LeetCode 69. Sqrt(x)

Description

Implement int sqrt(int x).

Compute and return the square root of x, where x is guaranteed to be a non-negative integer.

Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.

Example 1:

Input: 4
Output: 2

Example 2:

Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since 
             the decimal part is truncated, 2 is returned.

Explanation

We can convert square root from a math problem to a computer science problem. To find a square root is actually to search for a number among 0 to x. Just use binary search to solve the problem.

Java Solution

class Solution {
    public int mySqrt(int x) {
        long start = 0;
        long end = x;
        
        while (start + 1 < end) {
            long mid = start + (end - start) / 2;
            if (mid * mid == x) {
                return (int)mid;
            } else if (mid * mid < x) {
                start = mid;
            } else {
                end = mid;
            }
        }
        
        if (end * end == x) {
            return (int)end;
        }
        return (int)start;
    }
}

Python Solution

class Solution:
    def mySqrt(self, x: int) -> int:
        
        start = 0
        end = x
        
        while start + 1 < end:
            mid = start + (end - start) // 2
            
            if mid * mid == x:
                return mid
            elif mid * mid < x:
                start = mid
            else:
                end = mid
        
        if end * end == x:
            return end
        
        return start
  • Time complexity: O(log(N)).
  • Space complexity: O(1).

One Thought to “LeetCode 69. Sqrt(x)”

Leave a Reply

Your email address will not be published.