Problem Description

You have a city with n houses numbered from 1 to n. These houses are connected by streets in a linear fashion - house i connects to house i+1 for all houses from 1 to n-1. Additionally, there's one extra street that directly connects house x to house y.

Your task is to find, for each possible distance k (where 1 ≤ k ≤ n), how many pairs of houses have exactly k as their minimum distance.

The distance between two houses is the minimum number of streets you need to travel to get from one house to the other. Since the extra street between house x and house y creates a shortcut, you need to consider all possible paths:

• The direct path along the sequential streets
• Paths that use the shortcut from x to y

For each pair of houses (house₁, house₂), you need to:

1. Calculate the shortest path between them
2. Count this pair based on the minimum distance found

The output should be a 1-indexed array of length n where result[k] represents the total number of house pairs that have minimum distance k. Note that pairs (i, j) and (j, i) are counted as two separate pairs.

For example, if houses 3 and 5 have a minimum distance of 2 streets between them, this contributes 2 to the count for distance 2 (counting both (3,5) and (5,3)).

The challenge lies in correctly calculating the shortest path considering the shortcut street, which might provide a faster route than the sequential path for certain house pairs.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The problem explicitly involves houses (nodes) connected by streets (edges). We have houses numbered 1 to n with connections between consecutive houses, plus an additional edge between house x and house y. This forms a graph structure.

Is it a tree?

• No: A tree has no cycles, but our graph has a cycle due to the extra street connecting house x to house y (creating a shortcut/cycle in what would otherwise be a linear path).

Is the problem related to directed acyclic graphs?

• No: The graph is undirected (streets can be traveled in both directions) and contains a cycle.

Is the problem related to shortest paths?

• Yes: We need to find the minimum number of streets (shortest path) between each pair of houses.

Is the graph Weighted?

• No: All streets have the same weight/cost of 1 (each street counts as 1 in the distance calculation).

BFS

• Yes: Since we're dealing with an unweighted graph and need to find shortest paths, BFS is the appropriate algorithm.

Conclusion: The flowchart suggests using BFS (Breadth-First Search) for finding shortest paths in this unweighted graph. BFS is perfect for finding shortest paths in unweighted graphs because it explores nodes level by level, guaranteeing that when we first reach a node, we've found the shortest path to it.

For this specific problem, we can either:

1. Run BFS from each house to find distances to all other houses
2. Use the mathematical approach shown in the solution (which essentially computes what BFS would find) by considering all possible paths: direct sequential path, or paths using the x-y shortcut

Intuition

When we have houses connected in a line from 1 to n with an additional shortcut between houses x and y, we need to think about how this shortcut affects the distances between any two houses.

For any pair of houses (i, j), there are essentially three possible paths to consider:

1. Direct sequential path: Just walk along the houses in order. The distance is simply |i - j|.

2. Path using the shortcut from x to y: Go from house i to house x, take the shortcut to house y, then continue to house j. The total distance is |i - x| + 1 + |j - y|.

3. Path using the shortcut from y to x: Go from house i to house y, take the shortcut to house x, then continue to house j. The total distance is |i - y| + 1 + |j - x|.

The key insight is that the shortest path between any two houses must be one of these three options. We don't need to run BFS from every house because the graph structure is simple enough that we can directly calculate all possible paths mathematically.

Since we need to count pairs for each distance k, we can:

• Enumerate all possible pairs (i, j) where i < j
• Calculate the minimum distance using the three path options
• Increment the count for that distance by 2 (counting both (i, j) and (j, i))

This approach works because the graph has a very specific structure - it's essentially a line graph with one additional edge. This limited structure means we can exhaustively check all possible routing options without needing to explore the graph dynamically. The solution transforms what could be a graph traversal problem into a simple enumeration and calculation problem.

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

Solution Approach

The implementation follows a brute force enumeration approach where we examine every possible pair of houses and calculate their minimum distance.

Step 1: Index Adjustment

x, y = x - 1, y - 1

Convert from 1-indexed to 0-indexed for easier array manipulation.

Step 2: Initialize Result Array

ans = [0] * n

Create an array of size n to store the count of pairs for each distance from 1 to n.

Step 3: Enumerate All Pairs

for i in range(n):
    for j in range(i + 1, n):

We iterate through all unique pairs (i, j) where i < j. This ensures we don't count the same pair twice and don't consider a house paired with itself.

Step 4: Calculate Three Possible Distances For each pair (i, j), we calculate:

• a = j - i: The direct sequential distance (since j > i, this is always positive)
• b = abs(i - x) + 1 + abs(j - y): Distance going from i to x, taking the shortcut to y, then to j
• c = abs(i - y) + 1 + abs(j - x): Distance going from i to y, taking the shortcut to x, then to j

The +1 in paths b and c accounts for the shortcut street itself between houses x and y.

