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Find K such that changing all elements of the Array greater than K to K will make array sum N

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  • Difficulty Level : Medium
  • Last Updated : 30 Jan, 2022

Given an array arr[] and an integer N, the task is to find a number K from the given array such that if all the elements in the arr greater than K are changed to K then the sum of all the elements in the resulting array will be N. Print -1 if it is not possible. 
Examples: 
 

Input: arr[] = {3, 1, 10, 8, 4}, N = 16 
Output:
If all the elements greater than 4 are changed to 4 then the resulting array will be {3, 1, 4, 4, 4} 
Hence the sum will be 16 which is the required value of N.
Input: arr[] = {3, 1, 10, 8, 4}, N = 11 
Output: -1 
 

 

Approach: The idea is to use sorting to solve the above problem. 
 

  • First, sort the array in ascending order.
  • Then traverse the array and for each element arr[i].
  • Assume that all the elements after it has been changed to arr[i] and calculate the sum.
  • If it is equal to N then return arr[i].
  • If the end of the array is reached then print -1.

Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to return K such that changing
// all elements greater than K to K will
// make array sum N otherwise return -1
int findK(int arr[], int size, int N)
{
 
    // Sorting the array in increasing order
    sort(arr, arr + size);
    int temp_sum = 0;
 
    // Loop through all the elements of the array
    for (int i = 0; i < size; i++) {
        temp_sum += arr[i];
 
        // Checking if sum of array equals N
        if (N - temp_sum == arr[i] * (size - i - 1)) {
            return arr[i];
        }
    }
    return -1;
}
 
// Driver code
int main()
{
    int arr[] = { 3, 1, 10, 4, 8 };
    int size = sizeof(arr) / sizeof(int);
    int N = 16;
 
    cout << findK(arr, size, N);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
     
    // Function to return K such that changing
    // all elements greater than K to K will
    // make array sum N otherwise return -1
    static int findK(int arr[], int size, int N)
    {
     
        // Sorting the array in increasing order
        Arrays.sort(arr);
        int temp_sum = 0;
     
        // Loop through all the elements of the array
        for (int i = 0; i < size; i++)
        {
            temp_sum += arr[i];
     
            // Checking if sum of array equals N
            if (N - temp_sum == arr[i] * (size - i - 1))
            {
                return arr[i];
            }
        }
        return -1;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int []arr = { 3, 1, 10, 4, 8 };
        int size = arr.length;
        int N = 16;
     
        System.out.print(findK(arr, size, N));
    }
}
 
// This code is contributed by AnkitRai01

Python




# Python3 implementation of the approach
 
# Function to return K such that changing
# all elements greater than K to K will
# make array sum N otherwise return -1
def findK(arr, size, N):
 
    # Sorting the array in increasing order
    arr = sorted(arr)
    temp_sum = 0
 
    # Loop through all the elements of the array
    for i in range(size):
        temp_sum += arr[i]
 
        # Checking if sum of array equals N
        if (N - temp_sum == arr[i] * (size - i - 1)):
            return arr[i]
    return -1
 
# Driver code
arr = [3, 1, 10, 4, 8]
size = len(arr)
N = 16
 
print(findK(arr, size, N))
 
# This code is contributed by mohit kumar 29

C#




// C# implementation of the above approach
using System;
 
class GFG
{
     
    // Function to return K such that changing
    // all elements greater than K to K will
    // make array sum N otherwise return -1
    static int findK(int []arr, int size, int N)
    {
     
        // Sorting the array in increasing order
        Array.Sort(arr);
        int temp_sum = 0;
     
        // Loop through all the elements of the array
        for (int i = 0; i < size; i++)
        {
            temp_sum += arr[i];
     
            // Checking if sum of array equals N
            if (N - temp_sum == arr[i] * (size - i - 1))
            {
                return arr[i];
            }
        }
        return -1;
    }
     
    // Driver code
    public static void Main()
    {
        int []arr = { 3, 1, 10, 4, 8 };
        int size = arr.Length;
        int N = 16;
     
        Console.Write(findK(arr, size, N));
    }
}
 
// This code is contributed by AnkitRai01

Javascript




<script>
    // Javascript implementation of the approach
     
    // Function to return K such that changing
    // all elements greater than K to K will
    // make array sum N otherwise return -1
    function findK(arr, size, N)
    {
       
        // Sorting the array in increasing order
        arr.sort(function(a, b){return a - b});
        let temp_sum = 0;
       
        // Loop through all the elements of the array
        for (let i = 0; i < size; i++)
        {
            temp_sum += arr[i];
       
            // Checking if sum of array equals N
            if (N - temp_sum == arr[i] * (size - i - 1))
            {
                return arr[i];
            }
        }
        return -1;
    }
     
    let arr = [ 3, 1, 10, 4, 8 ];
    let size = arr.length;
    let N = 16;
 
    document.write(findK(arr, size, N));
 
// This code is contributed by divyesh07019.
</script>

Output: 

4

 

Time Complexity: O(N * log N)

Auxiliary Space: O(1)


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