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Check whether the number can be made perfect square after adding K

  • Last Updated : 08 Apr, 2021

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 K
void 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 code
int 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 K
import 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 K
using 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 K
function 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 code
var n = 7, k = 2;
isPerfectSquare(n + k);
 
</script>
Output: 
Yes

 

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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