Minimum Skips to Arrive at Meeting On Time — LeetCode 1883 Python Solution

The problem

You are given an integer hoursBefore, the number of hours you have to travel to your meeting. To arrive at your meeting, you have to travel through n roads.

Example

Input

dist = [1,3,2], speed = 4, hoursBefore = 2

Output

1

Explanation

Without skipping any rests, you will arrive in (1/4 + 3/4) + (3/4 + 1/4) + (2/4) = 2.5 hours.

Python solution

class Solution:
    def minSkips(self, dist: List[int], speed: int, hoursBefore: int) -> int:
        n = len(dist)
        f = [[inf] * (n + 1) for _ in range(n + 1)]
        f[0][0] = 0
        eps = 1e-8
        for i, x in enumerate(dist, 1):
            for j in range(i + 1):
                if j < i:
                    f[i][j] = min(f[i][j], ceil(f[i - 1][j] + x / speed - eps))
                if j:
                    f[i][j] = min(f[i][j], f[i - 1][j - 1] + x / speed)
        for j in range(n + 1):
            if f[n][j] <= hoursBefore + eps:
                return j
        return -1

Complexity

Pattern: Dynamic Programming

Define a state, write the transition, and stop recomputing the same subproblem. LeetCode 1883. Minimum Skips to Arrive at Meeting On Time is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Dynamic Programming.

The dynamic programming guide has the Python template for the pattern and the 481 LeetCode problems that use it.

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