What is the time complexity of Count Number of Ways to Place Houses?

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

What data structures are used to solve Count Number of Ways to Place Houses?

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

What are the key insights for solving Count Number of Ways to Place Houses?

Insight 1: The problem can be solved using dynamic programming. We can calculate the number of ways to arrange houses on one side of the street and then square the result (modulo 10^9 + 7) because the arrangements on each side are independent. Insight 2: We can define two states for each plot: 'house' (a house is placed) and 'space' (the plot is empty). The number of valid arrangements for each state can be calculated iteratively based on the previous states.

Which companies ask LeetCode 2320. Count Number of Ways to Place Houses in interviews?

This problem is frequently asked at Nagarro.

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Count Number of Ways to Place Houses - LeetCode 2320 Solution

Q: How to solve Count Number of Ways to Place Houses?

The solution uses dynamic programming to calculate the number of ways to arrange houses on one side of the street. It maintains two variables: 'house' and 'space'. 'house' represents the number of valid arrangements ending with a house, and 'space' represents the number of valid arrangements ending with a space. The solution iterates from 2 to n, updating the 'house' and 'space' values based on the previous values. The total number of arrangements for one side is the sum of 'house' and 'space'. Finally, it squares the total number of arrangements (modulo 10^9 + 7) to account for both sides of the street.

Count Number of Ways to Place Houses - Complete Solution Guide

Count Number of Ways to Place Houses is LeetCode problem 2320, a Medium 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 street with n * 2 plots , where there are n plots on each side of the street. The plots on each side are numbered from 1 to n . On each plot, a house can be placed. Return the number of ways houses can be placed such that no two houses are adjacent to each other on the same side of the street . Since the answer may be very large, return it modulo 10 9 + 7 . Note that if a house is placed on the i th plot on one side of the street, a house can also be placed on the i th plot on the oth

Detailed Explanation

The problem asks us to calculate the number of ways to place houses on both sides of a street with 'n' plots on each side, such that no two houses are adjacent on the same side. We need to return the answer modulo 10^9 + 7. Each plot can either have a house or be empty, and the choices on one side of the street are independent of the other side. We are given that 1 <= n <= 10^4.

Solution Approach

The solution uses dynamic programming to calculate the number of ways to arrange houses on one side of the street. It maintains two variables: 'house' and 'space'. 'house' represents the number of valid arrangements ending with a house, and 'space' represents the number of valid arrangements ending with a space. The solution iterates from 2 to n, updating the 'house' and 'space' values based on the previous values. The total number of arrangements for one side is the sum of 'house' and 'space'. Finally, it squares the total number of arrangements (modulo 10^9 + 7) to account for both sides of the street.

Step-by-Step Algorithm

Step 1: Initialize 'house' and 'space' to 1. This represents the base case when n = 1. There's one way to place a house and one way to leave the plot empty.
Step 2: Iterate from 2 to n (exclusive). In each iteration, update 'house' and 'space'.
Step 3: The new 'house' value is the previous 'space' value because we can only place a house if the previous plot was empty.
Step 4: The new 'space' value is the sum of the previous 'house' and 'space' values (modulo 10^9 + 7) because we can always put a space regardless of the previous plot.
Step 5: After the loop, calculate the total number of ways for one side by summing 'house' and 'space' (modulo 10^9 + 7).
Step 6: Calculate the total number of ways for both sides by squaring the result from the previous step (modulo 10^9 + 7) and return the result.

Key Insights

Insight 1: The problem can be solved using dynamic programming. We can calculate the number of ways to arrange houses on one side of the street and then square the result (modulo 10^9 + 7) because the arrangements on each side are independent.
Insight 2: We can define two states for each plot: 'house' (a house is placed) and 'space' (the plot is empty). The number of valid arrangements for each state can be calculated iteratively based on the previous states.
Insight 3: The modulo operation is crucial to prevent integer overflow, especially since we are multiplying the result of one side by itself.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Dynamic Programming.

Companies

Asked at: Nagarro.

