Leetcode #568: Maximum Vacation Days
In this guide, we solve Leetcode #568 Maximum Vacation Days 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
LeetCode wants to give one of its best employees the option to travel among n cities to collect algorithm problems. But all work and no play makes Jack a dull boy, you could take vacations in some particular cities and weeks.
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Array, Dynamic Programming, Matrix
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: flights = [[0,1,1],[1,0,1],[1,1,0]], days = [[1,3,1],[6,0,3],[3,3,3]]
Output: 12
Explanation:
One of the best strategies is:
1st week : fly from city 0 to city 1 on Monday, and play 6 days and work 1 day.
(Although you start at city 0, we could also fly to and start at other cities since it is Monday.)
2nd week : fly from city 1 to city 2 on Monday, and play 3 days and work 4 days.
3rd week : stay at city 2, and play 3 days and work 4 days.
Ans = 6 + 3 + 3 = 12.
Python Solution
class Solution:
def maxVacationDays(self, flights: List[List[int]], days: List[List[int]]) -> int:
n = len(flights)
K = len(days[0])
f = [[-inf] * n for _ in range(K + 1)]
f[0][0] = 0
for k in range(1, K + 1):
for j in range(n):
f[k][j] = f[k - 1][j]
for i in range(n):
if flights[i][j]:
f[k][j] = max(f[k][j], f[k - 1][i])
f[k][j] += days[j][k - 1]
return max(f[-1][j] for j in range(n))
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
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.