based on certain conditions.

  • Arithmetic Operations: Performing basic mathematical operations like subtraction and addition.

  1. Difficulty Classification:

    • Variable Initialization (Easy): This is one of the basic concepts in any programming language and is typically learned early in the programming journey. It involves declaring variables and assigning initial values to them.

    • Arithmetic Operations (Easy): These operations are straightforward as they directly relate to basic mathematical operations learned in school.

    • Iteration (Intermediate): This concept is more complex than the above two as it requires understanding the flow of the program and how it changes with each iteration. A common challenge is deciding when to start and stop the iteration.

    • Conditional Checking (Intermediate): Conditionals require understanding logical statements and making the program perform different actions based on different conditions. This can be tricky as it requires careful thinking about the conditions and what the code should do in each case.

  2. Problem-Solving Approach:

    • Understanding the Problem Statement: The first step is always to understand what is being asked. In this case, it’s to find the kth missing number in a sorted list. Understanding the constraints, such as the list being sorted and distinct, is also crucial.

    • Identifying the Coding Concepts Needed: Based on the problem statement, we identify that we will need iteration (to go through the list), conditional checking (to find the missing number), and arithmetic operations (to calculate the missing number).

    • Initializing Variables: We start by setting up our variables. In this case, we need a variable to keep track of the kth missing number.

    • Setting Up the Iteration: Next, we set up a loop to go through each pair of numbers in the list.

    • Implementing Conditional Checks: Within the loop, we implement conditional checks to find out if the kth missing number is within the current pair of numbers.

    • Updating Variables: If the kth missing number is not within the current pair, we update the k value by subtracting the number of missing numbers in the current pair. If it is within the pair, we calculate the missing number and update our variable.

    • Returning the Result: Finally, we return the calculated kth missing number.

By mastering each of these individual coding drills, one can easily assemble them into a final solution for this problem.

Targeted Drills in Python

Alright, let’s break this down into simpler Python coding drills. The identified concepts were:

  1. Variable Initialization
  2. Arithmetic Operations
  3. Iteration
  4. Conditional Checking

1. Variable Initialization Drill:

Here’s a simple Python exercise that shows how to initialize a variable.

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# Initialize a variable called 'num'
num = 10

# Print the value of 'num'
print(num)

2. Arithmetic Operations Drill:

This Python exercise demonstrates how to perform basic arithmetic operations.

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# Initialize two variables
a = 5
b = 2

# Perform and print their sum, difference, product, and division
print(a + b)  # Sum
print(a - b)  # Difference
print(a * b)  # Product
print(a / b)  # Division

3. Iteration Drill:

This Python exercise shows how to use a for loop to iterate over a list of numbers.

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# Initialize a list of numbers
numbers = [1, 2, 3, 4, 5]

# Iterate over the list and print each number
for num in numbers:
    print(num)

4. Conditional Checking Drill:

This Python exercise demonstrates how to use conditional statements.

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# Initialize a variable
num = 10

# Check if the number is greater than, equal to, or less than 5
if num > 5:
    print("Number is greater than 5")
elif num == 5:
    print("Number is equal to 5")
else:
    print("Number is less than 5")

Problem-Specific Concept: Difference between numbers

This problem-specific concept is needed to calculate the number of missing numbers between each pair of numbers in the list. Here’s a simple Python drill for this:

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# Initialize two numbers
num1 = 5
num2 = 10

# Calculate and print the difference between the numbers
print(num2 - num1)

Integration of Drills:

Now that we’ve covered all the basic and problem-specific drills, let’s see how they would fit together to solve the original problem:

  1. Variable Initialization: We start by initializing our variables: the input list, the kth number, and a variable to keep track of the kth missing number.

  2. Iteration: We set up a loop to iterate over the input list.

  3. Arithmetic Operations and Conditional Checking: Inside the loop, we use our problem-specific concept drill to calculate the number of missing numbers between each pair. Then we use conditional checks to see if the kth missing number is within this range. If it is, we calculate the kth missing number using arithmetic operations. If it isn’t, we update k by subtracting the number of missing numbers in the current pair.

By breaking down the problem into these drills, we can gradually build up the understanding and solution for the problem in a more manageable way.

Q&A

Similar Problems

Here are 10 problems from LeetCode that require similar problem-solving strategies or use similar underlying concepts as the problem we’ve just solved:

  1. Two Sum (LeetCode 1): This problem involves iterating over an array and checking certain conditions, similar to our original problem. It involves iterating over the list and then looking for a pair that sums up to a particular value.

  2. Find First and Last Position of Element in Sorted Array (LeetCode 34): In this problem, you’re asked to find the starting and ending position of a given target value in a sorted array, which is similar to finding a particular missing number in our original problem.

  3. Search in Rotated Sorted Array (LeetCode 33): This problem requires similar strategies of searching in an array with certain constraints, much like our problem where we need to find the kth missing integer.

  4. Longest Substring Without Repeating Characters (LeetCode 3): This problem is similar in the sense that you’re iterating over a list (in this case, a string) and checking conditions to update a certain variable (the longest substring).

  5. Container With Most Water (LeetCode 11): Here, you’re iterating over an array and keeping track of the maximum area, which is somewhat similar to tracking the kth missing number in our original problem.

  6. Jump Game (LeetCode 55): This problem requires you to check if you can reach the last index from the first, given that the maximum jump length at each position is determined by the number at that position. It’s similar to our problem because you’re checking if a certain condition can be met as you iterate over the array.

  7. Rotate Array (LeetCode 189): The need to iterate over an array and perform certain operations on the elements is a shared strategy with our original problem.

  8. Majority Element (LeetCode 169): In this problem, you’re asked to find the majority element in an array, which requires iterating over the array and checking conditions to update a certain variable, much like our original problem.

  9. Product of Array Except Self (LeetCode 238): This problem is about iterating over an array and updating a resultant array based on certain conditions, similar to our original problem.

  10. Find the Duplicate Number (LeetCode 287): This problem involves finding a duplicate number in an array, where you have to iterate over the array and use certain conditions to find the duplicate number.

While each of these problems has unique aspects, they all involve similar strategies to our original problem: iterating over an array, updating variables based on certain conditions, and/or finding specific elements in the array.