LeetCode 1478: Allocate Mailboxes Solution

Master LeetCode problem 1478 (Allocate Mailboxes), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1478. Allocate Mailboxes

Hard

Array Math Dynamic Programming Sorting

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming. We will iterate over all possible ranges of houses to consider for placing a mailbox. For each range, we will calculate the cost of placing a mailbox at each possible position within that range. We will then keep track of the minimum total distance for each subproblem.

Sort the houses array in ascending order.
Initialize a 2D array dp[][] where dp[i][k] represents the minimum total distance for the first i houses with k mailboxes.
Calculate the cost of placing a mailbox at each possible position within the range of houses.
Update dp[i][k] by considering all possible partition points for the last mailbox placement.
The final answer will be dp[n][k] where n is the number of houses.

Time Complexity:

The time complexity of this approach is O(n^2 * k), where n is the number of houses.

Space Complexity:

The space complexity of this approach is O(n * k).

Solution Code

class Solution {
    public int minDistance(int[] houses, int k) {
        Arrays.sort(houses);
        int n = houses.length;
        int[][] dp = new int[n + 1][k + 1];
        
        for (int i = 0; i <= n; i++) {
            Arrays.fill(dp[i], Integer.MAX_VALUE);
        }
        
        for (int i = 1; i <= n; i++) {
            dp[i][1] = calculateCost(houses, 0, i - 1);
        }
        
        for (int i = 2; i <= n; i++) {
            for (int j = 2; j <= k; j++) {
                for (int x = 1; x < i; x++) {
                    dp[i][j] = Math.min(dp[i][j], dp[x][j - 1] + calculateCost(houses, x, i - 1));
                }
            }
        }
        
        return dp[n][k];
    }
    
    private int calculateCost(int[] houses, int start, int end) {
        int median = (start + end) / 2;
        int cost = 0;
        
        for (int i = start; i <= end; i++) {
            cost += Math.abs(houses[i] - houses[median]);
        }
        
        return cost;
    }
}

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