# 69. Sqrt(x)

## Description

Given a non-negative integer x, return the square root of x rounded down to the nearest integer. The returned integer should be non-negative as well.

You must not use any built-in exponent function or operator.

• For example, do not use pow(x, 0.5) in c++ or x ** 0.5 in python.

Example 1:

Input: x = 4
Output: 2
Explanation: The square root of 4 is 2, so we return 2.


Example 2:

Input: x = 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since we round it down to the nearest integer, 2 is returned.


Constraints:

• 0 <= x <= 231 - 1

## Solutions

Solution 1: Binary Search

We define the left boundary of the binary search as $l = 0$ and the right boundary as $r = x$, then we search for the square root within the range $[l, r]$.

In each step of the search, we find the middle value $mid = (l + r + 1) / 2$. If $mid > x / mid$, it means the square root is within the range $[l, mid - 1]$, so we set $r = mid - 1$. Otherwise, it means the square root is within the range $[mid, r]$, so we set $l = mid$.

After the search ends, we return $l$.

The time complexity is $O(\log x)$, and the space complexity is $O(1)$.

• class Solution {
public int mySqrt(int x) {
int l = 0, r = x;
while (l < r) {
int mid = (l + r + 1) >>> 1;
if (mid > x / mid) {
r = mid - 1;
} else {
l = mid;
}
}
return l;
}
}

• class Solution {
public:
int mySqrt(int x) {
int l = 0, r = x;
while (l < r) {
int mid = (l + r + 1ll) >> 1;
if (mid > x / mid) {
r = mid - 1;
} else {
l = mid;
}
}
return l;
}
};

• class Solution:
def mySqrt(self, x: int) -> int:
l, r = 0, x
while l < r:
mid = (l + r + 1) >> 1
if mid > x // mid:
r = mid - 1
else:
l = mid
return l


• func mySqrt(x int) int {
return sort.Search(x+1, func(i int) bool { return i*i > x }) - 1
}

• /**
* @param {number} x
* @return {number}
*/
var mySqrt = function (x) {
let [l, r] = [0, x];
while (l < r) {
const mid = (l + r + 1) >> 1;
if (mid > x / mid) {
r = mid - 1;
} else {
l = mid;
}
}
return l;
};


• public class Solution {
public int MySqrt(int x) {
int l = 0, r = x;
while (l < r) {
int mid = (l + r + 1) >>> 1;
if (mid > x / mid) {
r = mid - 1;
} else {
l = mid;
}
}
return l;
}
}

• impl Solution {
pub fn my_sqrt(x: i32) -> i32 {
let mut l = 0;
let mut r = x;

while l < r {
let mid = (l + r + 1) / 2;

if mid > x / mid {
r = mid - 1;
} else {
l = mid;
}
}

l
}
}