LeetCode #1473 — HARD

Paint House III

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

The Problem

Problem Statement

There is a row of m houses in a small city, each house must be painted with one of the n colors (labeled from 1 to n), some houses that have been painted last summer should not be painted again.

A neighborhood is a maximal group of continuous houses that are painted with the same color.

For example: houses = [1,2,2,3,3,2,1,1] contains 5 neighborhoods [{1}, {2,2}, {3,3}, {2}, {1,1}].

Given an array houses, an m x n matrix cost and an integer target where:

houses[i]: is the color of the house i, and 0 if the house is not painted yet.
cost[i][j]: is the cost of paint the house i with the color j + 1.

Return the minimum cost of painting all the remaining houses in such a way that there are exactly target neighborhoods. If it is not possible, return -1.

Example 1:

Input: houses = [0,0,0,0,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3
Output: 9
Explanation: Paint houses of this way [1,2,2,1,1]
This array contains target = 3 neighborhoods, [{1}, {2,2}, {1,1}].
Cost of paint all houses (1 + 1 + 1 + 1 + 5) = 9.

Example 2:

Input: houses = [0,2,1,2,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3
Output: 11
Explanation: Some houses are already painted, Paint the houses of this way [2,2,1,2,2]
This array contains target = 3 neighborhoods, [{2,2}, {1}, {2,2}]. 
Cost of paint the first and last house (10 + 1) = 11.

Example 3:

Input: houses = [3,1,2,3], cost = [[1,1,1],[1,1,1],[1,1,1],[1,1,1]], m = 4, n = 3, target = 3
Output: -1
Explanation: Houses are already painted with a total of 4 neighborhoods [{3},{1},{2},{3}] different of target = 3.

Constraints:

m == houses.length == cost.length
n == cost[i].length
1 <= m <= 100
1 <= n <= 20
1 <= target <= m
0 <= houses[i] <= n
1 <= cost[i][j] <= 10⁴

Step 01

Brute Force Baseline

Problem summary: There is a row of m houses in a small city, each house must be painted with one of the n colors (labeled from 1 to n), some houses that have been painted last summer should not be painted again. A neighborhood is a maximal group of continuous houses that are painted with the same color. For example: houses = [1,2,2,3,3,2,1,1] contains 5 neighborhoods [{1}, {2,2}, {3,3}, {2}, {1,1}]. Given an array houses, an m x n matrix cost and an integer target where: houses[i]: is the color of the house i, and 0 if the house is not painted yet. cost[i][j]: is the cost of paint the house i with the color j + 1. Return the minimum cost of painting all the remaining houses in such a way that there are exactly target neighborhoods. If it is not possible, return -1.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming

Example 1

[0,0,0,0,0]
[[1,10],[10,1],[10,1],[1,10],[5,1]]
5
2
3

Example 2

[0,2,1,2,0]
[[1,10],[10,1],[10,1],[1,10],[5,1]]
5
2
3

Example 3

[3,1,2,3]
[[1,1,1],[1,1,1],[1,1,1],[1,1,1]]
4
3
3

Core Insight

What unlocks the optimal approach

Use Dynamic programming.
Define dp[i][j][k] as the minimum cost where we have k neighborhoods in the first i houses and the i-th house is painted with the color j.

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 #1473: Paint House III
class Solution {
    public int minCost(int[] houses, int[][] cost, int m, int n, int target) {
        int[][][] f = new int[m][n + 1][target + 1];
        final int inf = 1 << 30;
        for (int[][] g : f) {
            for (int[] e : g) {
                Arrays.fill(e, inf);
            }
        }
        if (houses[0] == 0) {
            for (int j = 1; j <= n; ++j) {
                f[0][j][1] = cost[0][j - 1];
            }
        } else {
            f[0][houses[0]][1] = 0;
        }
        for (int i = 1; i < m; ++i) {
            if (houses[i] == 0) {
                for (int j = 1; j <= n; ++j) {
                    for (int k = 1; k <= Math.min(target, i + 1); ++k) {
                        for (int j0 = 1; j0 <= n; ++j0) {
                            if (j == j0) {
                                f[i][j][k] = Math.min(f[i][j][k], f[i - 1][j][k] + cost[i][j - 1]);
                            } else {
                                f[i][j][k]
                                    = Math.min(f[i][j][k], f[i - 1][j0][k - 1] + cost[i][j - 1]);
                            }
                        }
                    }
                }
            } else {
                int j = houses[i];
                for (int k = 1; k <= Math.min(target, i + 1); ++k) {
                    for (int j0 = 1; j0 <= n; ++j0) {
                        if (j == j0) {
                            f[i][j][k] = Math.min(f[i][j][k], f[i - 1][j][k]);
                        } else {
                            f[i][j][k] = Math.min(f[i][j][k], f[i - 1][j0][k - 1]);
                        }
                    }
                }
            }
        }
        int ans = inf;
        for (int j = 1; j <= n; ++j) {
            ans = Math.min(ans, f[m - 1][j][target]);
        }
        return ans >= inf ? -1 : ans;
    }
}

