#3017

Hard

LeetCode Problem Workspace

Count the Number of Houses at a Certain Distance II

Count the Number of Houses at a Certain Distance II is a graph problem that requires efficient pair counting with an additional edge between two houses.

graph prefix sum

Problem Statement

You are given three integers n, x, and y. There are n houses numbered 1 to n, connected by n-1 streets, with an additional street connecting house x to house y.

For each integer k from 1 to n, you must compute how many pairs of houses (house1, house2) exist such that the minimum number of streets between them is exactly k.

Examples

Example 1

Input: n = 3, x = 1, y = 3

Output: [6,0,0]

Let's look at each pair of houses:

For the pair (1, 2), we can go from house 1 to house 2 directly.
For the pair (2, 1), we can go from house 2 to house 1 directly.
For the pair (1, 3), we can go from house 1 to house 3 directly.
For the pair (3, 1), we can go from house 3 to house 1 directly.
For the pair (2, 3), we can go from house 2 to house 3 directly.
For the pair (3, 2), we can go from house 3 to house 2 directly.

Example 2

Input: n = 5, x = 2, y = 4

Output: [10,8,2,0,0]

For each distance k the pairs are:

For k == 1, the pairs are (1, 2), (2, 1), (2, 3), (3, 2), (2, 4), (4, 2), (3, 4), (4, 3), (4, 5), and (5, 4).
For k == 2, the pairs are (1, 3), (3, 1), (1, 4), (4, 1), (2, 5), (5, 2), (3, 5), and (5, 3).
For k == 3, the pairs are (1, 5), and (5, 1).
For k == 4 and k == 5, there are no pairs.

Example 3

Input: n = 4, x = 1, y = 1

Output: [6,4,2,0]

For each distance k the pairs are:

For k == 1, the pairs are (1, 2), (2, 1), (2, 3), (3, 2), (3, 4), and (4, 3).
For k == 2, the pairs are (1, 3), (3, 1), (2, 4), and (4, 2).
For k == 3, the pairs are (1, 4), and (4, 1).
For k == 4, there are no pairs.

Constraints

2 <= n <= 105
1 <= x, y <= n

Solution Approach

Graph Representation

Represent the houses and streets as a graph, where each house is a node and each street is an edge. The extra street between houses x and y adds complexity, requiring a specific handling for distances between those two nodes.

Prefix Sum Usage

Prefix sum can be used to quickly calculate the number of pairs of houses at each distance. Without the extra edge, the number of pairs at a distance k is easily determined by 2 * (n - k). This helps avoid recalculating pair distances multiple times.

Efficient Counting

Combine graph traversal with prefix sum to efficiently compute the number of valid pairs for each distance. Special care is needed to count pairs affected by the additional street between houses x and y, adjusting distances accordingly.

Complexity Analysis

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

The time and space complexities depend on the approach chosen. A direct approach using graph traversal could result in O(n^2) complexity, while optimizing with prefix sums or dynamic programming can reduce this significantly.

What Interviewers Usually Probe

Look for a clear understanding of graph traversal and its integration with prefix sums.
Evaluate the candidate's ability to handle edge cases, especially the influence of the additional street.
Check if the candidate can optimize their approach to avoid unnecessary recalculations.

Common Pitfalls or Variants

Common pitfalls

Ignoring the impact of the additional street between houses x and y, which can alter the pair distances.
Incorrectly handling edge cases where there are no valid pairs for certain distances.
Failing to optimize the approach to handle larger inputs within the problem's constraints.

Follow-up variants

Modify the problem to ask for the number of pairs at exactly k distance for a given subset of houses.
Extend the problem to handle multiple additional streets between different pairs of houses.
Change the problem to ask for the distance between a specific pair of houses, rather than counting all pairs.

FAQ

What is the main pattern used to solve the Count the Number of Houses at a Certain Distance II problem?

The problem is based on Graph plus Prefix Sum techniques to calculate the number of house pairs at each distance efficiently.

How does the additional street between houses x and y affect the problem?

The extra street modifies the pair distances, requiring an adjusted calculation for the number of valid pairs for certain distances.

Can this problem be solved without using graph traversal?

While graph traversal is the most intuitive approach, optimizations using prefix sums can help avoid recalculating distances repeatedly, speeding up the solution.

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

The time complexity depends on the method used. A brute force approach could have a time complexity of O(n^2), while optimizations may reduce it.

What are the key challenges in this problem?

The main challenges are handling the additional street between houses x and y and efficiently counting valid pairs without excessive recalculations.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def countOfPairs(self, n: int, x: int, y: int) -> List[int]:
        if abs(x - y) <= 1:
            return [2 * x for x in reversed(range(n))]
        cycle_len = abs(x - y) + 1
        n2 = n - cycle_len + 2
        res = [2 * x for x in reversed(range(n2))]
        while len(res) < n:
            res.append(0)
        res2 = [cycle_len * 2] * (cycle_len >> 1)
        if not cycle_len & 1:
            res2[-1] = cycle_len
        res2[0] -= 2
        for i in range(len(res2)):
            res[i] += res2[i]
        if x > y:
            x, y = y, x
        tail1 = x - 1
        tail2 = n - y
        for tail in (tail1, tail2):
            if not tail:
                continue
            i_mx = tail + (cycle_len >> 1)
            val_mx = 4 * min((cycle_len - 3) >> 1, tail)
            i_mx2 = i_mx - (1 - (cycle_len & 1))
            res3 = [val_mx] * i_mx
            res3[0] = 0
            res3[1] = 0
            if not cycle_len & 1:
                res3[-1] = 0
            for i, j in enumerate(range(4, val_mx, 4)):
                res3[i + 2] = j
                res3[i_mx2 - i - 1] = j
            for i in range(1, tail + 1):
                res3[i] += 2
            if not cycle_len & 1:
                mn = cycle_len >> 1
                for i in range(mn, mn + tail):
                    res3[i] += 2
            for i in range(len(res3)):
                res[i] += res3[i]
        return res

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