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 arbitrary 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.
- Traverse the array and count the number of 1.
- If K >= (size of array i.e N – count) then return true, Else return false.
Below is the implementation of the above approach:
C++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// Function that will tell// whether it is possible or Notint 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 codeint 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;} |
Java
//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 } |
Python3
# Python implementation of # above approach# Function that will tell# whether it is possible or Notdef 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 codeArr = [5, 1, 2]N = len(Arr)K = 2result = Series(Arr, N, K)if result == 1: print("Possible")else: print("Not Possible")# This code is contributed# by Shrikant13 |
C#
//C# implementation of above approachusing 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} |
PHP
<?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?> |
Javascript
<script>// Javascript implementation of above approach// Function that will tell// whether it is possible or Notfunction Series(Arr, N, K){ var count = 0; for(var i = 0; i < N; i++) if (Arr[i] == 1) count++; if (K >= (N - count)) return 1; else return 0;}// Driver codevar Arr = [ 5, 1, 2 ];var N = Arr.length;var K = 2;// Calling function.var result = Series(Arr, N, K);if (result == 1) document.write("Possible");else document.write("Not Possible"); // This code is contributed by rutvik_56</script> |
Output
Possible
Complexity Analysis:
- Time Complexity: O(N)
- Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!




