69. Sqrt(x)

LeetCode 69. Sqrt(x)

Description

Given a non-negative integer x, compute and return the square root of x.

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: x = 4
Output: 2

Example 2:

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

Constraints:

• 0 <= x <= 2^31 - 1

Tags

Math, Binary Search

Solution

Perform Binary Search between 0 and x/2+1 (for x=1). If x is not a perfect square number, i.e., we cannot find mid^2 == x, we evaluate if the square of the right pointer is smaller than x first, then check the left one. The reason is we have to find the maximum value whose square is no greater than x.

Complexity

• Time complexity: $O(\log(x))$

• Space complexity: $O(1)$

Code

func mySqrt(x int) int {
	l, r := 0, x/2+1
	for l+1 < r {
		mid := l + (r-l)/2
		square := mid * mid
		if square == x {
			return mid
		}
		if square < x {
			l = mid
		} else {
			r = mid
		}
	}
	if r*r <= x {
		return r
	}
	return l
}