Given two numbers N and K, the task is to check whether the given number N can be made a perfect square after adding K to it.
Examples:
Input: N = 7, K = 2
Output: Yes
Explanation:
7 + 2 = 9 which is a perfect square.Input: N = 5, K = 3
Output: No
Explanation:
5 + 3 = 8 which is not a perfect square.
Approach: The idea is to compute the value of N + K and check if it is a perfect square or not. In order to check if a number is a perfect square or not, refer to this article.
Below is the implementation of the above approach:
C++
// C++ program to check whether the number// can be made perfect square after adding K#include <bits/stdc++.h>using namespace std;// Function to check whether the number// can be made perfect square after adding Kvoid isPerfectSquare(long long int x){ // Computing the square root of // the number long double sr = round(sqrt(x)); // Print Yes if the number // is a perfect square if (sr * sr == x) cout << "Yes"; else cout << "No";}// Driver codeint main(){ int n = 7, k = 2; isPerfectSquare(n + k); return 0;} |
Java
// Java program to check whether the number// can be made perfect square after adding Kimport java.util.*;class GFG{ // Function to check whether the number // can be made perfect square after adding K static void isPerfectSquare(int x) { // Computing the square root of // the number int sr = (int)Math.sqrt(x); // Print Yes if the number // is a perfect square if (sr * sr == x) System.out.println("Yes"); else System.out.println("No"); } // Driver code public static void main(String args[]) { int n = 7, k = 2; isPerfectSquare(n + k); }}// This code is contributed by Yash_R |
Python3
# Python3 program to check whether the number # can be made perfect square after adding K from math import sqrt# Function to check whether the number # can be made perfect square after adding K def isPerfectSquare(x) : # Computing the square root of # the number sr = int(sqrt(x)); # Print Yes if the number # is a perfect square if (sr * sr == x) : print("Yes"); else : print("No"); # Driver code if __name__ == "__main__" : n = 7; k = 2; isPerfectSquare(n + k); # This code is contributed by Yash_R |
C#
// C# program to check whether the number// can be made perfect square after adding Kusing System;class GFG{ // Function to check whether the number // can be made perfect square after adding K static void isPerfectSquare(int x) { // Computing the square root of // the number int sr = (int)Math.Sqrt(x); // Print Yes if the number // is a perfect square if (sr * sr == x) Console.WriteLine("Yes"); else Console.WriteLine("No"); } // Driver code public static void Main(String []args) { int n = 7; int k = 2; isPerfectSquare(n + k); }}// This code is contributed by Yash_R |
Javascript
<script>// Javascript program to check whether the number// can be made perfect square after adding K// Function to check whether the number// can be made perfect square after adding Kfunction isPerfectSquare(x){ // Computing the square root of // the number var sr = Math.round(Math.sqrt(x)); // Print Yes if the number // is a perfect square if (sr * sr == x) document.write("Yes"); else document.write("No");}// Driver codevar n = 7, k = 2;isPerfectSquare(n + k);</script> |
Output:
Yes
Time complexity: O(log(N)), where N is sum of given n and k.
Auxiliary space: O(1)
Note: Similar, it can be checked whether (N – K) can be a perfect square or not, by replacing ‘+’ with ‘-‘ sign in the above function.
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