Sqrtx leetcode 69 python

### problem statement

given a non-negative integer `x`, return the square root of `x` rounded down to the nearest integer. implement the function `int mysqrt(int x)`.

### constraints
- `0 = x = 2^31 - 1`

### approach

there are several ways to approach this problem, but we will cover two common methods: **binary search** and **newton's method**.

#### method 1: binary search

1. **initialize** two pointers: `left = 0` and `right = x`.
2. **binary search**:
- while `left = right`:
- calculate the `mid` point: `mid = (left + right) // 2`.
- check if `mid * mid` is equal to `x`. if it is, return `mid`.
- if `mid * mid` is less than `x`, move the `left` pointer to `mid + 1`.
- if `mid * mid` is greater than `x`, move the `right` pointer to `mid - 1`.
3. the answer will be `right` at the end, as it will be the largest integer where `right * right = x`.

#### code example: binary search

here is the implementation using the binary search approach:

### explanation of the code
- we first handle the base cases where `x` is less than 2, returning `x` directly.
- we then set up our binary search with `left` and `right` pointers.
- in each iteration, we calculate the `mid` and check if its square is equal to `x`, less than `x`, or greater than `x`.
- depending on the comparison, we adjust our `left` or `right` pointers accordingly.
- finally, we return `right`, which will be the largest integer whose square is less than or equal to `x`.

### method 2: newton's method

newton's method can also be used to find the square root. it relies on the idea of successive approximations.

here's a brief outline of th ...

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