What is the time complexity of Paint House III?

The optimal solution has a time complexity of O(m*target*n*n).

What is the space complexity of Paint House III?

The space complexity is O(target*n).

What data structures are used to solve Paint House III?

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

What are the key insights for solving Paint House III?

Insight 1: Dynamic programming is suitable for this problem because we're looking for the minimum cost based on previous sub-problems (painting previous houses) and making choices (painting current house with different colors) Insight 2: The state of the DP should include the house index, number of neighborhoods formed up to that house, and the color of the last house painted. This allows for efficient transition calculations.

Which companies ask LeetCode 1473. Paint House III in interviews?

This problem is frequently asked at PayPal.

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Paint House III - LeetCode 1473 Solution

Paint House III - Complete Solution Guide

Paint House III is LeetCode problem 1473, 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

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

Detailed Explanation

The problem asks us to find the minimum cost to paint a row of `m` houses with `n` different colors such that there are exactly `target` neighborhoods. A neighborhood is defined as a maximal group of continuous houses painted with the same color. Some houses may already be painted, and we are given the cost to paint each unpainted house with each of the `n` colors. If it's not possible to achieve the desired `target` neighborhoods, return -1.

Solution Approach

The problem is solved using dynamic programming. We maintain a `dp` table where `dp[k][c]` represents the minimum cost to paint houses up to the current house with `k` neighborhoods, and the last house being painted with color `c+1`. The algorithm iterates through each house, each possible number of neighborhoods, and each color to calculate the minimum cost. The key is to determine whether painting the current house will continue the existing neighborhood or start a new one. If it continues the existing one, we simply add the cost of painting the current house with the same color. If it starts a new one, we consider the minimum cost of forming `k-1` neighborhoods up to the previous house with colors other than the current color.

Step-by-Step Algorithm

Step 1: Initialize a 2D DP table `dp[target+1][n]` with infinity, where `dp[k][c]` stores the minimum cost to paint up to the current house with k neighborhoods ending with color c+1.
Step 2: Initialize the base case. If the first house is unpainted (houses[0] == 0), then set `dp[1][c] = cost[0][c]` for all colors. If the first house is already painted, set `dp[1][houses[0] - 1] = 0`.
Step 3: Iterate through each house from the second house (i=1) up to the last house (m-1).
Step 4: For each house `i`, iterate through each possible number of neighborhoods `k` from 1 to `min(target, i + 1)`.
Step 5: For each house `i` and neighborhood count `k`, pre-compute the two smallest values in `dp[k-1]` and their indices. This allows for O(1) access of min cost for different neighborhood ending color.
Step 6: For each house `i`, neighborhood `k` and color `c`, check if the current house is already painted and whether the current color `c` matches the pre-painted color of the current house. If they do not match, `continue` to the next color as this will not lead to a valid solution. Also add cost of coloring the current house.
Step 7: Calculate the cost of continuing the existing neighborhood by setting the current house's color to the previous house's color.
Step 8: Calculate the cost of starting a new neighborhood. This will involve using the minimum possible cost to construct `k-1` neighborhoods up to the previous house, with any color that's different from the current house's color. Since we pre-computed the two smallest costs in the previous step, this can be calculated efficiently.
Step 9: Store the minimum of continuing or starting a new neighborhood in `dp[k][c]`. If painting house `i` with color `c` produces a lower cost, update the entry in the dp table.
Step 10: After iterating through all houses, colors, and neighborhood counts, find the minimum value in `dp[target]` array. If the minimum value is infinity, return -1; otherwise, return the minimum value.

Key Insights

Insight 1: Dynamic programming is suitable for this problem because we're looking for the minimum cost based on previous sub-problems (painting previous houses) and making choices (painting current house with different colors)
Insight 2: The state of the DP should include the house index, number of neighborhoods formed up to that house, and the color of the last house painted. This allows for efficient transition calculations.
Insight 3: The crucial part is to keep track of the minimum cost when adding a new neighborhood or continuing an existing one. Pre-calculating the two minimum costs of the previous state speeds up the calculation within the loops.

