#2320

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Count Number of Ways to Place Houses

This problem asks you to calculate the number of ways to place houses along a street while preventing adjacent houses on the same side.

dynamic programming

Problem Statement

You are given a street with 2n plots, where each side of the street has n plots. Houses can be placed on each plot, but no two houses should be adjacent on the same side of the street. The goal is to calculate the number of possible configurations of houses under these constraints, considering that the solution must be returned modulo 10^9 + 7.

Additionally, it is allowed to place a house on the ith plot on one side of the street, and also place a house on the same ith plot on the opposite side. The problem asks you to return the total number of valid house placements for all the plots combined.

Examples

Example 1

Input: n = 1

Output: 4

Possible arrangements:

All plots are empty.
A house is placed on one side of the street.
A house is placed on the other side of the street.
Two houses are placed, one on each side of the street.

Example 2

Input: n = 2

Output: 9

The 9 possible arrangements are shown in the diagram above.

Constraints

1 <= n <= 104

Solution Approach

Dynamic Programming Approach

The problem can be solved using dynamic programming (DP). First, solve for one side of the street, using a state transition to track the possible ways to place houses on that side. Then, apply the same logic to the other side of the street and combine the results. The transition depends on whether a house is placed on a plot or not, ensuring adjacent plots are not both occupied.

State Transition Analysis

By setting up the problem for one side of the street, you can compute the number of ways to place houses by considering the current plot and deciding whether to place a house or leave it empty, avoiding adjacent placements. This can be modeled as a state transition problem, where each state tracks the number of ways to place houses up to a certain plot.

Final Solution Calculation

Once the DP solution is computed for one side, the result for both sides combined can be derived by considering the independent placements for each side. This final calculation ensures that all constraints are respected, and the result is taken modulo 10^9 + 7 to handle large numbers.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity depend on the DP approach implemented for calculating the number of valid placements on one side of the street. The time complexity is O(n), where n is the number of plots, as we iterate through the plots. The space complexity can be optimized to O(1) by storing only the previous state of the DP array.

What Interviewers Usually Probe

The candidate demonstrates an understanding of state transition dynamic programming and can apply it to solve the problem.
The candidate breaks down the problem into manageable subproblems, such as solving for one side of the street first.
The candidate correctly handles the modulo operation to ensure that the final result fits within the required limits.

Common Pitfalls or Variants

Common pitfalls

Overcomplicating the solution by not simplifying the problem to solving for one side of the street first.
Neglecting the modulo operation or forgetting to return the result modulo 10^9 + 7.
Failing to handle the case where both sides of the street need to be considered independently but with overlapping constraints.

Follow-up variants

Increase the number of plots n to test the scalability of the DP approach.
Consider constraints such as not allowing houses to be placed on the first or last plot of the street.
Extend the problem to include additional constraints, such as a maximum number of houses that can be placed on the street.

FAQ

What is the optimal approach to solve Count Number of Ways to Place Houses?

The optimal approach involves dynamic programming, first solving for one side of the street and then extending it to both sides while ensuring no adjacent houses.

How do I handle large numbers in this problem?

You should return the result modulo 10^9 + 7 to handle large numbers and avoid overflow.

What is the time complexity of solving this problem?

The time complexity of the solution is O(n), where n is the number of plots, as the algorithm iterates through the plots once.

Can this problem be solved with a greedy approach?

A greedy approach may fail because it does not consider the possibility of placing houses on both sides of the street in a way that maximizes the solution.

What is state transition dynamic programming?

State transition dynamic programming is a method of solving problems by breaking them down into simpler subproblems and transitioning between states based on decisions made at each step.

terminal

Solution

Solution 1: Dynamic Programming

Since the placement of houses on both sides of the street does not affect each other, we can consider the placement on one side only, and then square the number of ways for one side to get the final result modulo.

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class Solution:
    def countHousePlacements(self, n: int) -> int:
        mod = 10**9 + 7
        f = [1] * n
        g = [1] * n
        for i in range(1, n):
            f[i] = g[i - 1]
            g[i] = (f[i - 1] + g[i - 1]) % mod
        v = f[-1] + g[-1]
        return v * v % mod

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