LeetCode #3429 — MEDIUM

Paint House IV

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an even integer n representing the number of houses arranged in a straight line, and a 2D array cost of size n x 3, where cost[i][j] represents the cost of painting house i with color j + 1.

The houses will look beautiful if they satisfy the following conditions:

No two adjacent houses are painted the same color.
Houses equidistant from the ends of the row are not painted the same color. For example, if n = 6, houses at positions (0, 5), (1, 4), and (2, 3) are considered equidistant.

Return the minimum cost to paint the houses such that they look beautiful.

Example 1:

Input: n = 4, cost = [[3,5,7],[6,2,9],[4,8,1],[7,3,5]]

Output: 9

Explanation:

The optimal painting sequence is [1, 2, 3, 2] with corresponding costs [3, 2, 1, 3]. This satisfies the following conditions:

No adjacent houses have the same color.
Houses at positions 0 and 3 (equidistant from the ends) are not painted the same color (1 != 2).
Houses at positions 1 and 2 (equidistant from the ends) are not painted the same color (2 != 3).

The minimum cost to paint the houses so that they look beautiful is 3 + 2 + 1 + 3 = 9.

Example 2:

Input: n = 6, cost = [[2,4,6],[5,3,8],[7,1,9],[4,6,2],[3,5,7],[8,2,4]]

Output: 18

Explanation:

The optimal painting sequence is [1, 3, 2, 3, 1, 2] with corresponding costs [2, 8, 1, 2, 3, 2]. This satisfies the following conditions:

No adjacent houses have the same color.
Houses at positions 0 and 5 (equidistant from the ends) are not painted the same color (1 != 2).
Houses at positions 1 and 4 (equidistant from the ends) are not painted the same color (3 != 1).
Houses at positions 2 and 3 (equidistant from the ends) are not painted the same color (2 != 3).

The minimum cost to paint the houses so that they look beautiful is 2 + 8 + 1 + 2 + 3 + 2 = 18.

Constraints:

2 <= n <= 10⁵
n is even.
cost.length == n
cost[i].length == 3
0 <= cost[i][j] <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given an even integer n representing the number of houses arranged in a straight line, and a 2D array cost of size n x 3, where cost[i][j] represents the cost of painting house i with color j + 1. The houses will look beautiful if they satisfy the following conditions: No two adjacent houses are painted the same color. Houses equidistant from the ends of the row are not painted the same color. For example, if n = 6, houses at positions (0, 5), (1, 4), and (2, 3) are considered equidistant. Return the minimum cost to paint the houses such that they look beautiful.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming

Example 1

4
[[3,5,7],[6,2,9],[4,8,1],[7,3,5]]

Example 2

6
[[2,4,6],[5,3,8],[7,1,9],[4,6,2],[3,5,7],[8,2,4]]

Core Insight

What unlocks the optimal approach

Use dynamic programming to calculate the minimum cost while ensuring that the adjacency and equidistant constraints are satisfied.
Try all 9 combinations of colors for equidistant pairs to get the minimum cost.

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

Upper-end input sizes

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 #3429: Paint House IV
class Solution {
  public long minCost(int n, int[][] costs) {
    final int INVALID_COLOR = 3;
    Long[][][] mem = new Long[n / 2][4][4];
    return minCost(costs, 0, INVALID_COLOR, INVALID_COLOR, mem);
  }

  private long minCost(int[][] costs, int i, int prevLeftColor, int prevRightColor,
                       Long[][][] mem) {
    if (i == costs.length / 2)
      return 0;
    if (mem[i][prevLeftColor][prevRightColor] != null)
      return mem[i][prevLeftColor][prevRightColor];

    long res = Long.MAX_VALUE;

    for (final int leftColor : getValidColors(prevLeftColor))
      for (final int rightColor : getValidColors(prevRightColor)) {
        if (leftColor == rightColor)
          continue;
        final long leftCost = costs[i][leftColor];
        final long rightCost = costs[costs.length - 1 - i][rightColor];
        res =
            Math.min(res, leftCost + rightCost + minCost(costs, i + 1, leftColor, rightColor, mem));
      }

    return mem[i][prevLeftColor][prevRightColor] = res;
  }

  private List<Integer> getValidColors(int prevColor) {
    List<Integer> validColors = new ArrayList<>();
    for (int color = 0; color < 3; ++color)
      if (color != prevColor)
        validColors.add(color);
    return validColors;
  }
}

// Accepted solution for LeetCode #3429: Paint House IV
package main

import (
	"math"
	"slices"
)

func minCost(n int, cost [][]int) int64 {
	f := make([][3][3]int, n/2+1)
	for i, row := range cost[:n/2] {
		row2 := cost[n-1-i]
		for preJ := range 3 {
			for preK := range 3 {
				res := math.MaxInt
				for j, c1 := range row {
					if j == preJ {
						continue
					}
					for k, c2 := range row2 {
						if k != preK && k != j {
							res = min(res, f[i][j][k]+c1+c2)
						}
					}
				}
				f[i+1][preJ][preK] = res
			}
		}
	}

