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Suppose we have an array that represents elements of arithmetic progression in order. One element is missing. We have to find the missing element. So if arr = [2, 4, 8, 10, 12, 14], output is 6, as 6 is missing.

Using binary search, we can solve this problem. We will go to the middle element, then check whether the difference between middle and next to the middle is same as diff 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 AP, then missing element lies in the right half, otherwise to the left half.

#include <iostream> using namespace std; class Progression { public: int missingUtil(int arr[], int left, int right, int diff) { if (right <= left) return INT_MAX; int mid = left + (right - left) / 2; if (arr[mid + 1] - arr[mid] != diff) return (arr[mid] + diff); if (mid > 0 && arr[mid] - arr[mid - 1] != diff) return (arr[mid - 1] + diff); if (arr[mid] == arr[0] + mid * diff) return missingUtil(arr, mid + 1, right, diff); return missingUtil(arr, left, mid - 1, diff); } int missingElement(int arr[], int n) { int diff = (arr[n - 1] - arr[0]) / n; return missingUtil(arr, 0, n - 1, diff); } }; int main() { Progression pg; int arr[] = {2, 4, 8, 10, 12, 14}; int n = sizeof(arr) / sizeof(arr[0]); cout << "The missing element is: " << pg.missingElement(arr, n); }

The missing element is: 6

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