# Kth Missing Positive Number Problem

## Description

LeetCode Problem 1539. Given an array arr of positive integers sorted in a strictly increasing order, and an integer k.

Find the kth positive integer that is missing from this array.

Example 1:

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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:

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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.

## Solution

We can scan through the array. During the scanning, we use a variable *i* to keep track of the index of array element. We use a variable *j* to keep track of how many positive numbers are missing. When *j* reaches *k*, we can stop scanning. We also use a variable *num* to enumerate all positive numbers, starting from 1 to the largest number in the array.

During the scanning, if *num* is in the array, which means *num = arr[i]*, we increase *i* to point to the next element in the array. If *num* is not in the array, we increase *j* which means one more positive number is missing in the array. We stop the search when *j* reaches *k*.

The time complexity is O(n), and the space complexity is O(1).

## Sample C++ Code

This is a C++ implementation of the above approach.

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#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int findKthPositive(vector<int>& arr, int k) {
int i = 0, j = 1, num = 0;
while (j <= k) {
num += 1;
if ((i < arr.size()) && (num == arr[i])) {
i ++;
} else {
j ++;
}
}
return num;
}
int main() {
vector<int> arr = {2, 3, 4, 7, 11};
int k = 5;
cout << findKthPositive(arr, k) << endl;
}