# Find the missing number in Geometric Progression in C++

Suppose we have an array that represents elements of geometric progression in order. One element is missing. We have to find the missing element. So if arr = [1, 3, 27, 81], output is 9, as 9 is missing.

Using binary search, we can solve this problem. We will go to the middle element, then check whether the ratio between middle and next to the middle is same as common ratio or not. If not, then missing element is present between indices mid and mid + 1. If the middle element is the n/2th element in the GP, then missing element lies in the right half, otherwise to the left half.

## Example

#include <iostream>
#include <cmath>
using namespace std;
class Progression {
public:
int missingUtil(int arr[], int left, int right, int ratio) {
if (right <= left)
return INT_MAX;
int mid = left + (right - left) / 2;
if (arr[mid + 1] - arr[mid] != ratio)
return (arr[mid] * ratio);
if (mid > 0 && arr[mid] / arr[mid - 1] != ratio)
return (arr[mid - 1] * ratio);
if (arr[mid] == arr * pow(ratio, mid))
return missingUtil(arr, mid + 1, right, ratio);
return missingUtil(arr, left, mid - 1, ratio);
}
int missingElement(int arr[], int n) {
int ratio = pow(arr[n-1]/arr, 1.0/n);
return missingUtil(arr, 0, n - 1, ratio);
}
};
int main() {
Progression pg;
int arr[] = {1, 3, 27, 81};
int n = sizeof(arr) / sizeof(arr);
cout << "The missing element is: " << pg.missingElement(arr, n);
}

## Output

The missing element is: 9