Introduction

In this blog post, we will explore how to solve the LeetCode problem "1776" 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

There are n cars traveling at different speeds in the same direction along a one-lane road. You are given an array cars of length n, where cars[i] = [positioni, speedi] represents: positioni is the distance between the ith car and the beginning of the road in meters. It is guaranteed that positioni < positioni+1. speedi is the initial speed of the ith car in meters per second. For simplicity, cars can be considered as points moving along the number line. Two cars collide when they occupy the same position. Once a car collides with another car, they unite and form a single car fleet. The cars in the formed fleet will have the same position and the same speed, which is the initial speed of the slowest car in the fleet. Return an array answer, where answer[i] is the time, in seconds, at which the ith car collides with the next car, or -1 if the car does not collide with the next car. Answers within 10-5 of the actual answers are accepted. Example 1: Input: cars = [[1,2],[2,1],[4,3],[7,2]] Output: [1.00000,-1.00000,3.00000,-1.00000] Explanation: After exactly one second, the first car will collide with the second car, and form a car fleet with speed 1 m/s. After exactly 3 seconds, the third car will collide with the fourth car, and form a car fleet with speed 2 m/s. Example 2: Input: cars = [[3,4],[5,4],[6,3],[9,1]] Output: [2.00000,1.00000,1.50000,-1.00000] Constraints: 1 <= cars.length <= 105 1 <= positioni, speedi <= 106 positioni < positioni+1

Explanation

Here's a breakdown of the approach, complexities, and the Python code:

• Approach:

  • Iterate through the cars from right to left using a stack to keep track of potential collisions.
  • For each car, calculate the collision time with the car at the top of the stack. If the current car collides sooner than the top of the stack, pop the stack.
  • If the stack is empty after the above step, there's no collision. Otherwise, record the collision time.

• Complexity:

  • Runtime: O(n) - Each car is visited and potentially pushed/popped from the stack once.
  • Storage: O(n) - In the worst case, the stack can contain all the cars.

Code

    def getCollisionTimes(cars):
    n = len(cars)
    answer = [-1.0] * n
    stack = []  # Store indices of cars

    for i in range(n - 1, -1, -1):
        while stack:
            # Calculate collision time with the car at the top of the stack
            pos1, speed1 = cars[i]
            pos2, speed2 = cars[stack[-1]]

            if speed1 <= speed2:
                # No collision possible as the first car is slower
                stack.pop()
                continue

            collision_time = (pos2 - pos1) / (speed1 - speed2)

            # Check if the collision time is valid
            if answer[stack[-1]] == -1 or collision_time <= answer[stack[-1]]:
                answer[i] = collision_time
                break
            else:
                # The current car at the top of stack, will collide with another car sooner.  Discard this current car from stack.
                stack.pop()

        if not stack:
            answer[i] = -1.0
        stack.append(i)

    return answer