Problem Description

The problem presents a scenario simulating a city with houses numbered from 1 to n, all connected sequentially by a street, making the distance between houses i and i+1 just one street. Additionally, there is another direct street connecting house x to house y. The objective is to calculate the number of pairs of houses such that the minimum number of streets required to travel from one house to another is k, for all k between 1 and n, inclusive.

The challenge is set within the boundaries of these constraints to find an efficient way to perform the computations for all k. The result is an array where each index k (1-indexed) contains the total number of pairs of houses at k distance apart.

An important note is that houses x and y may be the same, making the additional street a cycle with one house.

Flowchart Walkthrough

To determine the most appropriate algorithm for solving Leetcode problem 3017, "Count the Number of Houses at a Certain Distance II", let's step through the problem using the algorithm flowchart available at the Flowchart:

Is it a graph?

• Yes: The problem features trees that can be considered as a graph representation where nodes are the dwellings and the edges are the roads connecting them.

Is it a tree?

• Yes: Although the primary data structure is trees, it's important to note that there are multiple trees situated within a virtual grid (or graph). Each tree is considered a sub-graph on its own.

Is the problem related to directed acyclic graphs (DAGs)?

• No: Trees by nature are undirected and acyclic, but the question doesn't involve the acyclic properties in its problem-solving. Thus, we shift away from DAG-specific solutions.

Since it's a tree and we're not looking into DAG properties further, and the problem is indeed looking towards connectivity (as distances between nodes in trees concern which nodes are connected and reachable within a given distance):

• We skip deeper exploration into shortest paths (related more to weighted graphs), directed issues, or backtracking that are alternative pathways on the chart.

Is the problem related to shortest paths?

• Yes: While shortest paths usually suggest minimum path considerations, here the applicable notion is ensuring correct distance measurement which aligns with BFS capabilities in an unweighted graph.

Is the graph weighted?

• No: The problem doesn't specify weights on the distances between houses; only the integer count for distance matters, typically aligning with BFS use in unweighted scenarios.

Conclusion: Following the decision path in the flowchart steers us towards utilizing Breadth-First Search (BFS). BFS is ideally suited for traversing trees (a part of the graph category) to explore level-by-level (or layer-by-layer). This fits perfectly for finding all nodes at a certain exact distance from a given node in unweighted and undirected graphs like trees, especially when no specific direction or weighted criteria influence the connectivity.

Using BFS would allow handling each node exactly once and maintaining a queue to explore each node's neighbors ensures we can count all dwellings within the desired distance efficiently.

Intuition

When approaching this problem, it's essential to consider the pattern and symmetry in the distance between houses. The existence of the additional street between houses x and y potentially creates shortcuts for some pairs of houses, affecting the total count differently, depending on the distance.

First, we observe that without the additional street, the number of pairs of houses that are k streets apart follows a predictable pattern: it starts high when k is 1 and linearly decreases to 0 as k approaches n.

With the added street, this pattern gets disrupted. If x and y are right next to each other or the same, it effectively does not change the count for any k, so we simply use the initial pattern.

However, if there's a gap between x and y, this creates a cycle that allows for shortcuts. The solution's strategy is to calculate the effect of this cycle on the count of pairs for each k.

To calculate these effects, the solution breaks down the problem into different segments: between the two endpoints x, y of the street establishing the cycle, within the cycle itself, and from the cycle endpoints to the ends of the range from 1 to n.

We then meticulously calculate the effects of distance k being within these different segments and how they add up to the total number of pairs. Special care is taken when the cycle length is odd or even, as it affects the symmetry.

The solution constructs the result progressively:

• It starts with the base count without considering the additional street.
• It then adds the effects of the cycle.
• Finally, it adjusts the counts for pairs that use the parts of the city beyond the cycle, considering both sides of the cycle endpoints up to houses numbered 1 and n.

The solution employs a methodical approach to map out all possible pairs without missing any edge cases, creating a comprehensive and correct count for each k.

Learn more about Breadth-First Search, Graph and Prefix Sum patterns.

Solution Approach

The implementation of the solution focuses on calculating the number of pairs (house_1, house_2) for each distance k. The primary data structure used is a list, res, which is initialized to reflect the number of pairs when no additional street is considered. These calculations start from the farthest distance n - 1 and move down to the shortest. The reason for this initialization is that at maximum distance n - 1, there are only two pairs of houses, and at a distance of n - 2, there are four pairs, and so on.

Once the res list is initialized, the algorithm considers the existence of the additional street between x and y, creating a cycle with a certain length, named cycle_len. This cycle introduces new pairs at various distances.

The first step in analyzing the cycle's impact is to consider the distances within the cycle itself, for which the algorithm initializes another list, res2. The number of pairs is more considerable in the middle of the cycle (at distance cycle_len // 2) and decreases towards the ends of the cycle. The algorithm makes adjustments for the case when the cycle length is odd or even, as an even cycle has a single longest distance, while an odd cycle has two equal longest distances.

