How do you solve Allocate Mailboxes on LeetCode?

Allocate Mailboxes (LeetCode #1478) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n² * k) time complexity and O(n * k) space complexity. Key insight: The optimal mailbox positions often align with the median of the houses.

What is the brute force solution for Allocate Mailboxes?

The brute force approach for Allocate Mailboxes is Brute Force, with O(n² * C(n, k)) time complexity and O(1) space complexity. The brute-force approach involves generating all possible combinations of k mailboxes and calculating the total distance for each combination. This method is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n² * k).

What algorithm or data structure is used to solve Allocate Mailboxes?

Allocate Mailboxes uses the following concepts: Array, Math, Dynamic Programming, Sorting. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Allocate Mailboxes?

Always explain your thought process clearly when discussing your approach. Consider edge cases, such as when k equals the number of houses. Practice dynamic programming problems to become familiar with breaking down complex problems.

What are common mistakes when solving Allocate Mailboxes?

Not considering the impact of sorting the houses before processing. Failing to correctly implement the distance calculation for ranges of houses.

#1478

Allocate Mailboxes

Hard

ArrayMathDynamic ProgrammingSortingDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n² * C(n, k))	O(n² * k)
Space	O(1)	O(n * k)

💡

Intuition

Time O(n² * k)Space O(n * k)

The optimal solution uses dynamic programming to efficiently compute the minimum distance by breaking the problem into smaller subproblems. It builds on the idea that the best position for a mailbox is often near the median of the houses.

⚙️

Algorithm

4 steps

1Step 1: Sort the houses array.
2Step 2: Create a DP table where dp[i][j] represents the minimum distance for the first i houses with j mailboxes.
3Step 3: For each house, calculate the minimum distance for all possible mailbox positions and update the DP table.
4Step 4: Return dp[n][k] where n is the number of houses.

solution.py20 lines

1# Full working Python code
2import sys
3
4def minDistance(houses, k):
5    houses.sort()
6    n = len(houses)
7    dp = [[0] * (k + 1) for _ in range(n + 1)]
8    for i in range(1, n + 1):
9        for j in range(1, k + 1):
10            dp[i][j] = sys.maxsize
11            for p in range(i):
12                distance = calculateDistance(houses, p, i - 1)
13                dp[i][j] = min(dp[i][j], dp[p][j - 1] + distance)
14    return dp[n][k]
15
16
17def calculateDistance(houses, left, right):
18    mid = (left + right) // 2
19    median = houses[mid]
20    return sum(abs(house - median) for house in houses[left:right + 1])

ℹ

Complexity note: The time complexity is significantly improved because we avoid checking all combinations directly and instead use dynamic programming to build solutions incrementally.

1The optimal mailbox positions often align with the median of the houses.
2Dynamic programming allows us to break down the problem into manageable pieces, reducing the need for exhaustive combinations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.