Syntax error Missing Number in an array in O(1) Time Complexity

Missing Number in an array in O(1) Time Complexity

Problem Description

We are given an array that contains numbers from 1 to n, and one number is missing between the given numbers from 1 to n. We have to return the missing number in O(1) time complexity. In this problem, we are going to discuss how we can find the missing number in an array.

Example 1

  • Input: array = {1, 2, 3, 5, 6, 7}
  • Output: 4

Explanation

From 1 to 7, the number 4 is missing in the given array.

Example 2

  • Input: array = {1, 2, 3, 4, 5, 6, 7, 8, 10}
  • Output: 9

Explanation

From 1 to 10, the number 9 is missing in the given array.

Example 3

  • Input: array = {1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}
  • Output: 8

Explanation

From 1 to 20, the number 8 is missing in the given array.

Direct Formula Approach

This is a simple and direct approach to finding missing numbers in a given array in reduced time complexity of O(N). In this approach, we first find the sum of the first n numbers from 1 to n using the formula for the sum of the first n natural numbers: n * (n + 1) / 2. Now, we find the sum of all elements in the given array. Subtract the sum of n numbers from the sum of all elements in the given array to get the missing number of the array.

C++ Implementation

#include <iostream>
#include <vector>
using namespace std;

int missingNumber(vector<int>& arr) {
    int n = arr.size() + 1;
    int totalSum = n * (n + 1) / 2;
    int arraySum = 0;

    for(int i : arr) {
        arraySum += i;
    }

    return totalSum - arraySum;
}

int main() {
    vector<int> arr = {1, 2, 3, 4, 6, 7};
    int ans = missingNumber(arr);
    cout << ans << endl;
    return 0;
}

Output

5

Java Implementation

import java.util.*;

public class MissingNumber {
    public static int missingNumber(List<Integer> arr) {
        int n = arr.size() + 1;
        int totalSum = n * (n + 1) / 2;
        int arraySum = 0;

        for (int num : arr) {
            arraySum += num;
        }

        return totalSum - arraySum;
    }

    public static void main(String[] args) {
        List<Integer> arr = Arrays.asList(1, 2, 3, 4, 6, 7);
        int ans = missingNumber(arr);
        System.out.println(ans);
    }
}

Python Implementation

def missing_number(arr):
    n = len(arr) + 1
    total_sum = n * (n + 1) // 2
    array_sum = sum(arr)

    return total_sum - array_sum

if __name__ == "__main__":
    arr = [1, 2, 3, 4, 6, 7]
    ans = missing_number(arr)
    print(ans)

JavaScript Implementation

<script>
function missingNumber(arr) {
    const n = arr.length + 1;
    const totalSum = (n * (n + 1)) / 2;
    const arraySum = arr.reduce((sum, num) => sum + num, 0);

    return totalSum - arraySum;
}

const arr = [1, 2, 3, 4, 6, 7];
const ans = missingNumber(arr);
document.write(ans);
</script>

PHP Implementation

<?php
function missingNumber($arr) {
    $n = count($arr) + 1;
    $totalSum = $n * ($n + 1) / 2;
    $arraySum = array_sum($arr);
    return $totalSum - $arraySum;
}

$arr = [1, 2, 3, 4, 6, 7];
$ans = missingNumber($arr);
echo $ans;
?>

C# Implementation

using System;
using System.Collections.Generic;

class MissingNumber {
    public static int MissingNumberFunc(List<int> arr) {
        int n = arr.Count + 1;
        int totalSum = n * (n + 1) / 2;
        int arraySum = 0;

        foreach (int num in arr) {
            arraySum += num;
        }

        return totalSum - arraySum;
    }

    static void Main(string[] args) {
        List<int> arr = new List<int> { 1, 2, 3, 4, 6, 7 };
        int ans = MissingNumberFunc(arr);
        Console.WriteLine(ans);
    }
}

Time Complexity: O(n)

Space Complexity: O(1)

Updated on: 2024-12-30T19:22:27+05:30

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