<p>You are given three <strong>positive</strong> integers <code>n</code>, <code>x</code>, and <code>y</code>.</p>

<p>In a city, there exist houses numbered <code>1</code> to <code>n</code> connected by <code>n</code> streets. There is a street connecting the house numbered <code>i</code> with the house numbered <code>i + 1</code> for all <code>1 &lt;= i &lt;= n - 1</code> . An additional street connects the house numbered <code>x</code> with the house numbered <code>y</code>.</p>

<p>For each <code>k</code>, such that <code>1 &lt;= k &lt;= n</code>, you need to find the number of <strong>pairs of houses</strong> <code>(house<sub>1</sub>, house<sub>2</sub>)</code> such that the <strong>minimum</strong> number of streets that need to be traveled to reach <code>house<sub>2</sub></code> from <code>house<sub>1</sub></code> is <code>k</code>.</p>

<p>Return <em>a <strong>1-indexed</strong> array </em><code>result</code><em> of length </em><code>n</code><em> where </em><code>result[k]</code><em> represents the <strong>total</strong> number of pairs of houses such that the <strong>minimum</strong> streets required to reach one house from the other is </em><code>k</code>.</p>

<p><strong>Note</strong> that <code>x</code> and <code>y</code> can be <strong>equal</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/12/20/example2.png" style="width: 474px; height: 197px;" />
<pre>
<strong>Input:</strong> n = 3, x = 1, y = 3
<strong>Output:</strong> [6,0,0]
<strong>Explanation:</strong> Let&#39;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.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/12/20/example3.png" style="width: 668px; height: 174px;" />
<pre>
<strong>Input:</strong> n = 5, x = 2, y = 4
<strong>Output:</strong> [10,8,2,0,0]
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/12/20/example5.png" style="width: 544px; height: 130px;" />
<pre>
<strong>Input:</strong> n = 4, x = 1, y = 1
<strong>Output:</strong> [6,4,2,0]
<strong>Explanation:</strong> 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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= x, y &lt;= n</code></li>
</ul>

class Solution:
    def countOfPairs(self, n: int, x: int, y: int) -> List[int]:
        answer = [0] * n
        def add(array):
            for i, x in enumerate(array):
                answer[i] += x
        def add_line(length):
            add(range(2 * length, -1, -2))
        def add_loop(loop):
            if loop % 2 == 1:
                add([2 * loop] * (loop // 2))
            else:
                add([2 * loop] * (loop // 2 - 1) + [loop])
        def add_fork(prongs):
            a = [0]
            for i in range(2, sum(prongs) + 1):
                a.append(2 * min(min(prongs), i - 1, sum(prongs) + 1 - i))
            add(a)
        def add_stemmed_fork(prongs):
            a = [0, 0]
            for i in range(3, sum(prongs) + 2):
                a.append(2 * min(min(prongs), i - 2, sum(prongs) + 2 - i))
            add(a)
            
        x -= 1
        y -= 1
        x, y = min(x, y), max(x, y)
        tails = sorted((x, n - 1 - y))
        if x == y or x == y - 1:
            add_line(n - 1)
        elif tails[1] == 0:
            add_loop(n)
        elif tails[0] == 0:
            tail = tails[1]
            loop = n - tail
            add_line(tail)
            add_loop(loop)
            add_fork((tail, (loop - 1) // 2))
            add_fork((tail, loop // 2))
        else:
            loop = n - sum(tails)
            add_loop(loop)
            add_line(tails[0])
            add_line(tails[1])
            add_stemmed_fork(tails)
            add_fork((tails[0], (loop - 1) // 2))
            add_fork((tails[0], loop // 2))
            add_fork((tails[1], (loop - 1) // 2))
            add_fork((tails[1], loop // 2))
        return answer
1065