How do you solve Count the Number of Houses at a Certain Distance II on LeetCode?

Count the Number of Houses at a Certain Distance II (LeetCode #3017) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the structure of the graph helps to optimize the solution.

What is the time complexity of Count the Number of Houses at a Certain Distance II?

The optimal solution for Count the Number of Houses at a Certain Distance II runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only loop through the distances once. The space complexity is O(n) due to the result array.

What is the brute force solution for Count the Number of Houses at a Certain Distance II?

The brute force approach for Count the Number of Houses at a Certain Distance II is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we will check every possible pair of houses and calculate the distance between them. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count the Number of Houses at a Certain Distance II?

Count the Number of Houses at a Certain Distance II uses the following concepts: Graph Theory, Prefix Sum. The recommended patterns to study are: Graph Theory, Prefix Sum.

What are the interview tips for Count the Number of Houses at a Certain Distance II?

Always start with a brute-force solution to understand the problem better. Look for patterns in the problem that can help you optimize your solution. Be mindful of edge cases, especially when dealing with distances and connections.

What are common mistakes when solving Count the Number of Houses at a Certain Distance II?

Not considering the special case of the additional street correctly. Forgetting to handle the 1-based indexing of the result array.

#3017

Count the Number of Houses at a Certain Distance II

Hard

Graph TheoryPrefix SumGraph TheoryPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal approach leverages the structure of the problem and the properties of distances in a linear graph. We can calculate the number of pairs at each distance more efficiently without checking every pair.

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Algorithm

3 steps

1Step 1: Initialize a result array of size n with all zeros.
2Step 2: Calculate the total pairs for distance k without the additional street: result[k] = 2 * (n - k).
3Step 3: Adjust the counts for the special case of the additional street between x and y.

solution.py8 lines

1def countHouses(n, x, y):
2    result = [0] * n
3    for k in range(1, n + 1):
4        result[k - 1] = 2 * (n - k)
5    if abs(x - y) <= n:
6        distance = abs(x - y)
7        result[distance - 1] -= 2
8    return result

ℹ

Complexity note: The time complexity is O(n) because we only loop through the distances once. The space complexity is O(n) due to the result array.

1Understanding the structure of the graph helps to optimize the solution.
2Counting pairs based on distance rather than checking each pair directly saves time.

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