// Accepted solution for LeetCode #1473: Paint House III
func minCost(houses []int, cost [][]int, m int, n int, target int) int {
	f := make([][][]int, m)
	const inf = 1 << 30
	for i := range f {
		f[i] = make([][]int, n+1)
		for j := range f[i] {
			f[i][j] = make([]int, target+1)
			for k := range f[i][j] {
				f[i][j][k] = inf
			}
		}
	}
	if houses[0] == 0 {
		for j := 1; j <= n; j++ {
			f[0][j][1] = cost[0][j-1]
		}
	} else {
		f[0][houses[0]][1] = 0
	}
	for i := 1; i < m; i++ {
		if houses[i] == 0 {
			for j := 1; j <= n; j++ {
				for k := 1; k <= target && k <= i+1; k++ {
					for j0 := 1; j0 <= n; j0++ {
						if j == j0 {
							f[i][j][k] = min(f[i][j][k], f[i-1][j][k]+cost[i][j-1])
						} else {
							f[i][j][k] = min(f[i][j][k], f[i-1][j0][k-1]+cost[i][j-1])
						}
					}
				}
			}
		} else {
			j := houses[i]
			for k := 1; k <= target && k <= i+1; k++ {
				for j0 := 1; j0 <= n; j0++ {
					if j == j0 {
						f[i][j][k] = min(f[i][j][k], f[i-1][j][k])
					} else {
						f[i][j][k] = min(f[i][j][k], f[i-1][j0][k-1])
					}
				}
			}
		}
	}
	ans := inf
	for j := 1; j <= n; j++ {
		ans = min(ans, f[m-1][j][target])
	}
	if ans == inf {
		return -1
	}
	return ans
}

# Accepted solution for LeetCode #1473: Paint House III
class Solution:
    def minCost(
        self, houses: List[int], cost: List[List[int]], m: int, n: int, target: int
    ) -> int:
        f = [[[inf] * (target + 1) for _ in range(n + 1)] for _ in range(m)]
        if houses[0] == 0:
            for j, c in enumerate(cost[0], 1):
                f[0][j][1] = c
        else:
            f[0][houses[0]][1] = 0
        for i in range(1, m):
            if houses[i] == 0:
                for j in range(1, n + 1):
                    for k in range(1, min(target + 1, i + 2)):
                        for j0 in range(1, n + 1):
                            if j == j0:
                                f[i][j][k] = min(
                                    f[i][j][k], f[i - 1][j][k] + cost[i][j - 1]
                                )
                            else:
                                f[i][j][k] = min(
                                    f[i][j][k], f[i - 1][j0][k - 1] + cost[i][j - 1]
                                )
            else:
                j = houses[i]
                for k in range(1, min(target + 1, i + 2)):
                    for j0 in range(1, n + 1):
                        if j == j0:
                            f[i][j][k] = min(f[i][j][k], f[i - 1][j][k])
                        else:
                            f[i][j][k] = min(f[i][j][k], f[i - 1][j0][k - 1])

        ans = min(f[-1][j][target] for j in range(1, n + 1))
        return -1 if ans >= inf else ans