Step 5: Find Minimum and Update Count

ans[min(a, b, c) - 1] += 2

• Find the minimum of the three distances
• Subtract 1 to convert to 0-indexed array position
• Add 2 to the count because both (i, j) and (j, i) are valid pairs with the same distance

Time Complexity: O(n²) as we examine all pairs of houses Space Complexity: O(n) for the result array

This solution elegantly avoids the need for actual graph traversal by leveraging the specific structure of the problem - a linear arrangement with one additional edge.

Example Walkthrough

Let's walk through a concrete example with n = 5, x = 2, y = 4.

Setup:

• Houses: 1, 2, 3, 4, 5 connected sequentially
• Extra shortcut: between house 2 and house 4
• After converting to 0-indexed: x = 1, y = 3

Visual representation:

1 --- 2 --- 3 --- 4 --- 5
      |___________|
       (shortcut)

Step-by-step calculation for each pair:

Let's examine pair (1, 4) which is (0, 3) in 0-indexed:

• Direct path: 3 - 0 = 3 streets
• Path via x→y: |0-1| + 1 + |3-3| = 1 + 1 + 0 = 2 streets
• Path via y→x: |0-3| + 1 + |3-1| = 3 + 1 + 2 = 6 streets
• Minimum = 2, so we add 2 to ans[1]

Let's examine pair (2, 5) which is (1, 4) in 0-indexed:

• Direct path: 4 - 1 = 3 streets
• Path via x→y: |1-1| + 1 + |4-3| = 0 + 1 + 1 = 2 streets
• Path via y→x: |1-3| + 1 + |4-1| = 2 + 1 + 3 = 6 streets
• Minimum = 2, so we add 2 to ans[1]

Complete enumeration of all pairs:

Final result:

• Distance 1: 10 pairs (houses that are adjacent or connected by shortcut)
• Distance 2: 8 pairs (houses that benefit from the shortcut)
• Distance 3: 2 pairs (houses too far to benefit from shortcut)
• Distance 4: 0 pairs
• Distance 5: 0 pairs

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

The shortcut between houses 2 and 4 creates more pairs with distance 1 (including the direct shortcut connection) and allows some pairs that would have distance 3 or 4 to achieve distance 2 instead.

Solution Implementation

Time and Space Complexity

The time complexity is O(n^2), where n is the value given in the problem. This is because the code uses two nested loops - the outer loop runs from 0 to n-1, and the inner loop runs from i+1 to n-1. The total number of iterations is (n-1) + (n-2) + ... + 1 = n(n-1)/2, which simplifies to O(n^2). Inside the nested loops, all operations (calculating distances a, b, c, finding the minimum, and updating the answer array) take constant time O(1).

The space complexity is O(1) when excluding the space consumed by the answer array ans. The algorithm only uses a constant amount of extra space for variables x, y, i, j, a, b, and c, regardless of the input size. If we include the answer array in the space complexity analysis, it would be O(n) since the answer array has exactly n elements.

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

Common Pitfalls

1. Forgetting to Handle the Case When x = y

When the shortcut connects a house to itself (x = y), the shortcut effectively doesn't exist. The current solution handles this correctly since the distance calculations naturally work out, but it's easy to overlook this edge case.

Why it works: When x = y, the paths via shortcut become:

• path_via_x_to_y = abs(i - x) + 1 + abs(j - x)
• path_via_y_to_x = abs(i - x) + 1 + abs(j - x)

Both paths are identical and add an unnecessary +1, making them always longer than or equal to the direct path.

2. Index Conversion Errors

The problem uses 1-indexed houses while Python uses 0-indexing. A common mistake is:

• Forgetting to convert x and y to 0-indexed at the beginning
• Converting the wrong way (adding 1 instead of subtracting 1)
• Forgetting to subtract 1 when updating the result array

Solution: Always convert immediately at the start and be consistent:

x, y = x - 1, y - 1  # Convert to 0-indexed
# ...
result[min_distance - 1] += 2  # min_distance is 1-indexed, array is 0-indexed

3. Counting Pairs Incorrectly

A frequent error is adding only 1 instead of 2 for each pair, forgetting that (i, j) and (j, i) are counted as separate pairs.

Wrong:

result[min_distance - 1] += 1  # Only counts one direction

Correct:

result[min_distance - 1] += 2  # Counts both (i,j) and (j,i)

4. Incorrect Distance Calculation for Shortcut Paths

Forgetting the +1 for the shortcut street itself when calculating paths through x-y:

Wrong:

path_via_x_to_y = abs(i - x) + abs(j - y)  # Missing the shortcut street

Correct:

path_via_x_to_y = abs(i - x) + 1 + abs(j - y)  # +1 for the shortcut street

5. Not Considering All Three Paths

Some might assume the shortcut always provides a shorter path or only consider two paths (direct and one via shortcut). You must check all three possibilities:

1. Direct sequential path
2. Path using shortcut from x to y
3. Path using shortcut from y to x (different when traveling in opposite direction)

6. Loop Boundary Issues

Using incorrect loop boundaries that either miss pairs or count duplicates:

Wrong:

for i in range(n):
    for j in range(n):  # Includes i=j and counts each pair twice

Correct:

for i in range(n):
    for j in range(i + 1, n):  # Only unique pairs where i < j