Count the Number of Houses at a Certain Distance II
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3000-3099/3017.Count%20the%20Number%20of%20Houses%20at%20a%20Certain%20Distance%20II/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3000-3099/3017.Count%20the%20Number%20of%20Houses%20at%20a%20Certain%20Distance%20II/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3000-3099/3017.Count%20the%20Number%20of%20Houses%20at%20a%20Certain%20Distance%20II/images/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>
Graph; Prefix Sum
func countOfPairs(n int, x int, y int) []int64 {
	if x > y {
		x, y = y, x
	}
	A := make([]int64, n)
	for i := 1; i <= n; i++ {
		A[0] += 2
		A[min(int64(i-1), int64(math.Abs(float64(i-y)))+int64(x))] -= 1
		A[min(int64(n-i), int64(math.Abs(float64(i-x)))+1+int64(n-y))] -= 1
		A[min(int64(math.Abs(float64(i-x))), int64(math.Abs(float64(y-i)))+1)] += 1
		A[min(int64(math.Abs(float64(i-x)))+1, int64(math.Abs(float64(y-i))))] += 1
		r := max(int64(x-i), 0) + max(int64(i-y), 0)
		A[r+int64((y-x+0)/2)] -= 1
		A[r+int64((y-x+1)/2)] -= 1
	}
	for i := 1; i < n; i++ {
		A[i] += A[i-1]
	}

	return A
}
