What is the time complexity of Number of Ways to Paint N × 3 Grid?

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

What data structures are used to solve Number of Ways to Paint N × 3 Grid?

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

What are the key insights for solving Number of Ways to Paint N × 3 Grid?

Insight 1: The problem can be solved using dynamic programming. We don't need to explicitly explore all color combinations because the number of ways to color row 'i' depends only on row 'i-1'. Insight 2: We can categorize each row into two types: 'aba' (where the first and third colors are the same) and 'abc' (where all three colors are different).

Which companies ask LeetCode 1411. Number of Ways to Paint N × 3 Grid in interviews?

This problem is frequently asked at Fortinet.

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Number of Ways to Paint N × 3 Grid - LeetCode 1411 Solution

Q: How to solve Number of Ways to Paint N × 3 Grid?

The solution employs dynamic programming to efficiently calculate the number of valid grid colorings. It leverages the observation that the number of ways to color a row depends only on the previous row. We maintain two variables: 'aba' and 'abc', representing the number of valid colorings for a row with patterns 'aba' and 'abc', respectively. We initialize these variables for the first row (n=1). Then, we iterate from row 2 to 'n', updating 'aba' and 'abc' based on the number of ways they can be formed from the previous row's 'aba' and 'abc' configurations. Finally, we return the sum of 'aba' and 'abc' modulo 10^9 + 7.

Number of Ways to Paint N × 3 Grid - Complete Solution Guide

Number of Ways to Paint N × 3 Grid is LeetCode problem 1411, 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 have a grid of size n x 3 and you want to paint each cell of the grid with exactly one of the three colors: Red , Yellow, or Green while making sure that no two adjacent cells have the same color (i.e., no two cells that share vertical or horizontal sides have the same color). Given n the number of rows of the grid, return the number of ways you can paint this grid . As the answer may grow large, the answer must be computed modulo 10 9 + 7 . Example 1: Input: n = 1 Output: 12 Explanation: Th

Detailed Explanation

The problem asks us to find the number of ways to color an n x 3 grid using three colors (Red, Yellow, Green) such that no two adjacent cells (horizontally or vertically) have the same color. We need to return the result modulo 10^9 + 7 to handle potentially large numbers. The input is the number of rows, 'n', and the output is the number of valid colorings modulo 10^9 + 7. The constraint is 1 <= n <= 5000.

Solution Approach

The solution employs dynamic programming to efficiently calculate the number of valid grid colorings. It leverages the observation that the number of ways to color a row depends only on the previous row. We maintain two variables: 'aba' and 'abc', representing the number of valid colorings for a row with patterns 'aba' and 'abc', respectively. We initialize these variables for the first row (n=1). Then, we iterate from row 2 to 'n', updating 'aba' and 'abc' based on the number of ways they can be formed from the previous row's 'aba' and 'abc' configurations. Finally, we return the sum of 'aba' and 'abc' modulo 10^9 + 7.

Step-by-Step Algorithm

Step 1: Initialize 'MOD' to 10^9 + 7. This constant will be used for modulo operations to prevent integer overflow.
Step 2: Initialize 'aba' and 'abc' to 6. For n=1, there are 6 'aba' patterns (e.g., RGR, RBR, GRG, GBG, YRY, YBY) and 6 'abc' patterns (e.g., RGB, RBG, GRB, GBR, YRB, YBR).
Step 3: Iterate from 2 to 'n' (exclusive). In each iteration, calculate the new values for 'aba' and 'abc' based on the previous values.
Step 4: The number of ways to form an 'aba' pattern in the current row is (aba * 3 + abc * 2) % MOD, using the previous row's 'aba' and 'abc' counts.
Step 5: The number of ways to form an 'abc' pattern in the current row is (aba * 2 + abc * 2) % MOD, using the previous row's 'aba' and 'abc' counts.
Step 6: Update 'aba' and 'abc' with the newly calculated values for the current row.
Step 7: After the loop finishes, return (aba + abc) % MOD. This is the total number of ways to paint the n x 3 grid.

Key Insights

Insight 1: The problem can be solved using dynamic programming. We don't need to explicitly explore all color combinations because the number of ways to color row 'i' depends only on row 'i-1'.
Insight 2: We can categorize each row into two types: 'aba' (where the first and third colors are the same) and 'abc' (where all three colors are different).
Insight 3: Instead of storing the entire grid, we only need to keep track of the number of ways to color the last row as 'aba' and 'abc'. This reduces the space complexity significantly.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Dynamic Programming.

Companies

Asked at: Fortinet.

Number of Ways to Paint N × 3 Grid | LeetCode 1411

Hard

Dynamic Programming

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int numOfWays(int n) {
        int MOD = (int)1e9 + 7;
        
        // Let aba be the number of ways to paint a row with 2 distinct colors (e.g., RGR).
        // Let abc be the number of ways to paint a row with 3 distinct colors (e.g., RGY).
        
