Given an array of
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 Code
int 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 number public 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 Code if __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 number using 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 found if ( $x * $n == $sum )
echo ( $x );
else echo ( "-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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