Array Manipulation and Sum

Given an array arr[] of N integers and an integer S. The task is to find an element K in the array such that if all the elements from the array > K are made equal to K then the sum of all the elements of the resultant array becomes equal to S. If it is not possible to find such an element then print -1 .

Examples:

Input: M = 15, arr[] = {12, 3, 6, 7, 8} 
Output: 3 
Resultant array = {3, 3, 3, 3, 3} 
Sum of the array elements = 15 = S

Input: M = 5, arr[] = {1, 3, 2, 5, 8} 
Output: 1 

Approach: Sort the array. Traverse the array considering that the value of K is equal to the current element and then check if sum of the previous elements + (K * number of remaining elements) = S. If yes then print the value of K, if no such element found then print -1 in the end.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the required element
int getElement(int a[], int n, int S)
{
    // Sort the array
    sort(a, a + n);

    int sum = 0;

    for (int i = 0; i < n; i++) {

        // If current element
        // satisfies the condition
        if (sum + (a[i] * (n - i)) == S)
            return a[i];
        sum += a[i];
    }

    // No element found
    return -1;
}

// Driver code
int main()
{
    int S = 5;
    int a[] = { 1, 3, 2, 5, 8 };
    int n = sizeof(a) / sizeof(a[0]);

    cout << getElement(a, n, S);

    return 0;
}

Java

//Java implementation of the approach
import java.util.Arrays;

class GFG
{
    // Function to return the required element
    static int getElement(int a[], int n, int S)
    {
        // Sort the array
        Arrays.sort(a);

        int sum = 0;

        for (int i = 0; i < n; i++)
        {

            // If current element
            // satisfies the condition
            if (sum + (a[i] * (n - i)) == S)
                return a[i];
            sum += a[i];
        }

        // No element found
        return -1;
    }

    // Driver code
    public static void main(String[] args)
    {
        int S = 5;
        int a[] = { 1, 3, 2, 5, 8 };
        int n = a.length;

        System.out.println(getElement(a, n, S));
    }
}

// This code is contributed by Mukul singh.

Python 3

# Python3 implementation of the approach

# Function to return the required element
def getElement(a, n, S) :

    # Sort the array
    a.sort();

    sum = 0;

    for i in range(n) :

        # If current element
        # satisfies the condition
        if (sum + (a[i] * (n - i)) == S) :
            return a[i];

        sum += a[i];

    # No element found
    return -1;

# Driver Code
if __name__ == "__main__" :

    S = 5;
    a = [ 1, 3, 2, 5, 8 ];
    n = len(a) ;

    print(getElement(a, n, S)) ;

# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;

class GFG
{
    // Function to return the required element
    static int getElement(int[] a, int n, int S)
    {
        // Sort the array
        Array.Sort(a);

        int sum = 0;

        for (int i = 0; i < n; i++)
        {

            // If current element
            // satisfies the condition
            if (sum + (a[i] * (n - i)) == S)
                return a[i];
            sum += a[i];
        }

        // No element found
        return -1;
    }

    // Driver code
    public static void Main()
    {
        int S = 5;
        int[] a = { 1, 3, 2, 5, 8 };
        int n = a.Length;

        Console.WriteLine(getElement(a, n, S));
    }
}

// This code is contributed by Mukul singh.

PHP

<?php
// PHP implementation of the approach

// Function to return the required element
function getElement($a, $n, $S)
{
    // Sort the array
    sort($a, 0);

    $sum = 0;

    for ($i = 0; $i < $n; $i++)
    {

        // If current element
        // satisfies the condition
        if ($sum + ($a[$i] *
           ($n - $i)) == $S)
            return $a[$i];
        $sum += $a[$i];
    }

    // No element found
    return -1;
}

// Driver code
$S = 5;
$a = array(1, 3, 2, 5, 8);
$n = sizeof($a);

echo getElement($a, $n, $S);

// This code is contributed
// by Akanksha Rai
?>

Javascript

<script>
 //Javascript implementation of the approach

// Function to return the required element
function getElement(a, n, S)
{
    // Sort the array
    a.sort();

    var sum = 0;

    for (var i = 0; i < n; i++) {

        // If current element
        // satisfies the condition
        if (sum + (a[i] * (n - i)) == S)
            return a[i];
        sum += a[i];
    }

    // No element found
    return -1;
}

var S = 5;
var a = [ 1, 3, 2, 5, 8 ];
var n = a.length;
document.write(getElement(a, n, S));


// This code is contributed by SoumikMondal
</script>

Output

1

Complexity Analysis:

• Time Complexity: O(N*logN)
• Auxiliary Space: O(1)