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

Count the Number of Houses at a Certain Distance I (LeetCode #3015) 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 problem allows for more efficient counting of pairs.

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

The optimal solution for Count the Number of Houses at a Certain Distance I runs in O(n) time and uses O(n) space. The time complexity is O(n) because we iterate through the houses a limited number of times. 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 I?

The brute force approach for Count the Number of Houses at a Certain Distance I 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 I?

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

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

Always clarify the problem statement and constraints before diving into coding. Think about edge cases, such as when x equals y or when n is small. Discuss your thought process and approach with the interviewer to get feedback.

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

Not considering the special connection between x and y correctly. Failing to account for pairs that are the same house.

#3015

Count the Number of Houses at a Certain Distance I

Medium

Breadth-First SearchGraph TheoryPrefix SumGraph TheoryBreadth-First Search

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)

In the optimal approach, we can leverage the structure of the problem and the properties of distances in a linear arrangement of houses. We can count pairs directly based on their distances without checking each pair.

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Algorithm

3 steps

1Step 1: Initialize a result array of size n with all zeros.
2Step 2: Calculate the number of pairs for distance 1, which includes all adjacent houses and the special connection between x and y.
3Step 3: For distances greater than 1, calculate pairs based on the distance formula and the total number of houses.

solution.py16 lines

1# Full working Python code
2
3def countHouses(n, x, y):
4    result = [0] * n
5    # Count pairs for distance 1
6    result[0] = (n - 1) * 2  # All adjacent pairs + (x, y) and (y, x) if they are adjacent
7    if abs(x - y) == 1:
8        result[0] += 2
9    # Count pairs for distance 2 and more
10    for k in range(2, n + 1):
11        if k <= n:
12            result[k - 1] = (n - k) * 2
13    return result
14
15# Example usage
16print(countHouses(3, 1, 3))  # Output: [6, 0, 0]

ℹ

Complexity note: The time complexity is O(n) because we iterate through the houses a limited number of times. The space complexity is O(n) due to the result array.

1Understanding the structure of the problem allows for more efficient counting of pairs.
2Utilizing properties of distances in a linear arrangement can simplify calculations.

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