# LeetCode 69. Sqrt(x)

## Description

Implement `int sqrt(int x)`.

Compute and return the square root of x, where x is guaranteed to be a non-negative integer.

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: 4
Output: 2
```

Example 2:

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

## Explanation

We can convert square root from a math problem to a computer science problem. To find a square root is actually to search for a number among 0 to x. Just use binary search to solve the problem.

## Java Solution

``````class Solution {
public int mySqrt(int x) {
long start = 0;
long end = x;

while (start + 1 < end) {
long mid = start + (end - start) / 2;
if (mid * mid == x) {
return (int)mid;
} else if (mid * mid < x) {
start = mid;
} else {
end = mid;
}
}

if (end * end == x) {
return (int)end;
}
return (int)start;
}
}``````

## Python Solution

``````class Solution:
def mySqrt(self, x: int) -> int:

start = 0
end = x

while start + 1 < end:
mid = start + (end - start) // 2

if mid * mid == x:
return mid
elif mid * mid < x:
start = mid
else:
end = mid

if end * end == x:
return end

return start``````
• Time complexity: O(log(N)).
• Space complexity: O(1).