What is the time complexity of Painting the Walls?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Painting the Walls?

This problem uses Array, Dynamic Programming. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Painting the Walls?

Insight 1: Dynamic Programming is suitable because the problem exhibits optimal substructure. The minimum cost to paint `n` walls can be built from the minimum costs of painting fewer walls. Insight 2: The problem is essentially a knapsack-like problem where we decide whether to include each wall using the paid painter or let the free painter handle some walls. The `time` array affects how many walls can be painted for free while a paid painter is working.

Which companies ask LeetCode 2742. Painting the Walls in interviews?

This problem is frequently asked at DE Shaw, Media.net, Snowflake.

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Painting the Walls - LeetCode 2742 Solution

Painting the Walls - Complete Solution Guide

Painting the Walls is LeetCode problem 2742, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 wal

Detailed Explanation

The problem asks us to find the minimum cost to paint `n` walls, given an array `cost` representing the cost of painting each wall by a paid painter, and an array `time` representing the time it takes for the paid painter to paint each wall. There's also a free painter that can paint any wall in 1 unit of time at no cost, but only when the paid painter is occupied. The goal is to minimize the total cost of painting all walls using a combination of the paid and free painters.

Solution Approach

The solution employs dynamic programming to determine the minimum cost. `dp[i]` stores the minimum cost to paint `i` walls. We iterate through each wall. For each wall, we have the option to paint it using the paid painter. If we choose to paint it with the paid painter, we need to consider how many walls the free painter can paint during that time. The core logic is to update the `dp` array based on whether the paid painter paints the current wall or not, considering the impact of the free painter.

Step-by-Step Algorithm

Step 1: Initialize a DP array `dp` of size `n+1`, where `dp[i]` represents the minimum cost to paint `i` walls. Set `dp[0] = 0` because it costs 0 to paint 0 walls.
Step 2: Initialize the rest of the `dp` array with a large value (infinity) to represent an unreachable state.
Step 3: Iterate through each wall `i` from 0 to `n-1`.
Step 4: For each wall `i`, iterate through the `dp` array from `n` down to 1. This reverse iteration is crucial for DP updates to ensure we're using values from the previous iteration.
Step 5: Inside the inner loop, update `dp[j]` as follows: `dp[j] = min(dp[j], dp[max(0, j - 1 - time[i])] + cost[i])`. This formula represents the decision of whether to paint the current wall using the paid painter. `dp[j]` is the current minimum cost to paint `j` walls. If we paint the current wall using the paid painter, it costs `cost[i]`. While the paid painter paints wall `i`, the free painter paints `time[i]` walls. Therefore, to paint `j` walls, we needed to have painted `j - 1 - time[i]` walls previously. We take the `max(0, ...)` because if `1 + time[i]` is greater than or equal to `j`, we only need to consider the case where we painted 0 walls initially.
Step 6: After iterating through all the walls, `dp[n]` will contain the minimum cost to paint all `n` walls.
Step 7: Return `dp[n]`.

Key Insights

Insight 1: Dynamic Programming is suitable because the problem exhibits optimal substructure. The minimum cost to paint `n` walls can be built from the minimum costs of painting fewer walls.
Insight 2: The problem is essentially a knapsack-like problem where we decide whether to include each wall using the paid painter or let the free painter handle some walls. The `time` array affects how many walls can be painted for free while a paid painter is working.
Insight 3: The crucial part is the state transition in DP. `dp[j]` represents the minimum cost to paint `j` walls. The free painter can paint walls when paid painter is busy, which allows us to reduce the number of walls painted using paid painter.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: DE Shaw, Media.net, Snowflake.

