Find the Missing Number

Medium Arrays

An array holds the numbers 1 to n with exactly one missing. Find it using the sum formula.

Sample input

[1, 2, 4, 5], n = 5

Sample output

Solution

arr = [1, 2, 4, 5]
n = 5
expected = n * (n + 1) // 2
print(expected - sum(arr))

const arr = [1, 2, 4, 5];
const n = 5;
const expected = (n * (n + 1)) / 2;
const actual = arr.reduce((a, b) => a + b, 0);
console.log(expected - actual);

public class Main {
    public static void main(String[] args) {
        int[] arr = {1, 2, 4, 5};
        int n = 5;
        int expected = n * (n + 1) / 2;
        int actual = 0;
        for (int x : arr) actual += x;
        System.out.println(expected - actual);
    }
}

fun main() {
    val arr = intArrayOf(1, 2, 4, 5)
    val n = 5
    val expected = n * (n + 1) / 2
    println(expected - arr.sum())
}

let arr = [1, 2, 4, 5]
let n = 5
let expected = n * (n + 1) / 2
print(expected - arr.reduce(0, +))

void main() {
  List<int> arr = [1, 2, 4, 5];
  int n = 5;
  int expected = n * (n + 1) ~/ 2;
  int actual = arr.reduce((a, b) => a + b);
  print(expected - actual);
}

#include <iostream>
using namespace std;

int main() {
    int arr[] = {1, 2, 4, 5};
    int n = 5, size = 4;
    int expected = n * (n + 1) / 2;
    int actual = 0;
    for (int i = 0; i < size; i++) actual += arr[i];
    cout << expected - actual << endl;
    return 0;
}

#include <stdio.h>

int main() {
    int arr[] = {1, 2, 4, 5};
    int n = 5, size = 4;
    int expected = n * (n + 1) / 2;
    int actual = 0;
    for (int i = 0; i < size; i++) actual += arr[i];
    printf("%d\n", expected - actual);
    return 0;
}