# 1539. Kth Missing Positive Number

## 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

Could you solve this problem in less than O(n) complexity?

## Solutions

Binary search.

• class Solution {
public int findKthPositive(int[] arr, int k) {
if (arr[0] > k) {
return k;
}
int left = 0, right = arr.length;
while (left < right) {
int mid = (left + right) >> 1;
if (arr[mid] - mid - 1 >= k) {
right = mid;
} else {
left = mid + 1;
}
}
return arr[left - 1] + k - (arr[left - 1] - (left - 1) - 1);
}
}

• class Solution {
public:
int findKthPositive(vector<int>& arr, int k) {
if (arr[0] > k) return k;
int left = 0, right = arr.size();
while (left < right) {
int mid = (left + right) >> 1;
if (arr[mid] - mid - 1 >= k)
right = mid;
else
left = mid + 1;
}
return arr[left - 1] + k - (arr[left - 1] - (left - 1) - 1);
}
};

• class Solution:
def findKthPositive(self, arr: List[int], k: int) -> int:
if arr[0] > k:
return k
left, right = 0, len(arr)
while left < right:
mid = (left + right) >> 1
if arr[mid] - mid - 1 >= k:
right = mid
else:
left = mid + 1
return arr[left - 1] + k - (arr[left - 1] - (left - 1) - 1)


• func findKthPositive(arr []int, k int) int {
if arr[0] > k {
return k
}
left, right := 0, len(arr)
for left < right {
mid := (left + right) >> 1
if arr[mid]-mid-1 >= k {
right = mid
} else {
left = mid + 1
}
}
return arr[left-1] + k - (arr[left-1] - (left - 1) - 1)
}