Move from brute-force thinking to an efficient approach using dynamic programming strategy.
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 109 + 7.
Note that if a house is placed on the ith plot on one side of the street, a house can also be placed on the ith plot on the other side of the street.
Example 1:
Input: n = 1 Output: 4 Explanation: Possible arrangements: 1. All plots are empty. 2. A house is placed on one side of the street. 3. A house is placed on the other side of the street. 4. Two houses are placed, one on each side of the street.
Example 2:
Input: n = 2 Output: 9 Explanation: The 9 possible arrangements are shown in the diagram above.
Constraints:
1 <= n <= 104Problem summary: 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 109 + 7. Note that if a house is placed on the ith plot on one side of the street, a house can also be placed on the ith plot on the other side of the street.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming
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climbing-stairs)house-robber)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2320: Count Number of Ways to Place Houses
class Solution {
public int countHousePlacements(int n) {
final int mod = (int) 1e9 + 7;
int[] f = new int[n];
int[] g = new int[n];
f[0] = 1;
g[0] = 1;
for (int i = 1; i < n; ++i) {
f[i] = g[i - 1];
g[i] = (f[i - 1] + g[i - 1]) % mod;
}
long v = (f[n - 1] + g[n - 1]) % mod;
return (int) (v * v % mod);
}
}
// Accepted solution for LeetCode #2320: Count Number of Ways to Place Houses
func countHousePlacements(n int) int {
const mod = 1e9 + 7
f := make([]int, n)
g := make([]int, n)
f[0], g[0] = 1, 1
for i := 1; i < n; i++ {
f[i] = g[i-1]
g[i] = (f[i-1] + g[i-1]) % mod
}
v := f[n-1] + g[n-1]
return v * v % mod
}
# Accepted solution for LeetCode #2320: Count Number of Ways to Place Houses
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
// Accepted solution for LeetCode #2320: Count Number of Ways to Place Houses
/**
* [2320] Count Number of Ways to Place Houses
*
* 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 other side of the street.
*
* Example 1:
*
* Input: n = 1
* Output: 4
* Explanation:
* Possible arrangements:
* 1. All plots are empty.
* 2. A house is placed on one side of the street.
* 3. A house is placed on the other side of the street.
* 4. Two houses are placed, one on each side of the street.
*
* Example 2:
* <img alt="" src="https://assets.leetcode.com/uploads/2022/05/12/arrangements.png" style="width: 500px; height: 500px;" />
* Input: n = 2
* Output: 9
* Explanation: The 9 possible arrangements are shown in the diagram above.
*
*
* Constraints:
*
* 1 <= n <= 10^4
*
*/
pub struct Solution {}
// problem: https://leetcode.com/problems/count-number-of-ways-to-place-houses/
// discuss: https://leetcode.com/problems/count-number-of-ways-to-place-houses/discuss/?currentPage=1&orderBy=most_votes&query=
// submission codes start here
impl Solution {
pub fn count_house_placements(n: i32) -> i32 {
0
}
}
// submission codes end
#[cfg(test)]
mod tests {
use super::*;
#[test]
#[ignore]
fn test_2320_example_1() {
let n = 1;
let result = 4;
assert_eq!(Solution::count_house_placements(n), result);
}
#[test]
#[ignore]
fn test_2320_example_2() {
let n = 2;
let result = 9;
assert_eq!(Solution::count_house_placements(n), result);
}
}
// Accepted solution for LeetCode #2320: Count Number of Ways to Place Houses
function countHousePlacements(n: number): number {
const f = new Array(n);
const g = new Array(n);
f[0] = g[0] = 1n;
const mod = BigInt(10 ** 9 + 7);
for (let i = 1; i < n; ++i) {
f[i] = g[i - 1];
g[i] = (f[i - 1] + g[i - 1]) % mod;
}
const v = f[n - 1] + g[n - 1];
return Number(v ** 2n % mod);
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.