Introduction

In this blog post, we will explore how to solve the LeetCode problem "780" 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 four integers sx, sy, tx, and ty, return true if it is possible to convert the point (sx, sy) to the point (tx, ty) through some operations, or false otherwise. The allowed operation on some point (x, y) is to convert it to either (x, x + y) or (x + y, y). Example 1: Input: sx = 1, sy = 1, tx = 3, ty = 5 Output: true Explanation: One series of moves that transforms the starting point to the target is: (1, 1) -> (1, 2) (1, 2) -> (3, 2) (3, 2) -> (3, 5) Example 2: Input: sx = 1, sy = 1, tx = 2, ty = 2 Output: false Example 3: Input: sx = 1, sy = 1, tx = 1, ty = 1 Output: true Constraints: 1 <= sx, sy, tx, ty <= 109

Explanation

Here's the breakdown of the problem and the Python solution:

• High-Level Approach:

  • Work backward from the target (tx, ty) towards the source (sx, sy). This avoids exploring unnecessary branches.
  • At each step, determine which operation could have led to the current point and reverse it.
  • If we reach (sx, sy), it's possible; otherwise, it's not.

• Complexity:

  • Runtime: O(log(max(tx, ty))), Storage: O(1)

Code

    def reachingPoints(sx: int, sy: int, tx: int, ty: int) -> bool:
    while tx >= sx and ty >= sy:
        if tx == sx and ty == sy:
            return True
        if tx == ty:
            return False
        if tx > ty:
            if ty > sy:
                tx %= ty
            else:
                return (tx - sx) % ty == 0
        else:
            if tx > sx:
                ty %= tx
            else:
                return (ty - sy) % tx == 0
    return False