# Solving Leetcode Interviews in Seconds with AI: Allocate Mailboxes


	# 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](https://www.chatmagic.app), 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(n<sup>2</sup>k), where n is the number of houses and k is the number of mailboxes. **Storage Complexity:** O(n<sup>2</sup>)

	
	# Code
	```python
	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]
	```