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 falsebool 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 codeint 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 notpublic 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 falsedef 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 codeif __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 notusing 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 falsefunction 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 codeif(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.neveropen.co.za/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 contributed by Utkarsh Trivedi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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