Least number to be added to or subtracted from N to make it a Perfect Square

Given a number N, find the minimum number that needs to be added to or subtracted from N, to make it a perfect square. If the number is to be added, print it with a + sign, else if the number is to be subtracted, print it with a – sign.

Examples:

Input: N = 14
Output: 2
Nearest perfect square before 14 = 9
Nearest perfect square after 14 = 16
Therefore 2 needs to be added to 14 to get the closest perfect square

Input: N = 18
Output: -2
Nearest perfect square before 18 = 16
Nearest perfect square after 18 = 25
Therefore 2 needs to be subtracted from 18 to get the closest perfect square

Approach:



  1. Get the number.
  2. Find the square root of the number and convert the result as an integer.
  3. After converting the double value to integer, this will contain the root of the perfect square before N, i.e. floor(square root(N)).
  4. Then find the square of this number, which will be the perfect square before N.
  5. Find the root of the perfect square after N, i.e. the ceil(square root(N)).
  6. Then find the square of this number, which will be the perfect square after N.
  7. Check whether the square of floor value is nearest to N or the ceil value.
  8. If the square of floor value is nearest to N, print the difference with a -sign. Else print the difference between the square of the ceil value and N with a + sign.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the Least number
int nearest(int n)
{
  
    // Get the perfect square
    // before and after N
    int prevSquare = sqrt(n);
    int nextSquare = prevSquare + 1;
    prevSquare = prevSquare * prevSquare;
    nextSquare = nextSquare * nextSquare;
  
    // Check which is nearest to N
    int ans
        = (n - prevSquare) < (nextSquare - n)
              ? (prevSquare - n)
              : (nextSquare - n);
  
    // return the result
    return ans;
}
  
// Driver code
int main()
{
    int n = 14;
    cout << nearest(n) << endl;
  
    n = 16;
    cout << nearest(n) << endl;
  
    n = 18;
    cout << nearest(n) << endl;
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG {
          
    // Function to return the Least number
    static int nearest(int n)
    {
      
        // Get the perfect square
        // before and after N
        int prevSquare = (int)Math.sqrt(n);
        int nextSquare = prevSquare + 1;
        prevSquare = prevSquare * prevSquare;
        nextSquare = nextSquare * nextSquare;
      
        // Check which is nearest to N
        int ans = (n - prevSquare) < (nextSquare - n)? (prevSquare - n): (nextSquare - n);
      
        // return the result
        return ans;
    }
      
    // Driver code
    public static void main (String[] args)
    {
        int n = 14;
        System.out.println(nearest(n));
      
        n = 16;
        System.out.println(nearest(n)) ;
      
        n = 18;
        System.out.println(nearest(n)) ;
      
    }
}
  
// This code is contributed by AnkitRai01

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Python3

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# Python3 implementation of the approach 
from math import sqrt
  
# Function to return the Least number 
def nearest(n) : 
  
    # Get the perfect square 
    # before and after N 
    prevSquare = int(sqrt(n)); 
    nextSquare = prevSquare + 1
    prevSquare = prevSquare * prevSquare; 
    nextSquare = nextSquare * nextSquare; 
  
    # Check which is nearest to N 
    ans    = (prevSquare - n) if (n - prevSquare) < (nextSquare - n) else (nextSquare - n); 
  
    # return the result 
    return ans; 
  
# Driver code 
if __name__ == "__main__"
  
    n = 14
    print(nearest(n)) ; 
  
    n = 16
    print(nearest(n)); 
  
    n = 18
    print(nearest(n)); 
  
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the approach
using System;
  
class GFG {
          
    // Function to return the Least number
    static int nearest(int n)
    {
      
        // Get the perfect square
        // before and after N
        int prevSquare = (int)Math.Sqrt(n);
        int nextSquare = prevSquare + 1;
        prevSquare = prevSquare * prevSquare;
        nextSquare = nextSquare * nextSquare;
      
        // Check which is nearest to N
        int ans = (n - prevSquare) < (nextSquare - n)? (prevSquare - n): (nextSquare - n);
      
        // return the result
        return ans;
    }
      
    // Driver code
    public static void Main (string[] args)
    {
        int n = 14;
        Console.WriteLine(nearest(n));
      
        n = 16;
        Console.WriteLine(nearest(n)) ;
      
        n = 18;
        Console.WriteLine(nearest(n)) ;
      
    }
}
  
// This code is contributed by AnkitRai01

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Output:

2
0
-2

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