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)
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