Count Number of Ways to Place Houses | LeetCode 2320

Medium

Dynamic Programming

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

Solution Code

Available in:

class Solution {
    public int countHousePlacements(int n) {
        int MOD = (int)1e9 + 7;
        
        // house: number of valid arrangements for i plots ending with a house
        // space: number of valid arrangements for i plots ending with a space
        
        // Base case for n=1 plot
        long house = 1;
        long space = 1;
        
        // Iterate from 2 to n plots
        for (int i = 0; i < n - 1; i++) {
            // To place a house at the current plot, the previous must be a space.
            // To place a space at the current plot, the previous can be a house or a space.
            // We use tuple assignment to update simultaneously without a temp variable.
            long temp = space;
            space = (house + space) % MOD;
            house = temp;
            
        }
        
        // Total ways for one side is the sum of possibilities
        long total_one_side = (house + space) % MOD;
        
        // The two sides of the street are independent, so we square the result
        return (int)((total_one_side * total_one_side) % MOD);
    }
}

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Problem Info

Description

Problem 2320: Count Number of Ways to Place Houses - Detailed explanation and solution approaches.

Topics

Dynamic Programming

Companies

Nagarro

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2320

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

6 steps

1
Step 1: Initialize 'house' and 'space' to 1. This represents the base case when n = 1. There's one way to place a house and one way to leave the plot empty.
2
Step 2: Iterate from 2 to n (exclusive). In each iteration, update 'house' and 'space'.
3
Step 3: The new 'house' value is the previous 'space' value because we can only place a house if the previous plot was empty.
4
Step 4: The new 'space' value is the sum of the previous 'house' and 'space' values (modulo 10^9 + 7) because we can always put a space regardless of the previous plot.
5
Step 5: After the loop, calculate the total number of ways for one side by summing 'house' and 'space' (modulo 10^9 + 7).
6
Step 6: Calculate the total number of ways for both sides by squaring the result from the previous step (modulo 10^9 + 7) and return the result.

💡

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. The space complexity is O(1) because we only use a few constant variables (house, space, MOD) regardless of the input size 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

Using Fibonacci Sequence

The number of ways to arrange houses on one side such that no two are adjacent is related to the Fibonacci sequence. The total number of arrangements is fib(n+2), where fib(x) is the x-th Fibonacci number.

Time: O(n) for iterative Fibonacci calculation or O(log n) for matrix exponentiation method to calculate Fibonacci numbers.Space: O(1)

Trade-offs: This approach leverages the Fibonacci sequence but conceptually is similar. The standard iterative Fibonacci method has same time complexity as the DP approach, while matrix exponentiation method is faster but requires more complex implementation.

✨

Key Insights

1Insight 1: The problem can be solved using dynamic programming. We can calculate the number of ways to arrange houses on one side of the street and then square the result (modulo 10^9 + 7) because the arrangements on each side are independent.
2Insight 2: We can define two states for each plot: 'house' (a house is placed) and 'space' (the plot is empty). The number of valid arrangements for each state can be calculated iteratively based on the previous states.
3Insight 3: The modulo operation is crucial to prevent integer overflow, especially since we are multiplying the result of one side by itself.

📚

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 house = 1 and space = 1, resulting in 4 possible arrangements (as shown in example 1).
Edge case 2: n = 10^4. The modulo operation prevents integer overflow and ensures that the result remains within the required range.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow. Failing to use the modulo operator at each step can lead to incorrect results.

✗Common mistake 2: Incorrect base case initialization. If the base case is not handled correctly, the subsequent calculations will be wrong.

✗Common mistake 3: Forgetting to square the result. It's crucial to remember that the arrangements on each side of the street are independent, so we need to multiply the number of ways for one side by itself.

✅

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 where resources cannot be adjacent. For example, placing servers in a data center such that adjacent slots cannot both be occupied due to power or cooling constraints.

🏦Real-world scenario 2: Combinatorial optimization problems where certain constraints limit the placement of objects. Consider placement of facilities where certain facilities can't be placed next to each other due to safety reasons

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: November 8, 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

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

@DevExpert92Jul 14

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

@AlgoMasterJul 7

❓

Frequently Asked Questions

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."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Count Number of Ways to Place Houses"?

For this medium-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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