Change K elements so that (a1^2 + a2^2 + …+ aN^2 ) <= (a1 + a2 +…+ aN) becomes true

Given an array Arr of size N. The task is to tell whether it is possible to change at most K elements of this sequence to arbitrary positive integers in such a way that the below condition holds.

Examples:



Input:N = 2, Arr[] = {1, 2}, K = 2
Output: Possible
(As A[2] can be change to 1)

Input: N = 2, Arr[] = {5, 6}, K = 1
Output: Not Possible
(As we can only change 1 element to any arbitary number
and after changing it doesn't satisfy above condition) 

Approach: When all the elements of the array becomes equal to 1 then only the given equation can be satisfied, else not.

  1. Traverse the array and count the number of 1.
  2. If K >= (size of array i.e N – count) then return true, Else return false.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function that will tell
// whether it is possible or Not
int Series(int Arr[], int N, int K)
{
    int count = 0;
    for (int i = 0; i < N; i++)
        if (Arr[i] == 1)
            count++;
  
    if (K >= (N - count))
        return 1;
    else
        return 0;
}
  
// Driver code
int main()
{
    int Arr[] = { 5, 1, 2 };
    int N = sizeof(Arr) / sizeof(Arr[0]);
    int K = 2;
  
    // Calling function.
    int result = Series(Arr, N, K);
  
    if (result == 1)
        cout << "Possible";
    else
        cout << "Not Possible";
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

//Java  implementation of above approach 
  
import java.io.*;
  
class GFG {
      
// Function that will tell 
// whether it is possible or Not 
static int Series(int Arr[], int N, int K) 
    int count = 0
    for (int i = 0; i < N; i++) 
        if (Arr[i] == 1
            count++; 
  
    if (K >= (N - count)) 
        return 1
    else
        return 0
  
// Driver code 
    public static void main (String[] args) {
    int Arr[] = { 5, 1, 2 }; 
    int N = Arr.length; 
    int K = 2
    // Calling function. 
    int result = Series(Arr, N, K); 
    if (result == 1
            System.out.println ("Possible"); 
    else
            System.out.println( "Not Possible"); 
          
    }
//This Code is Contributed by ajit    
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python implementation of 
# above approach
  
# Function that will tell
# whether it is possible or Not
def Series(Arr, N, K):
    count = 0
    for i in range(N):
        if Arr[i] == 1:
            count += 1
    if K >= (N - count):
        return 1
    return 0
  
# Driver code
Arr = [5, 1, 2]
N = len(Arr)
K = 2
  
result = Series(Arr, N, K)
if result == 1:
    print("Possible")
else:
    print("Not Possible")
  
# This code is contributed
# by Shrikant13

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

//C# implementation of above approach
  
using System;
  
public class GFG{
      
          
// Function that will tell 
// whether it is possible or Not 
static int Series(int []Arr, int N, int K) 
    int count = 0; 
    for (int i = 0; i < N; i++) 
        if (Arr[i] == 1) 
            count++; 
  
    if (K >= (N - count)) 
        return 1; 
    else
        return 0; 
  
// Driver code 
      
    static public void Main (){
    int []Arr = { 5, 1, 2 }; 
    int N = Arr.Length; 
    int K = 2; 
    // Calling function. 
    int result = Series(Arr, N, K); 
    if (result == 1) 
            Console.WriteLine ("Possible"); 
    else
            Console.WriteLine( "Not Possible"); 
          
    
//This Code is Contributed by akt_mit

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP implementation of above approach 
  
// Function that will tell 
// whether it is possible or Not 
function Series($Arr, $N, $K
    $count = 0; 
    for ($i = 0; $i < $N; $i++) 
        if ($Arr[$i] == 1) 
            $count++; 
  
    if ($K >= ($N - $count)) 
        return 1; 
    else
        return 0; 
  
// Driver code 
$Arr = array( 5, 1, 2 ); 
$N = sizeof($Arr); 
$K = 2; 
  
// Calling function. 
$result = Series($Arr, $N, $K); 
  
if ($result == 1) 
    echo "Possible"
else
    echo "Not Possible"
  
// This code is contributed
// by Sach_Code
?>

chevron_right


Output:

Possible


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : jit_t, Sach_Code, shrikanth13