# LeetCode — Sqrt(x)

1 min readAug 22, 2021

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.

Note: You are not allowed to use any built-in exponent function or operator, such as `pow(x, 0.5)` or `x ** 0.5`.

Example 1:

`Input: x = 4Output: 2`

Example 2:

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

Constraints:

• `0 <= x <= 231 - 1`

Solution:
Making use of binary search to find out the number, whose power of 2 is mostly closed to x value. To assign the halved value of num & left smallest integer into mid_val variable, using its power of 2 to compare whether it is greater than x or not. If its value is greater, assign the middle value into ans variable and increment the value of left smallest number as well. If not, then decrement the value of num.