<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]:
        # . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . 
        #                 |_______________|
        # tL (tail left)   l (loop)              tR (tail right)

        # no effect (loop = 0)
        if x == y or abs(x - y) == 1:
            return [max(n - i - 1, 0) * 2 for i in range(n)]
        if not x < y:
            x, y = y, x
        ans = [0] * n
        l = y - x + 1
        # count pairs within the loop
        for i in range(l // 2):
            ans[i] += (l * 2) if (i + 1 < l // 2 or l % 2 == 1) else l
        tL = x - 1
        tR = n - y
        # print(ans)
        # count pairs in each tail
        for t in [tL, tR]:
            for i in range(max(t-1, 0)):
                ans[i] += (t - i - 1) * 2
        # print(ans)
        # count pairs that starts at a tail, and 
        # ends in the loop or the other tail 
        def process(t1, t2):
            # print(t1, t2)
            # print(ans)
            if t1 == 0:
                return

            base = [0] * n
            for i in range(2, t2 + 2):
                base[i] += 1
            base[0] = 2
            left = 2
            right = l
            cur = 1
            while left <= right:
                # print(left, right, "lr")
                if left != right:
                    base[cur] += 2 * 2
                else:
                    base[cur] += 1 * 2
                left += 1
                right -= 1
                cur += 1
            # print(base, "base")
            cumsum = 0
            for i in range(n):
                cumsum += base[i]
                if i >= t1:
                    cumsum -= base[i - t1]
                ans[i] += cumsum

        process(tL, tR)
        process(tR, tL)
        return ans

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