Valid Perfect Square
Created: September 19, 2019 by [lek-tin]
Last updated: May 9, 2020
Given a positive integer num, write a function which returns True if num is a perfect square else False.
Note
Do not use any built-in library function such as sqrt.
Example 1
Input: 16
Output: true
Example 2
Input: 14
Output: false
Solution (naive)
Java
class Solution {
public boolean isPerfectSquare(int num) {
if (num == 1) return true;
for (int i = 2; i < num; i++) {
if (i * i == num) {
return true;
}
if (i * i > num) {
break;
}
}
return false;
}
}
Solution (Binary Search version 1)
Let x be the square root, 1 < x < num/2 always holds true.
Python
class Solution:
def isPerfectSquare(self, num: int) -> bool:
if num == 1:
return True
left, right = 0, num
while left <= right:
mid = left + (right-left)//2
guess = mid*mid
if guess == num:
return True
if guess < num:
left = mid + 1
else:
right = mid - 1
return False
Java
class Solution {
public boolean isPerfectSquare(int num) {
if (num == 1) return true;
long left = 1;
long right = num;
while (left+1 < right) {
long mid = left + (right - left) / 2;
long guess = mid * mid;
if (guess == num) {
return true;
} else if (guess > num) {
right = mid - 1;
} else {
left = mid + 1;
}
}
return left*left == num || right*right==num;
}
}
Solution (Binary Search version 2)
Time: O(logN)
Space: O(1)
class Solution:
def isPerfectSquare(self, num: int) -> bool:
if num == 1:
return True
factor = num // 2
while factor*factor > num:
factor //= 2
for i in range(factor, factor*2+1):
if i*i == num:
return True
return False
Solution (Newton’s method)
Time: O(logN)
Space: O(1)
class Solution:
def isPerfectSquare(self, num: int) -> bool:
if num == 1:
return True
# newton's method
# f(x) = x^2 - num = 0
# x_next = x - f(x) / f'(x)
# f(x) = x^2 -num, f'(x) = 2x
# x_next = (x + num/x) / 2
x = num // 2
while x*x > num:
x = (x + num//x) // 2
return x*x == num