LeetCode 1478: Allocate Mailboxes
Problem Description
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).
Solutions
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;
}
}Loading editor...