Allocate Mailboxes - Leetcode Solution

Problem Statement:

You are given an integer array houses and an integer k. You must allocate k mailboxes in these houses in such a way that the maximum distance between any house and its nearest mailbox is minimized.

Return the minimum possible maximum distance between a house and its assigned mailbox.

Solution:

The main objective of the problem is to minimize the maximum distance between any house and its assigned mailbox. We can use binary search to solve this problem.

For each value of mid (minimum possible maximum distance), we can allocate the mailboxes based on the greedy approach.

We start by placing a mailbox at the first house and keep placing mailboxes at the farthest possible house until the maximum distance between any house and its nearest mailbox is greater than or equal to mid. Then we place the next mailbox at the farthest possible house from the last mailbox placed until all the mailboxes are allocated.

We can use dynamic programming to compute the cost matrix, where cost[i][j] represents the cost of placing a mailbox between house i and house j. The cost is the sum of distances between the houses and the mailbox.

The dp[i][j] represents the minimum possible maximum distance for allocating i mailboxes for houses 1 to j.

dp[i][j] = min(dp[i][j], dp[i-1][k] + cost[k+1][j]) for k from i-1 to j-1.

Finally, we return the value of dp[k][n], where n is the number of houses.

Time Complexity:

The time complexity of this solution is O(k*n^3), where n is the number of houses.

Code:

Here is the code for the solution:

def allocateMailboxes(houses: List[int], k: int) -> int: n = len(houses) houses.sort() cost = [[0 for _ in range(n)] for _ in range(n)] for i in range(n): for j in range(i, n): mid = (i + j) // 2 for x in range(i, j + 1): cost[i][j] += abs(houses[x] - houses[mid]) dp = [[float('inf') for _ in range(n)] for _ in range(k + 1)] for i in range(1, k + 1): for j in range(i - 1, n): if i == 1: dp[i][j] = cost[0][j] else: for k in range(i - 2, j): dp[i][j] = min(dp[i][j], dp[i - 1][k] + cost[k + 1][j]) return dp[k][n - 1]

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