Problem & Approach

Given all numbers between 1 and n except one, find the missing number.

Approach

• Key Observation
  • The sum of 1 to n is n * (n + 1) / 2.
  • Subtracting the sum of the given n-1 numbers gives the missing number.

Related Concepts

1. Math
2. Prefix Sum

Implementation with C++

#include <bits/stdc++.h>
using namespace std;

int main() {
    long long n;
    cin >> n;

    long long sum = n * (n + 1) / 2;
    for (int i = 0; i < n - 1; i++) {
        long long x;
        cin >> x;
        sum -= x;
    }

    cout << sum << "\n";
    return 0;
}