LeetCode #2742 — HARD

Painting the Walls

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

The Problem

Problem Statement

You are given two 0-indexed integer arrays, cost and time, of size n representing the costs and the time taken to paint n different walls respectively. There are two painters available:

A paid painter that paints the i^th wall in time[i] units of time and takes cost[i] units of money.
A free painter that paints any wall in 1 unit of time at a cost of 0. But the free painter can only be used if the paid painter is already occupied.

Return the minimum amount of money required to paint the n walls.

Example 1:

Input: cost = [1,2,3,2], time = [1,2,3,2]
Output: 3
Explanation: The walls at index 0 and 1 will be painted by the paid painter, and it will take 3 units of time; meanwhile, the free painter will paint the walls at index 2 and 3, free of cost in 2 units of time. Thus, the total cost is 1 + 2 = 3.

Example 2:

Input: cost = [2,3,4,2], time = [1,1,1,1]
Output: 4
Explanation: The walls at index 0 and 3 will be painted by the paid painter, and it will take 2 units of time; meanwhile, the free painter will paint the walls at index 1 and 2, free of cost in 2 units of time. Thus, the total cost is 2 + 2 = 4.

Constraints:

1 <= cost.length <= 500
cost.length == time.length
1 <= cost[i] <= 10⁶
1 <= time[i] <= 500

Step 01

Brute Force Baseline

Problem summary: You are given two 0-indexed integer arrays, cost and time, of size n representing the costs and the time taken to paint n different walls respectively. There are two painters available: A paid painter that paints the ith wall in time[i] units of time and takes cost[i] units of money. A free painter that paints any wall in 1 unit of time at a cost of 0. But the free painter can only be used if the paid painter is already occupied. Return the minimum amount of money required to paint the n walls.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming

Example 1

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

Example 2

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

Step 02

Core Insight

What unlocks the optimal approach

Can we break the problem down into smaller subproblems and use DP?
Paid painters will be used for a maximum of N/2 units of time. There is no need to use paid painter for a time greater than this.

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 #2742: Painting the Walls
class Solution {
    private int n;
    private int[] cost;
    private int[] time;
    private Integer[][] f;

    public int paintWalls(int[] cost, int[] time) {
        n = cost.length;
        this.cost = cost;
        this.time = time;
        f = new Integer[n][n << 1 | 1];
        return dfs(0, n);
    }

    private int dfs(int i, int j) {
        if (n - i <= j - n) {
            return 0;
        }
        if (i >= n) {
            return 1 << 30;
        }
        if (f[i][j] == null) {
            f[i][j] = Math.min(dfs(i + 1, j + time[i]) + cost[i], dfs(i + 1, j - 1));
        }
        return f[i][j];
    }
}

// Accepted solution for LeetCode #2742: Painting the Walls
func paintWalls(cost []int, time []int) int {
	n := len(cost)
	f := make([][]int, n)
	for i := range f {
		f[i] = make([]int, n<<1|1)
		for j := range f[i] {
			f[i][j] = -1
		}
	}
	var dfs func(i, j int) int
	dfs = func(i, j int) int {
		if n-i <= j-n {
			return 0
		}
		if i >= n {
			return 1 << 30
		}
		if f[i][j] == -1 {
			f[i][j] = min(dfs(i+1, j+time[i])+cost[i], dfs(i+1, j-1))
		}
		return f[i][j]
	}
	return dfs(0, n)
}

# Accepted solution for LeetCode #2742: Painting the Walls
class Solution:
    def paintWalls(self, cost: List[int], time: List[int]) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            if n - i <= j:
                return 0
            if i >= n:
                return inf
            return min(dfs(i + 1, j + time[i]) + cost[i], dfs(i + 1, j - 1))

        n = len(cost)
        return dfs(0, 0)

// Accepted solution for LeetCode #2742: Painting the Walls
impl Solution {
    #[allow(dead_code)]
    pub fn paint_walls(cost: Vec<i32>, time: Vec<i32>) -> i32 {
        let n = cost.len();
        let mut record_vec: Vec<Vec<i32>> = vec![vec![-1; n << 1 | 1]; n];
        Self::dfs(&mut record_vec, 0, n as i32, n as i32, &time, &cost)
    }

    #[allow(dead_code)]
    fn dfs(
        record_vec: &mut Vec<Vec<i32>>,
        i: i32,
        j: i32,
        n: i32,
        time: &Vec<i32>,
        cost: &Vec<i32>,
    ) -> i32 {
        if n - i <= j - n {
            // All the remaining walls can be printed at no cost
            // Just return 0
            return 0;
        }
        if i >= n {
            // No way this case can be achieved
            // Just return +INF
            return 1 << 30;
        }
        if record_vec[i as usize][j as usize] == -1 {
            // This record hasn't been written
            record_vec[i as usize][j as usize] = std::cmp::min(
                Self::dfs(record_vec, i + 1, j + time[i as usize], n, time, cost)
                    + cost[i as usize],
                Self::dfs(record_vec, i + 1, j - 1, n, time, cost),
            );
        }
        record_vec[i as usize][j as usize]
    }
}

// Accepted solution for LeetCode #2742: Painting the Walls
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2742: Painting the Walls
// class Solution {
//     private int n;
//     private int[] cost;
//     private int[] time;
//     private Integer[][] f;
// 
//     public int paintWalls(int[] cost, int[] time) {
//         n = cost.length;
//         this.cost = cost;
//         this.time = time;
//         f = new Integer[n][n << 1 | 1];
//         return dfs(0, n);
//     }
// 
//     private int dfs(int i, int j) {
//         if (n - i <= j - n) {
//             return 0;
//         }
//         if (i >= n) {
//             return 1 << 30;
//         }
//         if (f[i][j] == null) {
//             f[i][j] = Math.min(dfs(i + 1, j + time[i]) + cost[i], dfs(i + 1, j - 1));
//         }
//         return f[i][j];
//     }
// }

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.