        // For the first row (n=1):
        // There are 3 choices for the first color, 2 for the second.
        // aba patterns (C1, C2, C1): 3 * 2 = 6 ways.
        // abc patterns (C1, C2, C3): 3 * 2 * 1 = 6 ways.
        
        long aba = 6;
        long abc = 6;
        
        // For each subsequent row from 2 to n:
        for (int i = 0; i < n - 1; i++) {
            // Calculate the number of ways for the current row based on the previous row.
            // The right-hand side expressions use the values of 'aba' and 'abc' from the previous row.
            // The tuple assignment ensures that the old values are used for both calculations before updating.
            
            // To get an 'aba' pattern in the current row:
            // - From a previous 'aba' pattern, there are 3 ways.
            // - From a previous 'abc' pattern, there are 2 ways.
            // new_aba = (aba * 3 + abc * 2)
            
            // To get an 'abc' pattern in the current row:
            // - From a previous 'aba' pattern, there are 2 ways.
            // - From a previous 'abc' pattern, there are 2 ways.
            // new_abc = (aba * 2 + abc * 2)
            
            long new_aba = (aba * 3 + abc * 2) % MOD;
            long new_abc = (aba * 2 + abc * 2) % MOD;
            aba = new_aba;
            abc = new_abc;
            
        }
        
        return (int)((aba + abc) % MOD);
    }
}

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Problem Info

Description

Problem 1411: Number of Ways to Paint N × 3 Grid - Detailed explanation and solution approaches.

Topics

Dynamic Programming

Companies

Fortinet

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1411

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 'MOD' to 10^9 + 7. This constant will be used for modulo operations to prevent integer overflow.
2
Step 2: Initialize 'aba' and 'abc' to 6. For n=1, there are 6 'aba' patterns (e.g., RGR, RBR, GRG, GBG, YRY, YBY) and 6 'abc' patterns (e.g., RGB, RBG, GRB, GBR, YRB, YBR).
3
Step 3: Iterate from 2 to 'n' (exclusive). In each iteration, calculate the new values for 'aba' and 'abc' based on the previous values.
4
Step 4: The number of ways to form an 'aba' pattern in the current row is (aba * 3 + abc * 2) % MOD, using the previous row's 'aba' and 'abc' counts.
5
Step 5: The number of ways to form an 'abc' pattern in the current row is (aba * 2 + abc * 2) % MOD, using the previous row's 'aba' and 'abc' counts.
6
Step 6: Update 'aba' and 'abc' with the newly calculated values for the current row.
7
Step 7: After the loop finishes, return (aba + abc) % MOD. This is the total number of ways to paint the n x 3 grid.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) because we iterate from 2 to 'n' in a single loop. The space complexity is O(1) because we only use a fixed number of variables ('aba', 'abc', 'MOD') regardless of the input 'n'.

📖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

Matrix Exponentiation

The dynamic programming equations can be represented as a matrix transformation. We can use matrix exponentiation to calculate the nth power of the transformation matrix, resulting in a faster solution for very large 'n'.

Time: O(log n)Space: O(1)

Trade-offs: Matrix exponentiation provides better time complexity for very large 'n', but it has a higher constant factor and is more complex to implement. The O(n) solution is sufficient for the given constraint (n <= 5000).

✨

Key Insights

1Insight 1: The problem can be solved using dynamic programming. We don't need to explicitly explore all color combinations because the number of ways to color row 'i' depends only on row 'i-1'.
2Insight 2: We can categorize each row into two types: 'aba' (where the first and third colors are the same) and 'abc' (where all three colors are different).
3Insight 3: Instead of storing the entire grid, we only need to keep track of the number of ways to color the last row as 'aba' and 'abc'. This reduces the space complexity significantly.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: n = 1. The code correctly initializes 'aba' and 'abc' to 6, which are the correct values for the base case.
Edge case 2: Large values of 'n'. The code uses the modulo operator (%) to prevent integer overflow, ensuring that the result remains within the valid range.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using the modulo operator. The number of ways can become very large quickly, leading to integer overflow if the modulo operator is not used.

✗Common mistake 2: Incorrectly deriving the recurrence relations for 'aba' and 'abc'. The number of ways to transition from one row to the next needs to be carefully considered based on color constraints.

✅

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: Resource allocation problems where resources must be allocated to adjacent units with constraints that prevent using the same resource in adjacent units.

🏦Real-world scenario 2: DNA sequencing where different nucleotides (A, T, C, G) cannot be adjacent in certain configurations due to stability issues. The problem can be modeled as counting valid arrangements under adjacency constraints.

💡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

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

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: October 30, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

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

@InterviewProJul 7

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

@CodeNinjaJun 30

❓

Frequently Asked Questions

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 "Number of Ways to Paint N × 3 Grid"?

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

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