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Check perfect square using addition/subtraction

Given a positive integer n, check if it is perfect square or not using only addition/subtraction operations and in minimum time complexity.

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We can use the property of odd number for this purpose:

Addition of first n odd numbers is always perfect square 
1 + 3 = 4,      
1 + 3 + 5 = 9,     
1 + 3 + 5 + 7 + 9 + 11 = 36 ...

Below is the implementation of above idea :

C++

// C++ program to check if n is perfect square
// or not
#include <bits/stdc++.h>
  
using namespace std;
  
// This function returns true if n is
// perfect square, else false
bool isPerfectSquare(int n)
{
    // sum is sum of all odd numbers. i is
    // used one by one hold odd numbers
    for (int sum = 0, i = 1; sum < n; i += 2) {
        sum += i;
        if (sum == n)
            return true;
    }
    return false;
}
  
// Driver code
int main()
{
    isPerfectSquare(35) ? cout << "Yes " : cout << "No ";
    isPerfectSquare(49) ? cout << "Yes " : cout << "No ";
    return 0;
}

Java

// Java program to check if n
// is perfect square or not
  
public class GFG {
  
    // This function returns true if n
    // is perfect square, else false
    static boolean isPerfectSquare(int n)
    {
        // sum is sum of all odd numbers. i is
        // used one by one hold odd numbers
        for (int sum = 0, i = 1; sum < n; i += 2) {
            sum += i;
            if (sum == n)
                return true;
        }
        return false;
    }
  
    // Driver Code
    public static void main(String args[])
    {
  
        if (isPerfectSquare(35))
            System.out.println("Yes");
        else
            System.out.println("NO");
  
        if (isPerfectSquare(49))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
  
// This code is contributed by Sam007

Python3

# This function returns true if n is
# perfect square, else false
def isPerfectSquare(n):
  
    # the_sum is sum of all odd numbers. i is
    # used one by one hold odd numbers
    i = 1
    the_sum = 0
    while the_sum < n:
        the_sum += i
        if the_sum == n:
            return True
        i += 2
    return False
  
# Driver code
if __name__ == "__main__":
    print('Yes') if isPerfectSquare(35) else print('NO')
    print('Yes') if isPerfectSquare(49) else print('NO')
  
# This code works only in Python 3

/div>

C#

// C# program to check if n
// is perfect square or not
using System;
  
public class GFG {
  
    // This function returns true if n
    // is perfect square, else false
    static bool isPerfectSquare(int n)
    {
        // sum is sum of all odd numbers. i is
        // used one by one hold odd numbers
        for (int sum = 0, i = 1; sum < n; i += 2) {
            sum += i;
            if (sum == n)
                return true;
        }
        return false;
    }
  
    // Driver Code
    public static void Main(String[] args)
    {
  
        if (isPerfectSquare(35))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
  
        if (isPerfectSquare(49))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
  
// This code is contributed by Sam007.

PHP

<?php
// PHP program to check if n is 
// perfect square or not
  
// This function returns true if n is
// perfect square, else false
function isPerfectSquare($n)
{
    // sum is sum of all odd numbers.
    // i is used one by one hold odd
    // numbers
    for ( $sum = 0, $i = 1; $sum < $n;
                              $i += 2)
    {
        $sum += $i;
        if ($sum == $n)
            return true;
    }
      
    return false;
}
  
// Driver code
if(isPerfectSquare(35))
    echo "Yes ";
else
    echo "No ";
      
if(isPerfectSquare(49))
    echo "Yes ";
else
    echo "No ";
  
// This code is contributed by ajit.
?>


Output :



No
Yes

How does this work?
Below is explanation of above approach.

1 + 3 + 5 + ...  (2n-1) = &Sum;(2*i - 1) where 1<=i<=n
                        = 2*&Sum;i - &Sum;1  where 1<=i<=n
                        = 2n(n+1)/2 - n
                        = n(n+1) - n
                        = n2

Reference:
https://tutorialspoint.dev/slugresolver/sum-first-n-odd-numbers-o1-complexity/
http://blog.jgc.org/2008/02/sum-of-first-n-odd-numbers-is-always.html



This article is attributed to GeeksforGeeks.org

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