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Allocate Mailboxes

Difficulty: Hard

Problem Description

Given the array houses where houses[i] is the location of the i-th 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.

Key Insights

  • The problem requires minimizing the total distance from each house to its nearest mailbox.
  • A dynamic programming approach is suitable for this problem, where we can define subproblems based on the number of houses and mailboxes.
  • Sorting the houses helps in better calculating the distances as the mailbox allocation will follow the sorted order.
  • The problem can be broken down into choosing optimal positions for mailboxes to reduce the overall distance.

Space and Time Complexity

Time Complexity: O(n^2 * k), where n is the number of houses. The algorithm iterates through the houses for each mailbox allocation, leading to a quadratic relationship with respect to the number of houses multiplied by the number of mailboxes. Space Complexity: O(n * k), for storing the DP table that records the minimum distances for different combinations of houses and mailboxes.

Solution

The solution uses dynamic programming to find the minimum total distance. We define a DP table where dp[i][j] represents the minimum distance for the first i houses with j mailboxes. The key steps involve iterating through the houses and mailboxes and calculating the optimal distances based on previously computed values.

Code Solutions

def minDistance(houses, k):
    houses.sort()
    n = len(houses)
    
    # dp[i][j] will hold the minimum distance for first i houses with j mailboxes
    dp = [[float('inf')] * (k + 1) for _ in range(n + 1)]
    
    # Base case: 0 houses, 0 mailboxes has 0 distance
    dp[0][0] = 0
    
    # Precompute distances for all segments
    cost = [[0] * n for _ in range(n)]
    for i in range(n):
        for j in range(i, n):
            # Cost of placing a mailbox for houses[i] to houses[j]
            mid = (i + j) // 2
            for h in range(i, j + 1):
                cost[i][j] += abs(houses[h] - houses[mid])
    
    # Fill the DP table
    for j in range(1, k + 1):
        for i in range(1, n + 1):
            for p in range(i):
                dp[i][j] = min(dp[i][j], dp[p][j - 1] + cost[p][i - 1])
    
    return dp[n][k]
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