Leetcode #1478: Allocate Mailboxes
In this guide, we solve Leetcode #1478 Allocate Mailboxes 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
Given the array houses where houses[i] is the location of the ith house along a street and an integer k, allocate k mailboxes in the street. Return the minimum total distance between each house and its nearest mailbox.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Math, Dynamic Programming, Sorting
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: houses = [1,4,8,10,20], k = 3
Output: 5
Explanation: Allocate mailboxes in position 3, 9 and 20.
Minimum total distance from each houses to nearest mailboxes is |3-1| + |4-3| + |9-8| + |10-9| + |20-20| = 5
Python Solution
class Solution:
def minDistance(self, houses: List[int], k: int) -> int:
houses.sort()
n = len(houses)
g = [[0] * n for _ in range(n)]
for i in range(n - 2, -1, -1):
for j in range(i + 1, n):
g[i][j] = g[i + 1][j - 1] + houses[j] - houses[i]
f = [[inf] * (k + 1) for _ in range(n)]
for i in range(n):
f[i][1] = g[0][i]
for j in range(2, min(k + 1, i + 2)):
for p in range(i):
f[i][j] = min(f[i][j], f[p][j - 1] + g[p + 1][i])
return f[-1][k]
Complexity
The time complexity is , and the space complexity is , where is the number of houses. The space complexity is , where is the number of houses.
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.