	// 枚举所有初始颜色，取最小值
	res := math.MaxInt
	for _, row := range f[n/2] {
		res = min(res, slices.Min(row[:]))
	}
	return int64(res)
}

func minCost2(n int, cost [][]int) int64 {
	memo := make([][4][4]int, n/2)
	for i := range memo {
		for j := range memo[i] {
			for k := range memo[i][j] {
				memo[i][j][k] = -1
			}
		}
	}
	var dfs func(int, int, int) int
	dfs = func(i, preJ, preK int) (res int) {
		if i < 0 {
			return
		}
		p := &memo[i][preJ][preK]
		if *p != -1 {
			return *p
		}
		res = math.MaxInt
		for j, c1 := range cost[i] {
			if j == preJ {
				continue
			}
			for k, c2 := range cost[n-1-i] {
				if k != preK && k != j {
					res = min(res, dfs(i-1, j, k)+c1+c2)
				}
			}
		}
		*p = res
		return
	}
	return int64(dfs(n/2-1, 3, 3))
}

# Accepted solution for LeetCode #3429: Paint House IV
class Solution:
  def minCost(self, n: int, costs: list[list[int]]) -> int:
    INVALID_COLOR = 3

    def getValidColors(prevColor: int) -> list[int]:
      return [color for color in range(3) if color != prevColor]

    @functools.lru_cache(None)
    def minCost(i: int, prevLeftColor: int, prevRightColor: int) -> int:
      if i == len(costs) // 2:
        return 0
      res = math.inf
      for leftColor in getValidColors(prevLeftColor):
        for rightColor in getValidColors(prevRightColor):
          if leftColor == rightColor:
            continue
          leftCost = costs[i][leftColor]
          rightCost = costs[-1 - i][rightColor]
          res = min(res, leftCost + rightCost +
                    minCost(i + 1, leftColor, rightColor))
      return res

    return minCost(0, INVALID_COLOR, INVALID_COLOR)

// Accepted solution for LeetCode #3429: Paint House IV
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3429: Paint House IV
// class Solution {
//   public long minCost(int n, int[][] costs) {
//     final int INVALID_COLOR = 3;
//     Long[][][] mem = new Long[n / 2][4][4];
//     return minCost(costs, 0, INVALID_COLOR, INVALID_COLOR, mem);
//   }
// 
//   private long minCost(int[][] costs, int i, int prevLeftColor, int prevRightColor,
//                        Long[][][] mem) {
//     if (i == costs.length / 2)
//       return 0;
//     if (mem[i][prevLeftColor][prevRightColor] != null)
//       return mem[i][prevLeftColor][prevRightColor];
// 
//     long res = Long.MAX_VALUE;
// 
//     for (final int leftColor : getValidColors(prevLeftColor))
//       for (final int rightColor : getValidColors(prevRightColor)) {
//         if (leftColor == rightColor)
//           continue;
//         final long leftCost = costs[i][leftColor];
//         final long rightCost = costs[costs.length - 1 - i][rightColor];
//         res =
//             Math.min(res, leftCost + rightCost + minCost(costs, i + 1, leftColor, rightColor, mem));
//       }
// 
//     return mem[i][prevLeftColor][prevRightColor] = res;
//   }
// 
//   private List<Integer> getValidColors(int prevColor) {
//     List<Integer> validColors = new ArrayList<>();
//     for (int color = 0; color < 3; ++color)
//       if (color != prevColor)
//         validColors.add(color);
//     return validColors;
//   }
// }

// Accepted solution for LeetCode #3429: Paint House IV
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #3429: Paint House IV
// class Solution {
//   public long minCost(int n, int[][] costs) {
//     final int INVALID_COLOR = 3;
//     Long[][][] mem = new Long[n / 2][4][4];
//     return minCost(costs, 0, INVALID_COLOR, INVALID_COLOR, mem);
//   }
// 
//   private long minCost(int[][] costs, int i, int prevLeftColor, int prevRightColor,
//                        Long[][][] mem) {
//     if (i == costs.length / 2)
//       return 0;
//     if (mem[i][prevLeftColor][prevRightColor] != null)
//       return mem[i][prevLeftColor][prevRightColor];
// 
//     long res = Long.MAX_VALUE;
// 
//     for (final int leftColor : getValidColors(prevLeftColor))
//       for (final int rightColor : getValidColors(prevRightColor)) {
//         if (leftColor == rightColor)
//           continue;
//         final long leftCost = costs[i][leftColor];
//         final long rightCost = costs[costs.length - 1 - i][rightColor];
//         res =
//             Math.min(res, leftCost + rightCost + minCost(costs, i + 1, leftColor, rightColor, mem));
//       }
// 
//     return mem[i][prevLeftColor][prevRightColor] = res;
//   }
// 
//   private List<Integer> getValidColors(int prevColor) {
//     List<Integer> validColors = new ArrayList<>();
//     for (int color = 0; color < 3; ++color)
//       if (color != prevColor)
//         validColors.add(color);
//     return validColors;
//   }
// }

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 × m)

Space

O(n × m)

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.