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.
We strongly recommend you to minimize your browser and try this yourself first.
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\n" : cout << "No\n" ; isPerfectSquare(49) ? cout << "Yes\n" : cout << "No\n" ; 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 |
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\n" ; else echo "No\n" ; if (isPerfectSquare(49)) echo "Yes\n" ; else echo "No\n" ; // This code is contributed by ajit. ?> |
Javascript
<script> // JavaScript 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 (let sum = 0, i = 1; sum < n; i += 2) { sum += i; if (sum == n) return true ; } return false ; } // Driver Code if (isPerfectSquare(35)) document.write( "Yes" + "<br/>" ); else document.write( "NO" + "<br/>" ); if (isPerfectSquare(49)) document.write( "Yes" + "<br/>" ); else document.write( "No" + "<br/>" ); // This code is contributed by target_2. </script> |
Output :
No Yes
Time Complexity: O(n)
Auxiliary Space: O(1)
How does this work?
Below is explanation of above approach.
1 + 3 + 5 + ... (2n-1) = ?(2*i - 1) where 1<=i<=n = 2*?i - ?1 where 1<=i<=n = 2n(n+1)/2 - n = n(n+1) - n = n2
Reference:
https://www.w3wiki.net/sum-first-n-odd-numbers-o1-complexity/
http://blog.jgc.org/2008/02/sum-of-first-n-odd-numbers-is-always.html