# Sqrt(x) LeetCode Solution

Here, We see Sqrt(x) LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## Problem Statement

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.
```

## Sqrt(x) Leetcode Solution C++

``````class Solution {
public:
int mySqrt(int x) {
long long s=0,e=INT_MAX,ans=0;
while(s<=e){
long long m=s+(e-s)/2;
if(m*m<=x){
ans=m;
s=m+1;
}
else e=m-1;
}
return ans;
}
};```Code language: PHP (php)```

## Sqrt(x) Leetcode Solution Java

``````class Solution {
public int mySqrt(int x) {
long r = x;
while (r*r > x)
r = (r + x/r) / 2;
return (int) r;
}
}```Code language: PHP (php)```

## Sqrt(x) Leetcode Solution JavaScript

``````var mySqrt = function(x) {
r = x;
while (r*r > x)
r = ((r + x/r) / 2) | 0;
return r;
};```Code language: JavaScript (javascript)```

## Sqrt(x) Solution Python

``````class Solution(object):
def mySqrt(self, x):
r = x
while r*r > x:
r = (r + x/r) / 2
return r``````
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