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
class Solution {
    public long[] countOfPairs(int n, int x, int y) {
        --x;
        --y;
        if (x > y) {
            int temp = x;
            x = y;
            y = temp;
        }
        long[] diff = new long[n];
        for (int i = 0; i < n; ++i) {
            diff[0] += 1 + 1;
            ++diff[Math.min(Math.abs(i - x), Math.abs(i - y) + 1)];
            ++diff[Math.min(Math.abs(i - y), Math.abs(i - x) + 1)];
            --diff[Math.min(Math.abs(i - 0), Math.abs(i - y) + 1 + Math.abs(x - 0))];
            --diff[Math.min(Math.abs(i - (n - 1)), Math.abs(i - x) + 1 + Math.abs(y - (n - 1)))];
            --diff[Math.max(x - i, 0) + Math.max(i - y, 0) + ((y - x) + 0) / 2];
            --diff[Math.max(x - i, 0) + Math.max(i - y, 0) + ((y - x) + 1) / 2];
        }
        for (int i = 0; i + 1 < n; ++i) {
            diff[i + 1] += diff[i];
        }
        return diff;
    }
}