Complexity Analysis

Time Complexity: O(m*target*n*n)

Space Complexity: O(target*n)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: PayPal.

Paint House III | LeetCode 1473

Hard

Array Dynamic Programming

Time: O(m*target*n*n)Space: O(target*n)

Solution Code

Available in:

class Solution {
    public int minCost(int[] houses, int[][] cost, int m, int n, int target) {
        // dp[k][c]: min cost up to the previous house, with k neighborhoods, 
        // and the previous house painted color c+1
        int[][] dp = new int[target + 1][n];
        for (int i = 0; i <= target; i++) {
            for (int j = 0; j < n; j++) {
                dp[i][j] = Integer.MAX_VALUE;
            }
        }

        // Base case: for house 0
        if (houses[0] == 0) {
            for (int c = 0; c < n; c++) {
                dp[1][c] = cost[0][c];
            }
        } else {
            int c_idx = houses[0] - 1;
            dp[1][c_idx] = 0;
        }

        // Iterate through houses from 1 to m-1
        for (int i = 1; i < m; i++) {
            int[][] new_dp = new int[target + 1][n];
            for (int k = 0; k <= target; k++) {
                for (int l = 0; l < n; l++) {
                    new_dp[k][l] = Integer.MAX_VALUE;
                }
            }
            
            // k is the number of neighborhoods up to the current house i
            for (int k = 1; k <= Math.min(target, i + 1); k++) {
                
                // Pre-calculate min costs for k-1 neighborhoods from the previous state
                int min1 = Integer.MAX_VALUE, min2 = Integer.MAX_VALUE;
                int min1_idx = -1;
                
                for (int prev_c = 0; prev_c < n; prev_c++) {
                    int val = dp[k - 1][prev_c];
                    if (val < min1) {
                        min2 = min1;
                        min1 = val;
                        min1_idx = prev_c;
                    } else if (val < min2) {
                        min2 = val;
                    }
                }

                // Calculate costs for the current house i with color c
                for (int c = 0; c < n; c++) {
                    int current_color = c + 1;
                    if (houses[i] != 0 && houses[i] != current_color) {
                        continue;
                    }
                    
                    int paint_cost = (houses[i] == 0) ? cost[i][c] : 0;
                    
                    // Case 1: Continue the same neighborhood
                    int cost_same = new_dp[k][c];
                    
                    // Case 2: Start a new neighborhood
                    int cost_new = (c == min1_idx) ? min2 : min1;

                    int min_prev_cost = Math.min(cost_same == Integer.MAX_VALUE ? Integer.MAX_VALUE : cost_same, cost_new == Integer.MAX_VALUE ? Integer.MAX_VALUE : cost_new);

                    if (min_prev_cost != Integer.MAX_VALUE) {
                        new_dp[k][c] = min_prev_cost + paint_cost;
                    }
                }
            }
            dp = new_dp;
            
        }
        int result = Integer.MAX_VALUE;
        for(int val : dp[target]){
            result = Math.min(result, val);
        }
        
        return (result == Integer.MAX_VALUE) ? -1 : result;
    }
}

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Problem Info

Description

Problem 1473: Paint House III - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

