LeetCode #1478 — HARD

Allocate Mailboxes

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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.

The test cases are generated so that the answer fits in a 32-bit integer.

Example 1:

Input: houses = [1,4,8,10,20], k = 3
Output: 5
Explanation: Allocate mailboxes in position 3, 9 and 20.
Minimum total distance from each houses to nearest mailboxes is |3-1| + |4-3| + |9-8| + |10-9| + |20-20| = 5

Example 2:

Input: houses = [2,3,5,12,18], k = 2
Output: 9
Explanation: Allocate mailboxes in position 3 and 14.
Minimum total distance from each houses to nearest mailboxes is |2-3| + |3-3| + |5-3| + |12-14| + |18-14| = 9.

Constraints:

1 <= k <= houses.length <= 100
1 <= houses[i] <= 10⁴
All the integers of houses are unique.

Step 01

Brute Force Baseline

Problem summary: Given the array houses where houses[i] is the location of the ith 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. The test cases are generated so that the answer fits in a 32-bit integer.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Dynamic Programming

Example 1

[1,4,8,10,20]
3

Example 2

[2,3,5,12,18]
2

Step 02

Core Insight

What unlocks the optimal approach

If k =1, the minimum distance is obtained allocating the mailbox in the median of the array houses.
Generalize this idea, using dynamic programming allocating k mailboxes.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1478: Allocate Mailboxes
class Solution {
    public int minDistance(int[] houses, int k) {
        Arrays.sort(houses);
        int n = houses.length;
        int[][] g = new int[n][n];
        for (int i = n - 2; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                g[i][j] = g[i + 1][j - 1] + houses[j] - houses[i];
            }
        }
        int[][] f = new int[n][k + 1];
        final int inf = 1 << 30;
        for (int[] e : f) {
            Arrays.fill(e, inf);
        }
        for (int i = 0; i < n; ++i) {
            f[i][1] = g[0][i];
            for (int j = 2; j <= k && j <= i + 1; ++j) {
                for (int p = 0; p < i; ++p) {
                    f[i][j] = Math.min(f[i][j], f[p][j - 1] + g[p + 1][i]);
                }
            }
        }
        return f[n - 1][k];
    }
}

// Accepted solution for LeetCode #1478: Allocate Mailboxes
func minDistance(houses []int, k int) int {
	sort.Ints(houses)
	n := len(houses)
	g := make([][]int, n)
	f := make([][]int, n)
	const inf = 1 << 30
	for i := range g {
		g[i] = make([]int, n)
		f[i] = make([]int, k+1)
		for j := range f[i] {
			f[i][j] = inf
		}
	}
	for i := n - 2; i >= 0; i-- {
		for j := i + 1; j < n; j++ {
			g[i][j] = g[i+1][j-1] + houses[j] - houses[i]
		}
	}
	for i := 0; i < n; i++ {
		f[i][1] = g[0][i]
		for j := 2; j <= k && j <= i+1; j++ {
			for p := 0; p < i; p++ {
				f[i][j] = min(f[i][j], f[p][j-1]+g[p+1][i])
			}
		}
	}
	return f[n-1][k]
}

# Accepted solution for LeetCode #1478: Allocate Mailboxes
class Solution:
    def minDistance(self, houses: List[int], k: int) -> int:
        houses.sort()
        n = len(houses)
        g = [[0] * n for _ in range(n)]
        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]
        f = [[inf] * (k + 1) for _ in range(n)]
        for i in range(n):
            f[i][1] = g[0][i]
            for j in range(2, min(k + 1, i + 2)):
                for p in range(i):
                    f[i][j] = min(f[i][j], f[p][j - 1] + g[p + 1][i])
        return f[-1][k]

// Accepted solution for LeetCode #1478: Allocate Mailboxes
struct Solution;
use std::collections::HashMap;

impl Solution {
    fn min_distance(mut houses: Vec<i32>, k: i32) -> i32 {
        let n = houses.len();
        let k = k as usize;
        houses.sort_unstable();
        let mut memo: HashMap<(usize, usize), i32> = HashMap::new();
        Self::dp(0, k, &mut memo, &houses, n)
    }

    fn dp(
        start: usize,
        k: usize,
        memo: &mut HashMap<(usize, usize), i32>,
        houses: &[i32],
        n: usize,
    ) -> i32 {
        if let Some(&res) = memo.get(&(start, k)) {
            return res;
        }
        let res = if k == 1 {
            Self::distance(start, n, houses)
        } else {
            let mut min = std::i32::MAX;
            for i in start..=n - k {
                let end = i + 1;
                let dist = Self::distance(start, end, houses);
                min = min.min(dist + Self::dp(end, k - 1, memo, houses, n));
            }
            min
        };
        memo.insert((start, k), res);
        res
    }

    fn distance(start: usize, end: usize, houses: &[i32]) -> i32 {
        let mut sum = 0;
        let median = (houses[(start + end - 1) / 2] + houses[(start + end) / 2]) / 2;
        for i in start..end {
            sum += (houses[i] - median).abs();
        }
        sum
    }
}

#[test]
fn test() {
    let houses = vec![1, 4, 8, 10, 20];
    let k = 3;
    let res = 5;
    assert_eq!(Solution::min_distance(houses, k), res);
    let houses = vec![2, 3, 5, 12, 18];
    let k = 2;
    let res = 9;
    assert_eq!(Solution::min_distance(houses, k), res);
    let houses = vec![7, 4, 6, 1];
    let k = 1;
    let res = 8;
    assert_eq!(Solution::min_distance(houses, k), res);
    let houses = vec![3, 6, 14, 10];
    let k = 4;
    let res = 0;
    assert_eq!(Solution::min_distance(houses, k), res);
}

// Accepted solution for LeetCode #1478: Allocate Mailboxes
function minDistance(houses: number[], k: number): number {
    houses.sort((a, b) => a - b);
    const n = houses.length;
    const g: number[][] = Array.from({ length: n }, () => Array(n).fill(0));

    for (let i = n - 2; i >= 0; i--) {
        for (let j = i + 1; j < n; j++) {
            g[i][j] = g[i + 1][j - 1] + houses[j] - houses[i];
        }
    }

    const inf = Number.POSITIVE_INFINITY;
    const f: number[][] = Array.from({ length: n }, () => Array(k + 1).fill(inf));

    for (let i = 0; i < n; i++) {
        f[i][1] = g[0][i];
    }

    for (let j = 2; j <= k; j++) {
        for (let i = j - 1; i < n; i++) {
            for (let p = i - 1; p >= 0; p--) {
                f[i][j] = Math.min(f[i][j], f[p][j - 1] + g[p + 1][i]);
            }
        }
    }

    return f[n - 1][k];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2 × k)

Space

O(n^2)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.