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Description:

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

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Solution:

class Solution { public: int findKthPositive(vector<int>& arr, int k) { int l = 0; int r = arr.size(); // find the first index l s.t. nMissing(l) = A[l] - l - 1 >= k while (l < r) { const int m = (l + r) / 2; if (arr[m] - m - 1 >= k) r = m; else l = m + 1; } // the k-th missing positive // = A[l - 1] + k - nMissing(l - 1) // = A[l - 1] + k - (A[l - 1] - (l - 1) - 1) // = A[l - 1] + k - (A[l - 1] - l) // = l + k return l + k; } }; 

leetcode

challenge

Here is the link for the problem:
https://leetcode.com/problems/kth-missing-positive-number/