Sqrtx

Problem

https://leetcode.com/problems/sqrtx/

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 <= 2:sup:`31`- 1

Solution

Finding the square root of \(x\) is equivalent to finding \(y\) such that \(y^2 = x\). We can use binary search to find \(y\) in the range \([0, \sqrt{2^{32} - 1}] \approx [0, 2^{16}]\). After narrowing the range of possible square roots to within 1, within 1, we return the beginning of the range.

Pattern

Math, Binary Search

Code

def mySqrt(x: int) -> int:
    """Finds the largest integer :math:`y` such that :math:`y^2 \\leq x`.
    """
    start = 0
    end = 65536
    while start + 1 < end:
        mid = (start + end) // 2
        if mid**2 > x:
            end = mid
        else:
            start = mid

    return start

Test

>>> from Sqrtx import mySqrt
>>> mySqrt(4)
2
>>> mySqrt(8)
2

Sqrtx.mySqrt(x: int) → int

Finds the largest integer \(y\) such that \(y^2 \leq x\).