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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 
?> 

Output:

A = 2
B = 5

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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