Find the repeating and the missing number using two equations

Given an array arr[] of size N, each integer from the range [1, N] appears exactly once except A which appears twice and B which is missing. The task is to find the numbers A and B.
Examples:

Input: arr[] = {1, 2, 2, 3, 4} 
Output:
A = 2 
B = 5
Input: arr[] = {5, 3, 4, 1, 1} 
Output:
A = 1 
B = 2 

Approach: From the sum of first N natural numbers, 

SumN = 1 + 2 + 3 + … + N = (N * (N + 1)) / 2 
And, let the sum of all the array elements be Sum. Now, 
SumN = Sum – A + B 
A – B = Sum – SumN …(equation 1) 

And from the sum of the squares of first N natural numbers, 

SumSqN = 1² + 2² + 3² + … + N² = (N * (N + 1) * (2 * n + 1)) / 6 
And, let the sum of the squares of all the array elements be SumSq. Now, 
SumSq = SumSqN + A² – B²
SumSq – SumSqN = (A + B) * (A – B) …(equation 2) 

Put value of (A – B) from equation 1 in equation 2, 
SumSq – SumSqN = (A + B) * (Sum – SumN) 
A + B = (SumSq – SumSqN) / (Sum – SumN) …(equation 3) 
Solving equation 1 and equation 3 will give, 

B = (((SumSq – SumSqN) / (Sum – SumN)) + SumN – Sum) / 2
And, A = Sum – SumN + B

Below is the implementation of the above approach:

C++

//C++ implementation of the approach

#include <cmath>
#include<bits/stdc++.h>
#include <iostream>

using namespace std;

    // Function to print the required numbers
 void findNumbers(int arr[], int n)
    {

        // Sum of first n natural numbers
        int sumN = (n * (n + 1)) / 2;

        // Sum of squares of first n natural numbers
        int sumSqN = (n * (n + 1) * (2 * n + 1)) / 6;

        // To store the sum and sum of squares
        // of the array elements
        int sum = 0, sumSq = 0, i;

        for (i = 0; i < n; i++) {
            sum += arr[i];
            sumSq = sumSq + (pow(arr[i], 2));
        }

        int B = (((sumSq - sumSqN) / (sum - sumN)) + sumN - sum) / 2;
        int A = sum - sumN + B;
         cout << "A = " ;
         cout << A << endl;
         cout << "B = " ;
         cout << B << endl;
    }

    // Driver code
int main() {
        int arr[] = { 1, 2, 2, 3, 4 };
        int n = sizeof(arr)/sizeof(arr[0]);
        findNumbers(arr, n);
    return 0;
}

Java

// Java implementation of the approach
public class GFG {

    // Function to print the required numbers
    static void findNumbers(int arr[], int n)
    {

        // Sum of first n natural numbers
        int sumN = (n * (n + 1)) / 2;

        // Sum of squares of first n natural numbers
        int sumSqN = (n * (n + 1) * (2 * n + 1)) / 6;

        // To store the sum and sum of squares
        // of the array elements
        int sum = 0, sumSq = 0, i;

        for (i = 0; i < n; i++) {
            sum += arr[i];
            sumSq += Math.pow(arr[i], 2);
        }

        int B = (((sumSq - sumSqN) / (sum - sumN)) + sumN - sum) / 2;
        int A = sum - sumN + B;
        System.out.println("A = " + A + "\nB = " + B);
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 1, 2, 2, 3, 4 };
        int n = arr.length;

        findNumbers(arr, n);
    }
}

Python3

# Python3 implementation of the approach

import math
# Function to print the required numbers
def findNumbers(arr, n):


        # Sum of first n natural numbers
        sumN = (n * (n + 1)) / 2;

        # Sum of squares of first n natural numbers
        sumSqN = (n * (n + 1) * (2 * n + 1)) / 6;

        # To store the sum and sum of squares
        # of the array elements
        sum = 0;
        sumSq = 0;

        for i in range(0,n):
            sum = sum + arr[i];
            sumSq = sumSq + (math.pow(arr[i], 2));


        B = (((sumSq - sumSqN) / (sum - sumN)) + sumN - sum) / 2;
        A = sum - sumN + B;
        print("A = ",int(A)) ;
        print("B = ",int(B));


# Driver code

arr = [ 1, 2, 2, 3, 4 ];
n = len(arr);
findNumbers(arr, n);

#This code is contributed by Shivi_Aggarwal

C#

// C# implementation of the approach
using System;
public class GFG {

    // Function to print the required numbers
    static void findNumbers(int []arr, int n)
    {

        // Sum of first n natural numbers
        int sumN = (n * (n + 1)) / 2;

        // Sum of squares of first n natural numbers
        int sumSqN = (n * (n + 1) * (2 * n + 1)) / 6;

        // To store the sum and sum of squares
        // of the array elements
        int sum = 0, sumSq = 0, i;

        for (i = 0; i < n; i++) {
            sum += arr[i];
            sumSq += (int)Math.Pow(arr[i], 2);
        }

        int B = (((sumSq - sumSqN) / (sum - sumN)) + sumN - sum) / 2;
        int A = sum - sumN + B;
        Console.WriteLine("A = " + A + "\nB = " + B);
    }

    // Driver code
    public static void Main()
    {
        int []arr = { 1, 2, 2, 3, 4 };
        int n = arr.Length;

        findNumbers(arr, n);
    }
}
// This code is contributed by PrinciRaj1992

PHP

<?php
// PHP implementation of the approach

// Function to print the required numbers
function findNumbers($arr, $n)
{

    // Sum of first n natural numbers
    $sumN = ($n * ($n + 1)) / 2;

    // Sum of squares of first n
    // natural numbers
    $sumSqN = ($n * ($n + 1) *
                (2 * $n + 1)) / 6;

    // To store the sum and sum of
    // squares of the array elements
    $sum = 0 ;
    $sumSq = 0 ;

    for ($i = 0; $i < $n; $i++)
    {
        $sum += $arr[$i];
        $sumSq += pow($arr[$i], 2);
    }

    $B = ((($sumSq - $sumSqN) / ($sum - $sumN)) +
                                 $sumN - $sum) / 2;
    $A = $sum - $sumN + $B;
    echo "A = ", $A, "\nB = ", $B;
}

// Driver code
$arr = array( 1, 2, 2, 3, 4 );
$n = sizeof($arr) ;

findNumbers($arr, $n);

// This code is contributed by Ryuga
?>

Javascript

<script>

// Javascript implementation of the approach

// Function to print the required numbers
function findNumbers(arr, n)
{

    // Sum of first n natural numbers
    sumN = (n * (n + 1)) / 2;

    // Sum of squares of first n
    // natural numbers
    sumSqN = (n * (n + 1) *
                (2 * n + 1)) / 6;

    // To store the sum and sum of
    // squares of the array elements
    let sum = 0 ;
    let sumSq = 0 ;

    for (let i = 0;i < n; i++)
    {
        sum += arr[i];
        sumSq += Math.pow(arr[i], 2);
    }

    B = (((sumSq - sumSqN) / (sum - sumN)) +
                                sumN - sum) / 2;
    A = sum - sumN + B;
    document.write( "A = "+ A, "<br>B = ", B);
}

// Driver code
let arr = [ 1, 2, 2, 3, 4 ];
n = arr.length ;

findNumbers(arr, n);

// This code is contributed
// by bobby

</script>

Output:

A = 2
B = 5

Time Complexity: O(N), since the loop runs from 0 to (n – 1).
Auxiliary Space: O(1), since no extra space has been taken.