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Check whether the number can be made perfect square after adding K
  • Last Updated : 30 Mar, 2020

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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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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