Next, the algorithm handles distances that span outside the cycle, considering the two "tails" from each cycle endpoint (x and y) to the ends of the city (1 and n). Each "tail" is processed by another loop, providing a sub-list res3 representing the additional pairs formed by using the "tail" sections. The changes to the count of pairs are determined by the relative positions of the tail to the cycle and incrementally affect the counts as the distance k becomes larger than the tails themselves. Special cases occur when the cycle length is even, where adjustments are made accordingly.

Finally, each possible distance k's pairs are summed up in res by combining the base pairs, the pairs within the cycle, and the pairs utilizing the tails. The resulting list res is returned, and it reflects the complete count of pairs for each distance k, taking into account the direct street between houses x and y.

The entire approach requires careful consideration of symmetry and edge cases, mainly when the cycle's length affects the balance. Mathematical formulas and multiple passes over different parts of the city ensure that all pairs are counted correctly. The algorithm's multiple loops and conditional statements serve to navigate through the problem's space methodically and compute the required values for each case.

Example Walkthrough

Let's illustrate the solution approach with a small example where n = 5 (we have 5 houses), and there is an additional street between house x = 2 and house y = 5.

Initially, we initialize our result array res based on the natural sequence without considering the additional street:

• For k = 1, we have: (1,2), (2,3), (3,4), (4,5) – 4 pairs.
• For k = 2, we have: (1,3), (2,4), (3,5) – 3 pairs.
• For k = 3, we have: (1,4), (2,5) – 2 pairs.
• For k = 4, we have: (1,5) – 1 pair.

So, res = [0, 4, 3, 2, 1]. We're ignoring res[0] as our answers are 1-indexed.

Next, we consider the additional street which forms a cycle. The cycle length cycle_len is the distance from x to y which is 3 (from house 2 to house 5). Now, we look at pairs within the cycle (res2). There are two cases to think of: when k is less than cycle_len and when k is greater than cycle_len.

Starting from k = 1 up to k = cycle_len - 1, we notice the additional street doesn't create new pairs for k = 1 since it directly connects only houses 2 and 5, but it does introduce 1 new pair for k = 2: (2,5). So, res2 would be [0, 0, 1].

Furthermore, we analyze the "tails" from the ends of the cycle to the end of the city. For example, with k = 3, we can now go from house 1 to house 5 via houses 2 and 3 because of the additional street. So we add 1 to res[3]. Similarly, for k = 4, we can include pairs like (1,2) and (4,5) that are now at the distance because of the cycle, adding 2 to res[4].

Considering these effects, we update res by adding res2 and the contributions from the "tails":

• For k = 1, no change since no new pairs are created by the additional street.
• For k = 2, add one pair: (2,5).
• For k = 3, add one pair: (1,5).
• For k = 4, add two pairs: (1,2) and (4,5).

Updating res, our final answer becomes:

• res = [0, 4, 4, 3, 3] for k from 1 to 4, respectively.

So, our result array finally denotes how the existence of the additional street between houses 2 and 5 impacts the number of house pairs that are k streets apart for each k value.

Solution Implementation

Time and Space Complexity

The given code snippet is designed to calculate pairs in some specific context. Analyzing its computational complexity involves examining each section of the code and the loops that are present throughout.

Time Complexity

The time complexity can be broken down as follows:

• The check if abs(x - y) <= 1 is a constant-time operation, O(1).
• The calculation of cycle_len, n2, and initialization of res = [2 * x for x in reversed(range(n2))] has time complexity O(n), as it involves creating a list of size proportional to the input n.
• The while loop to fill the rest of res with zeros runs in O(n-cycle_len) time in the worst case. Since cycle_len is at most n, this can also be bounded by O(n).
• Initializing res2 is O(cycle_len) which can be, at most, O(n).
• The adjustment to the res2[0] and conditionally to res2[-1] is done in constant time, O(1).
• The for loop that updates res with elements of res2 runs at most cycle_len iterations, which can be bounded by O(n).
• The for loop to handle tail1 and tail2 runs twice. Within this loop:
  • Initializing res3's maximum length can be tail, which ultimately is less than n, hence O(n).
  • The inner for loop runs up to val_mx / 4 times, again limited by tail and hence O(n).
  • The final for loop updating res with res3 is bounded by O(n) as it iterates over the length of res3 which is less than or equal to n.

Hence, the overall time complexity is dominated by the number of times we run through operations proportional to n, which happens several times but not nested. Therefore, the cumulative time complexity is O(n).

Space Complexity

For space complexity, we consider the additional space used by the algorithm (without including the space used to store the input arguments):

• The space used by res, res2, res3 lists is the most significant part. Each list can at most have a length comparable to n since tail, tail1, and tail2 are derived from n.
• All other variables (cycle_len, n2, val_mx, i_mx, etc.) use constant space.

The overall space complexity of the method is O(n) due to the lists res, res2, and res3, which dominate the space requirements.

Learn more about how to find time and space complexity quickly using problem constraints.