Introduction

In this blog post, we will explore how to solve the LeetCode problem "69" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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. Constraints: 0 <= x <= 231 - 1

Explanation

Here's a solution to find the integer square root of a non-negative integer, optimized for efficiency, along with explanations and complexity analysis:

• Binary Search: The core idea is to use binary search on the range of possible square roots (0 to x). This is efficient because it quickly narrows down the search space.
• Midpoint Check: In each iteration, we calculate the midpoint and check if its square is less than or equal to x. If it is, we update our lower bound to potentially find a closer (larger) root; otherwise, we update the upper bound.

• Integer Overflow Prevention: When calculating the square of the midpoint, we should handle potential integer overflows carefully, especially for large values of x. We can mitigate this by using long long integer type.

• Runtime Complexity: O(log x). Storage Complexity: O(1).

Code

    def mySqrt(x):
    """
    Calculates the integer square root of a non-negative integer x.

    Args:
        x: A non-negative integer.

    Returns:
        The square root of x rounded down to the nearest integer.
    """
    if x == 0:
        return 0

    left = 1
    right = x
    ans = 0

    while left <= right:
        mid = left + (right - left) // 2  # Avoid potential overflow
        square = mid * mid

        if square == x:
            return mid
        elif square < x:
            ans = mid  # Potential answer, but look for a closer one
            left = mid + 1
        else:
            right = mid - 1

    return ans