Introduction

In this blog post, we will explore how to solve the LeetCode problem "1478" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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. The test cases are generated so that the answer fits in a 32-bit integer. Example 1: 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 Example 2: Input: houses = [2,3,5,12,18], k = 2 Output: 9 Explanation: Allocate mailboxes in position 3 and 14. Minimum total distance from each houses to nearest mailboxes is |2-3| + |3-3| + |5-3| + |12-14| + |18-14| = 9. Constraints: 1 <= k <= houses.length <= 100 1 <= houses[i] <= 104 All the integers of houses are unique.

Explanation

Here's the breakdown of the optimal solution:

• Sort the Houses: Sorting allows us to efficiently calculate the optimal mailbox location for a range of houses (the median).
• Dynamic Programming: Use DP to store and reuse solutions for subproblems. dp[i][j] stores the minimum total distance to place j mailboxes for the first i houses.

• Precompute Cost: Calculate the optimal cost of placing one mailbox within a range of houses beforehand to avoid redundant calculations.

• Runtime Complexity: O(n²k), where n is the number of houses and k is the number of mailboxes. Storage Complexity: O(n²)

Code

    def minDistance(houses, k):
    houses.sort()
    n = len(houses)
    cost = [[0] * n for _ in range(n)]

    for i in range(n):
        for j in range(i, n):
            median = houses[(i + j) // 2]
            for l in range(i, j + 1):
                cost[i][j] += abs(houses[l] - median)

    dp = [[float('inf')] * (k + 1) for _ in range(n + 1)]
    dp[0][0] = 0

    for i in range(1, n + 1):
        for j in range(1, min(i, k) + 1):
            for l in range(1, i + 1):
                dp[i][j] = min(dp[i][j], dp[l - 1][j - 1] + cost[l - 1][i - 1])

    return dp[n][k]