Leetcode #1883: Minimum Skips to Arrive at Meeting On Time
In this guide, we solve Leetcode #1883 Minimum Skips to Arrive at Meeting On Time in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
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.
You can skip the first rest to arrive in ((1/4 + 0) + (3/4 + 0)) + (2/4) = 1.5 hours.
Note that the second rest is shortened because you finish traveling the second road at an integer hour due to skipping the first rest.
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
The time complexity is , and the space complexity is , where is the number of roads. The space complexity is , where is the number of roads.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.