How do you solve Count Number of Ways to Place Houses on LeetCode?

Count Number of Ways to Place Houses (LeetCode #2320) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with Unknown time complexity and Unknown space complexity. Key insight: We can use dynamic programming to build the solution incrementally. The number of ways to place houses on one side of the street resembles the Fibonacci sequence, where the current state depends on the previous two states.

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

The optimal solution for Count Number of Ways to Place Houses runs in Unknown time and uses Unknown space.

What algorithm or data structure is used to solve Count Number of Ways to Place Houses?

Count Number of Ways to Place Houses uses the following concepts: Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Fibonacci Sequence.

What are the interview tips for Count Number of Ways to Place Houses?

Start with a brute-force approach to understand the problem better. Clearly define your base cases when using dynamic programming. Be mindful of edge cases, such as n = 1.

What are common mistakes when solving Count Number of Ways to Place Houses?

Not considering the modulo operation in large outputs. Overcomplicating the problem by not recognizing the Fibonacci-like pattern.

Count Number of Ways to Place Houses — LeetCode #2320 (Medium)

Tags: Dynamic Programming

Related patterns: Dynamic Programming, Fibonacci Sequence

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

We can explore all possible combinations of placing houses on both sides of the street. This means checking every possible arrangement of houses and ensuring that no two houses are adjacent on the same side.

The time complexity is O(n²) because we check each arrangement and validate it, leading to a nested loop scenario.

Step 1: Generate all possible combinations of house placements on one side of the street.
Step 2: For each combination, check if any two houses are adjacent.
Step 3: Count valid combinations and multiply by 2 (for both sides of the street).

For n = 2: 1. Arrangement: [0, 0] -> Valid 2. Arrangement: [0, 1] -> Valid 3. Arrangement: [1, 0] -> Valid 4. Arrangement: [1, 1] -> Invalid Count = 3 valid arrangements, total = 3 * 2 = 6.

Optimal Solution approach

Time complexity: Unknown. Space complexity: Unknown.

We can use dynamic programming to build the solution incrementally. The number of ways to place houses on one side of the street resembles the Fibonacci sequence, where the current state depends on the previous two states.

Step 1: Define a DP array where dp[i] represents the number of valid arrangements for i plots.
Step 2: Initialize base cases: dp[0] = 1 (no plots) and dp[1] = 2 (either empty or one house).
Step 3: Fill the DP array using the relation dp[i] = dp[i-1] + dp[i-2] for i >= 2.
Step 4: The final answer will be (dp[n] * dp[n]) % (10^9 + 7) since both sides are independent.

Common Mistakes

Not considering the modulo operation in large outputs.
Overcomplicating the problem by not recognizing the Fibonacci-like pattern.

Interview Tips

Start with a brute-force approach to understand the problem better.
Clearly define your base cases when using dynamic programming.
Be mindful of edge cases, such as n = 1.

#2320

Count Number of Ways to Place Houses

Medium

Dynamic Programming↗Dynamic Programming↗Fibonacci Sequence↗

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	Unknown
Space	O(1)	Unknown

💡

Intuition

Time UnknownSpace Unknown

⚙️

Algorithm

4 steps

1Step 1: Define a DP array where dp[i] represents the number of valid arrangements for i plots.
2Step 2: Initialize base cases: dp[0] = 1 (no plots) and dp[1] = 2 (either empty or one house).
3Step 3: Fill the DP array using the relation dp[i] = dp[i-1] + dp[i-2] for i >= 2.
4Step 4: The final answer will be (dp[n] * dp[n]) % (10^9 + 7) since both sides are independent.

solution.py12 lines

1# Full working Python code
2def countWaysOptimal(n):
3    MOD = 10**9 + 7
4    if n == 1:
5        return 4
6    dp = [0] * (n + 1)
7    dp[0], dp[1] = 1, 2
8    for i in range(2, n + 1):
9        dp[i] = (dp[i - 1] + dp[i - 2]) % MOD
10    return (dp[n] * dp[n]) % MOD
11
12print(countWaysOptimal(2))

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.