# Sqrt(x) Leetcode Solution

In this post, we are going to solve the Sqrt(x) Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

## Problem

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 builtin exponent function or operator, such as `pow(x, 0.5)` or `x ** 0.5`.

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 <= 231 - 1`

Now, lets see the leetcode solution of Text Justification Leetcode Solution.

### Sqrt(x) Leetcode Solution in Python

```class Solution:
def mySqrt(self, x: int) -> int:
l = 1
r = x + 1

while l < r:
m = (l + r) // 2
if m * m > x:
r = m
else:
l = m + 1

# L: smallest number s.t. l * l > x
return l - 1
```

### Sqrt(x) Leetcode Solutionin CPP

```class Solution {
public:
int mySqrt(int x) {
unsigned l = 1;
unsigned r = x + 1u;

while (l < r) {
const unsigned m = (l + r) / 2;
if (m > x / m)
r = m;
else
l = m + 1;
}

// L: smallest number s.t. l * l > x
return l - 1;
}
};
```

### Sqrt(x) Leetcode Solution in Java

```class Solution {
public int mySqrt(long x) {
long l = 1;
long r = x + 1;

while (l < r) {
final long m = (l + r) / 2;
if (m > x / m)
r = m;
else
l = m + 1;
}

// L: smallest number s.t. l * l > x
return (int) l - 1;
}
}
```

Note: This problem Sqrt(x) is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.

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