PayPal

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1473

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

10 steps

1
Step 1: Initialize a 2D DP table `dp[target+1][n]` with infinity, where `dp[k][c]` stores the minimum cost to paint up to the current house with k neighborhoods ending with color c+1.
2
Step 2: Initialize the base case. If the first house is unpainted (houses[0] == 0), then set `dp[1][c] = cost[0][c]` for all colors. If the first house is already painted, set `dp[1][houses[0] - 1] = 0`.
3
Step 3: Iterate through each house from the second house (i=1) up to the last house (m-1).
4
Step 4: For each house `i`, iterate through each possible number of neighborhoods `k` from 1 to `min(target, i + 1)`.
5
Step 5: For each house `i` and neighborhood count `k`, pre-compute the two smallest values in `dp[k-1]` and their indices. This allows for O(1) access of min cost for different neighborhood ending color.
6
Step 6: For each house `i`, neighborhood `k` and color `c`, check if the current house is already painted and whether the current color `c` matches the pre-painted color of the current house. If they do not match, `continue` to the next color as this will not lead to a valid solution. Also add cost of coloring the current house.
7
Step 7: Calculate the cost of continuing the existing neighborhood by setting the current house's color to the previous house's color.
8
Step 8: Calculate the cost of starting a new neighborhood. This will involve using the minimum possible cost to construct `k-1` neighborhoods up to the previous house, with any color that's different from the current house's color. Since we pre-computed the two smallest costs in the previous step, this can be calculated efficiently.
9
Step 9: Store the minimum of continuing or starting a new neighborhood in `dp[k][c]`. If painting house `i` with color `c` produces a lower cost, update the entry in the dp table.
10
Step 10: After iterating through all houses, colors, and neighborhood counts, find the minimum value in `dp[target]` array. If the minimum value is infinity, return -1; otherwise, return the minimum value.

💡

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(m*target*n*n)

💾

Space Complexity

O(target*n)

📝

Why?

The time complexity arises from the nested loops for houses, target neighborhoods, and colors, and the additional loop to pre-calculate the minimum cost from previous colors. The space complexity is determined by the size of the DP table required to store the 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

Memoization (Top-Down Dynamic Programming)

Instead of the bottom-up DP approach, we could use a recursive function with memoization to achieve the same result. The function would take the current house index, number of neighborhoods, and previous color as parameters and return the minimum cost to paint the houses from that point onwards.

Time: O(m * target * n * n)Space: O(m * target * n) due to recursion stack and memoization table

Trade-offs: Memoization might be easier to understand for some, but it could potentially lead to stack overflow issues for very large inputs compared to the iterative approach.

✨

Key Insights

1Insight 1: Dynamic programming is suitable for this problem because we're looking for the minimum cost based on previous sub-problems (painting previous houses) and making choices (painting current house with different colors)
2Insight 2: The state of the DP should include the house index, number of neighborhoods formed up to that house, and the color of the last house painted. This allows for efficient transition calculations.
3Insight 3: The crucial part is to keep track of the minimum cost when adding a new neighborhood or continuing an existing one. Pre-calculating the two minimum costs of the previous state speeds up the calculation within the loops.

📚

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: `m == 0` or `n == 0` or `target == 0`. The code implicitly handles these by returning -1 if a valid solution is not found, because the `dp` table would remain at initial infinite values.
Edge case 2: `target > m`. It's impossible to have more neighborhoods than the number of houses. The inner loop `k <= min(target, i + 1)` implicitly addresses this case.
Edge case 3: If the houses are already painted in a way that the target is unreachable (e.g., `houses = [1,2,3]` and `target = 1`). The algorithm correctly returns `-1` because no valid cost will be derived that reaches target neighbourhoods within the set constraints.
Edge case 4: Some houses are already colored, but we still need to paint other houses while respecting the already colored houses. The solution directly considers the already colored houses; only paint houses that are not colored or those that match the existing color.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where the current house is already painted. Make sure to only consider colors that match the existing color when the house is already painted.

✗Common mistake 2: Not correctly calculating the cost of starting a new neighborhood. It's important to consider the minimum cost of painting the previous houses with k-1 neighborhoods, excluding the color being used for the current house.

✗Common mistake 3: Initialize dp table to avoid `Integer.MAX_VALUE` overflow during addition of costs.

✅

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: Optimizing the layout of items in a shelf where items of the same category must be together (neighborhoods) and there are costs associated with placing items in different locations.

🏦Real-world scenario 2: Minimizing the cost of assigning tasks to employees, where grouping similar tasks together reduces overhead and the number of groups is constrained.

💡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 4, 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

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

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Paint House III"?

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.

🤖 Need help with this problem?