Painting the Walls | LeetCode 2742

Hard

Array Dynamic Programming

Time: O(n^2)Space: O(n)

Solution Code

Available in:

class Solution {
    public int paintWalls(int[] cost, int[] time) {
        int n = cost.length;
        int[] dp = new int[n + 1];
        for (int i = 0; i <= n; i++) {
            dp[i] = Integer.MAX_VALUE;
        }
        dp[0] = 0;

        for (int i = 0; i < n; i++) {
            int c = cost[i];
            int t = time[i];
            for (int j = n; j > 0; j--) {
                dp[j] = Math.min(dp[j], dp[Math.max(0, j - 1 - t)] + c);
            }
        }

        return dp[n];
    }
}

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Problem Info

Description

Problem 2742: Painting the Walls - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

DE ShawMedia.netSnowflake

Stats

Difficulty:Hard

Languages:4

Companies:3

Problem #:2742

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize a DP array `dp` of size `n+1`, where `dp[i]` represents the minimum cost to paint `i` walls. Set `dp[0] = 0` because it costs 0 to paint 0 walls.
2
Step 2: Initialize the rest of the `dp` array with a large value (infinity) to represent an unreachable state.
3
Step 3: Iterate through each wall `i` from 0 to `n-1`.
4
Step 4: For each wall `i`, iterate through the `dp` array from `n` down to 1. This reverse iteration is crucial for DP updates to ensure we're using values from the previous iteration.
5
Step 5: Inside the inner loop, update `dp[j]` as follows: `dp[j] = min(dp[j], dp[max(0, j - 1 - time[i])] + cost[i])`. This formula represents the decision of whether to paint the current wall using the paid painter. `dp[j]` is the current minimum cost to paint `j` walls. If we paint the current wall using the paid painter, it costs `cost[i]`. While the paid painter paints wall `i`, the free painter paints `time[i]` walls. Therefore, to paint `j` walls, we needed to have painted `j - 1 - time[i]` walls previously. We take the `max(0, ...)` because if `1 + time[i]` is greater than or equal to `j`, we only need to consider the case where we painted 0 walls initially.
6
Step 6: After iterating through all the walls, `dp[n]` will contain the minimum cost to paint all `n` walls.
7
Step 7: Return `dp[n]`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity stems directly from the nested loops. The outer loop processes each wall once, and the inner loop updates the DP table for each wall. The space complexity is determined by the size of the DP table required to store intermediate results.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursion with Memoization

The problem can also be solved using a recursive approach with memoization. This approach essentially explores all possible combinations of using the paid and free painters and stores the results in a memo to avoid redundant calculations.

Time: O(n^2)Space: O(n^2) due to the memoization table.

Trade-offs: The recursive approach might be more intuitive for some, but it often involves higher constant factors due to function call overhead. Also requires O(n^2) extra space for the memoization array.

✨

Key Insights

1Insight 1: Dynamic Programming is suitable because the problem exhibits optimal substructure. The minimum cost to paint `n` walls can be built from the minimum costs of painting fewer walls.
2Insight 2: The problem is essentially a knapsack-like problem where we decide whether to include each wall using the paid painter or let the free painter handle some walls. The `time` array affects how many walls can be painted for free while a paid painter is working.
3Insight 3: The crucial part is the state transition in DP. `dp[j]` represents the minimum cost to paint `j` walls. The free painter can paint walls when paid painter is busy, which allows us to reduce the number of walls painted using paid painter.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: `time[i]` is very large. The `max(0, j - 1 - t)` handles this edge case correctly, ensuring the index to `dp` remains non-negative.
Edge case 2: When all the costs are high. The initialization of `dp` array ensures correct calculation regardless of how high the costs are.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrect DP state transition. Failing to account for the free painter's effect and updating `dp` array with incorrect formula. For example, not using max(0, j - 1 - t) when accessing dp table.

✗Common mistake 2: Not initializing the `dp` array with a large enough value (infinity). This can lead to incorrect results, especially when dealing with larger costs.

✗Common mistake 3: Iterating the inner loop in the wrong order. The inner loop needs to iterate in reverse to avoid overwriting previous results.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Project scheduling with limited resources. Consider each wall as a task, cost as the resource consumption for a paid team, and time as the duration of the task. A free resource (like an automated tool) can work in parallel but has limitations. The problem becomes finding the optimal allocation of resources to minimize the overall project cost.

🏦Real-world scenario 2: Optimizing cloud resource allocation. Imagine different virtual machines (VMs) with varying costs and processing speeds. A 'free' but limited service could be automatically utilized while more expensive VMs are running. The problem translates to minimizing costs while completing a set of tasks within a deadline.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 13, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Painting the Walls"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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