Solution Explanation:

This problem asks to find the minimum total distance between houses and their nearest mailboxes, given a number of mailboxes k. The solution uses dynamic programming.

Core Idea:

The solution leverages dynamic programming to build up the minimum distance for placing j mailboxes among the first i houses. The state transitions consider all possible positions for the j-th mailbox.

1. Preprocessing:

• The houses array is sorted to facilitate calculations.
• A 2D array g is created to store the sum of distances for a contiguous subarray. g[i][j] stores the total distance if mailboxes are placed between houses i and j (inclusive) at the median position. This is calculated efficiently using a cumulative approach.

2. Dynamic Programming:

• A 2D array f is created, where f[i][j] represents the minimum total distance to place j mailboxes among the first i + 1 houses.
• The base case is f[i][1] = g[0][i] for each i, meaning placing one mailbox optimally among the first i + 1 houses.
• For j > 1, the solution iterates through all possible positions (p) for the (j-1)th mailbox. The minimum total distance is the minimum of f[p][j-1] + g[p+1][i], representing the minimum cost to place j-1 mailboxes among the first p + 1 houses plus the cost to place the jth mailbox among houses p+1 to i.

3. Result:

• The final result is f[n-1][k], representing the minimum total distance when placing k mailboxes among all n houses.

Time Complexity Analysis:

The dominant part is the nested loops in the DP calculation. The outer loops iterate O(n*k) times, while the inner loop iterates at most O(n) times. This results in a time complexity of O(n²k).

Space Complexity Analysis:

The solution uses two n x n matrices (g) and n x k matrix (f). Thus the space complexity is O(n² + nk). Since k is usually much smaller than n, this can be approximated as O(n²).

Code Explanation (Python):

class Solution:
    def minDistance(self, houses: List[int], k: int) -> int:
        houses.sort()
        n = len(houses)
        g = [[0] * n for _ in range(n)] # Precompute cumulative distances
        for i in range(n - 2, -1, -1):
            for j in range(i + 1, n):
                g[i][j] = g[i + 1][j - 1] + houses[j] - houses[i]  #Efficiently calculate distances
        f = [[float('inf')] * (k + 1) for _ in range(n)] #DP Table
        for i in range(n):
            f[i][1] = g[0][i] #Base case: 1 mailbox
            for j in range(2, min(k + 1, i + 2)): #Iterate through mailbox numbers and house positions
                for p in range(i): #Iterate through possible placements of previous mailbox
                    f[i][j] = min(f[i][j], f[p][j - 1] + g[p + 1][i]) #DP Transition
        return f[-1][k] #Final Result
 

The code in other languages (Java, C++, Go) follows the same logic and algorithmic structure, differing only in syntax and data structure implementations. The core dynamic programming approach remains consistent across all versions.