How do you solve Car Fleet II on LeetCode?

Car Fleet II (LeetCode #1776) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Cars with higher speeds cannot collide with cars ahead of them.

What is the brute force solution for Car Fleet II?

The brute force approach for Car Fleet II is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we check each car against every other car to determine when they will collide. This is straightforward but inefficient since it involves a lot of redundant calculations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Car Fleet II?

Car Fleet II uses the following concepts: Array, Math, Stack, Heap (Priority Queue), Monotonic Stack. The recommended patterns to study are: Stack, Greedy.

What are the interview tips for Car Fleet II?

Always clarify the problem constraints and edge cases. Think about how to reduce the number of comparisons needed. Practice explaining your thought process as you code.

What are common mistakes when solving Car Fleet II?

Not considering the direction of movement correctly. Failing to handle edge cases like cars with the same speed.

#1776

Car Fleet II

Hard

ArrayMathStackHeap (Priority Queue)Monotonic StackStackGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

We can use a stack to keep track of the cars that are still in play, allowing us to efficiently determine when collisions will occur. This reduces the number of checks we need to perform.

⚙️

Algorithm

4 steps

1Step 1: Initialize a stack to keep track of the collision times.
2Step 2: Iterate through the cars from right to left.
3Step 3: For each car, calculate the time to collide with the next car in the stack if it's slower.
4Step 4: If the current car is faster, it will not collide, so we skip it.

solution.py12 lines

1def getCollisionTimes(cars):
2    n = len(cars)
3    answer = [-1] * n
4    stack = []
5    for i in range(n - 1, -1, -1):
6        while stack and cars[i][1] > cars[stack[-1]][1]:
7            stack.pop()
8        if stack:
9            time = (cars[stack[-1]][0] - cars[i][0]) / (cars[i][1] - cars[stack[-1]][1])
10            answer[i] = time
11        stack.append(i)
12    return answer

ℹ

Complexity note: This complexity is achieved because we only iterate through the cars once and use a stack to keep track of potential collisions, leading to linear time complexity.

1Cars with higher speeds cannot collide with cars ahead of them.
2Using a stack allows us to efficiently track potential collisions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.