// Accepted solution for LeetCode #1473: Paint House III
struct Solution;

use std::collections::HashMap;

impl Solution {
    fn min_cost(houses: Vec<i32>, cost: Vec<Vec<i32>>, m: i32, n: i32, target: i32) -> i32 {
        let m = m as usize;
        let target = target as usize;
        let mut memo: HashMap<(usize, i32, usize), Option<i32>> = HashMap::new();
        if let Some(res) = (1..=n)
            .filter_map(|i| Self::dp(m, i, target, &mut memo, &houses, &cost, n))
            .min()
        {
            res
        } else {
            -1
        }
    }

    fn dp(
        size: usize,
        color: i32,
        target: usize,
        memo: &mut HashMap<(usize, i32, usize), Option<i32>>,
        houses: &[i32],
        cost: &[Vec<i32>],
        n: i32,
    ) -> Option<i32> {
        if target < 1 {
            return None;
        }
        if size < target {
            return None;
        }
        if houses[size - 1] != 0 && color != houses[size - 1] {
            return None;
        }
        if size == 1 {
            if target == 1 {
                if houses[0] == 0 {
                    Some(cost[0][(color - 1) as usize])
                } else {
                    if houses[0] == color {
                        Some(0)
                    } else {
                        None
                    }
                }
            } else {
                None
            }
        } else {
            if let Some(&res) = memo.get(&(size, color, target)) {
                return res;
            }

            let mut min = std::i32::MAX;
            let x = if houses[size - 1] == 0 {
                cost[size - 1][(color - 1) as usize]
            } else {
                0
            };

            for i in 1..=n {
                if let Some(y) = Self::dp(
                    size - 1,
                    i,
                    if i == color { target } else { target - 1 },
                    memo,
                    houses,
                    cost,
                    n,
                ) {
                    min = min.min(x + y);
                }
            }
            let res = if min == std::i32::MAX {
                None
            } else {
                Some(min)
            };

            memo.insert((size, color, target), res);
            res
        }
    }
}

#[test]
fn test() {
    let houses = vec![0, 0, 0, 0, 0];
    let cost = vec_vec_i32![[1, 10], [10, 1], [10, 1], [1, 10], [5, 1]];
    let m = 5;
    let n = 2;
    let target = 3;
    let res = 9;
    assert_eq!(Solution::min_cost(houses, cost, m, n, target), res);
    let houses = vec![0, 2, 1, 2, 0];
    let cost = vec_vec_i32![[1, 10], [10, 1], [10, 1], [1, 10], [5, 1]];
    let m = 5;
    let n = 2;
    let target = 3;
    let res = 11;
    assert_eq!(Solution::min_cost(houses, cost, m, n, target), res);
}

// Accepted solution for LeetCode #1473: Paint House III
function minCost(houses: number[], cost: number[][], m: number, n: number, target: number): number {
    const inf = 1 << 30;
    const f: number[][][] = new Array(m)
        .fill(0)
        .map(() => new Array(n + 1).fill(0).map(() => new Array(target + 1).fill(inf)));
    if (houses[0] === 0) {
        for (let j = 1; j <= n; ++j) {
            f[0][j][1] = cost[0][j - 1];
        }
    } else {
        f[0][houses[0]][1] = 0;
    }
    for (let i = 1; i < m; ++i) {
        if (houses[i] === 0) {
            for (let j = 1; j <= n; ++j) {
                for (let k = 1; k <= Math.min(target, i + 1); ++k) {
                    for (let j0 = 1; j0 <= n; ++j0) {
                        if (j0 === j) {
                            f[i][j][k] = Math.min(f[i][j][k], f[i - 1][j][k] + cost[i][j - 1]);
                        } else {
                            f[i][j][k] = Math.min(f[i][j][k], f[i - 1][j0][k - 1] + cost[i][j - 1]);
                        }
                    }
                }
            }
        } else {
            const j = houses[i];
            for (let k = 1; k <= Math.min(target, i + 1); ++k) {
                for (let j0 = 1; j0 <= n; ++j0) {
                    if (j0 === j) {
                        f[i][j][k] = Math.min(f[i][j][k], f[i - 1][j][k]);
                    } else {
                        f[i][j][k] = Math.min(f[i][j][k], f[i - 1][j0][k - 1]);
                    }
                }
            }
        }
    }
    let ans = inf;
    for (let j = 1; j <= n; ++j) {
        ans = Math.min(ans, f[m - 1][j][target]);
    }
    return ans >= inf ? -1 : ans;
}

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(m × n^2 × \textittarget)

Space

O(m × n × \textittarget)

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.

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.