Introduction

In this blog post, we will explore how to solve the LeetCode problem "1539" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given an array arr of positive integers sorted in a strictly increasing order, and an integer k. Return the kth positive integer that is missing from this array. Example 1: Input: arr = [2,3,4,7,11], k = 5 Output: 9 Explanation: The missing positive integers are [1,5,6,8,9,10,12,13,...]. The 5th missing positive integer is 9. Example 2: Input: arr = [1,2,3,4], k = 2 Output: 6 Explanation: The missing positive integers are [5,6,7,...]. The 2nd missing positive integer is 6. Constraints: 1 <= arr.length <= 1000 1 <= arr[i] <= 1000 1 <= k <= 1000 arr[i] < arr[j] for 1 <= i < j <= arr.length Follow up: Could you solve this problem in less than O(n) complexity?

Explanation

Here's the breakdown of the solution:

• Binary Search: The core idea is to use binary search to efficiently find the position in the sorted array where the kth missing positive integer would be located.
• Count Missing Numbers: For any index mid in the array, we can determine how many positive integers are missing before arr[mid] by calculating arr[mid] - (mid + 1).

• Adjust Search Range: Based on whether the number of missing integers before arr[mid] is less than or greater than k, we adjust the search range (left/right pointers) accordingly.

• Runtime Complexity: O(log n), where n is the length of the input array.

• Storage Complexity: O(1)

Code

    def findKthPositive(arr, k):
    """
    Finds the kth positive integer that is missing from a sorted array.

    Args:
    arr (list): A list of positive integers sorted in strictly increasing order.
    k (int): The index of the missing positive integer to find.

    Returns:
    int: The kth missing positive integer.
    """
    left, right = 0, len(arr) - 1

    while left <= right:
        mid = (left + right) // 2
        missing = arr[mid] - (mid + 1)

        if missing < k:
            left = mid + 1
        else:
            right = mid - 1

    return left + k