# Missing Element in Sorted Array in C++

Suppose we have a sorted array A of unique numbers, we have to find the K-th missing number starting from the leftmost number of the array. So if the array is like [4,7,9,10], and k = 1, then the element will be 5.

To solve this, we will follow these steps −

• n := size of the array, set low := 0 and high := n – 1

• if nums[n - 1] – nums + 1 – n < k, then

• return nums[n - 1] + (k – (nums[n - 1] – nums + 1 – n))

• while low < high – 1

• mid := low + (high - low)/2

• present := mid – low + 1

• absent := nums[mid] – nums[low] + 1 – present

• if absent >= k then high := mid, otherwise decrease k by absent, low := mid

• return nums[low] + k

Let us see the following implementation to get better understanding −

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int missingElement(vector<int>& nums, int k) {
int n = nums.size();
int low = 0;
int high = n - 1;
if(nums[n - 1] - nums + 1 - n < k) return nums[n - 1] + (k
- ( nums[n - 1] - nums + 1 - n)) ;
while(low < high - 1){
int mid = low + (high - low) / 2;
int present = mid - low + 1;
int absent = nums[mid] - nums[low] + 1 - present;
if(absent >= k){
high = mid;
}else{
k -= absent;
low = mid;
}
}
return nums[low] + k;
}
};
main(){
vector<int> v = {4,7,9,10};
Solution ob;
cout << (ob.missingElement(v, 1));
}

### Input

[4,7,9,10]
1

## Output

5