Given an array of 
integers, The task is to find another integer 
such that, if all the elements of the array are subtracted individually from the number , then the sum of all the differences should add to 0. If any such integer exists, print the value of , else print 
.
Example
Input: arr[] = {1, 2, 3}
Output: 2
Explanation:
Subtracting all the elements of arrays from 2,
The sum of difference is:
1 + 0 + (-1) = 0.
Solution: The idea is to calculate the total sum of the array and divide it by the size of the array. If the sum of the array is perfectly divisible by its size then the quotient obtained from this division operation will be the required hidden number.
Implementation:
C++
// C++ Program to find the// hidden number#include <iostream>using namespace std; // Driver Codeint main() { // Getting the size of array int n = 3; // Getting the array of size n int a[] = { 1, 2, 3 }; // Solution int i = 0; // Finding sum of the array elements long sum = 0; for (i = 0; i < n; i++) { sum += a[i]; } // Dividing sum by size n long x = sum / n; // Print x, if found if (x * n == sum) cout<<x<<endl; else cout<<("-1")<<endl; return 0; // This code is contributed// by shashank} |
Java
// Java Program to find the// hidden numberpublic class GFG { // Driver Code public static void main(String args[]) { // Getting the size of array int n = 3; // Getting the array of size n int a[] = { 1, 2, 3 }; // Solution int i = 0; // Finding sum of the array elements long sum = 0; for (i = 0; i < n; i++) { sum += a[i]; } // Dividing sum by size n long x = sum / n; // Print x, if found if (x * n == sum) System.out.println(x); else System.out.println("-1"); }} |
Python 3
# Python 3 Program to find the# hidden number# Driver Codeif __name__ == "__main__": # Getting the size of array n = 3 # Getting the array of size n a = [ 1, 2, 3 ] # Solution i = 0 # Finding sum of the . # array elements sum = 0 for i in range(n): sum += a[i] # Dividing sum by size n x = sum // n # Print x, if found if (x * n == sum): print(x) else: print("-1")# This code is contributed # by ChitraNayal |
C#
// C# Program to find the// hidden numberusing System; class GFG{ // Driver Code public static void Main() { // Getting the size of array int n = 3; // Getting the array of size n int []a = { 1, 2, 3 }; // Solution int i = 0; // Finding sum of the // array elements long sum = 0; for (i = 0; i < n; i++) { sum += a[i]; } // Dividing sum by size n long x = sum / n; // Print x, if found if (x * n == sum) Console.WriteLine(x); else Console.WriteLine("-1"); } } // This code is contributed// by inder_verma |
PHP
<?php// PHP Program to find the// hidden number// Driver Code// Getting the size of array$n = 3;// Getting the array of size n$a = array( 1, 2, 3 );// Solution$i = 0;// Finding sum of the array elements$sum = 0;for ($i = 0; $i < $n; $i++){ $sum += $a[$i];}// Dividing sum by size n$x = $sum / $n;// Print x, if foundif ($x * $n == $sum)echo($x);elseecho("-1");// This code is contributed// by inder_verma?> |
Javascript
<script>// JavaScript Program to find the// hidden number // Driver Code // Getting the size of array let n = 3; // Getting the array of size n let a=[1, 2, 3]; // Solution let i = 0; // Finding sum of the array elements let sum = 0; for (i = 0; i < n; i++) { sum += a[i]; } // Dividing sum by size n let x = sum / n; // Print x, if found if (x * n == sum) document.write(x); else document.write("-1");// This code is contributed by rag2127</script> |
Output
2
Complexity Analysis:
- Time Complexity: O(n), where n is the length of the given array.
- Auxiliary space: O(1)
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