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Find K such that changing all elements of the Array greater than K to K will make array sum N
  • Difficulty Level : Medium
  • Last Updated : 25 Feb, 2020

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: 4
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:

CPP




// 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


Output:

4

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