Process Restricted Friend Requests
<p>You are given an integer <code>n</code> indicating the number of people in a network. Each person is labeled from <code>0</code> to <code>n - 1</code>.</p>

<p>You are also given a <strong>0-indexed</strong> 2D integer array <code>restrictions</code>, where <code>restrictions[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> means that person <code>x<sub>i</sub></code> and person <code>y<sub>i</sub></code> <strong>cannot </strong>become <strong>friends</strong>,<strong> </strong>either <strong>directly</strong> or <strong>indirectly</strong> through other people.</p>

<p>Initially, no one is friends with each other. You are given a list of friend requests as a <strong>0-indexed</strong> 2D integer array <code>requests</code>, where <code>requests[j] = [u<sub>j</sub>, v<sub>j</sub>]</code> is a friend request between person <code>u<sub>j</sub></code> and person <code>v<sub>j</sub></code>.</p>

<p>A friend request is <strong>successful </strong>if <code>u<sub>j</sub></code> and <code>v<sub>j</sub></code> can be <strong>friends</strong>. Each friend request is processed in the given order (i.e., <code>requests[j]</code> occurs before <code>requests[j + 1]</code>), and upon a successful request, <code>u<sub>j</sub></code> and <code>v<sub>j</sub></code> <strong>become direct friends</strong> for all future friend requests.</p>

<p>Return <em>a <strong>boolean array</strong> </em><code>result</code>,<em> where each </em><code>result[j]</code><em> is </em><code>true</code><em> if the </em><code>j<sup>th</sup></code><em> friend request is <strong>successful</strong> or </em><code>false</code><em> if it is not</em>.</p>

<p><strong>Note:</strong> If <code>u<sub>j</sub></code> and <code>v<sub>j</sub></code> are already direct friends, the request is still <strong>successful</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 3, restrictions = [[0,1]], requests = [[0,2],[2,1]]
<strong>Output:</strong> [true,false]
<strong>Explanation:
</strong>Request 0: Person 0 and person 2 can be friends, so they become direct friends. 
Request 1: Person 2 and person 1 cannot be friends since person 0 and person 1 would be indirect friends (1--2--0).
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 3, restrictions = [[0,1]], requests = [[1,2],[0,2]]
<strong>Output:</strong> [true,false]
<strong>Explanation:
</strong>Request 0: Person 1 and person 2 can be friends, so they become direct friends.
Request 1: Person 0 and person 2 cannot be friends since person 0 and person 1 would be indirect friends (0--2--1).
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 5, restrictions = [[0,1],[1,2],[2,3]], requests = [[0,4],[1,2],[3,1],[3,4]]
<strong>Output:</strong> [true,false,true,false]
<strong>Explanation:
</strong>Request 0: Person 0 and person 4 can be friends, so they become direct friends.
Request 1: Person 1 and person 2 cannot be friends since they are directly restricted.
Request 2: Person 3 and person 1 can be friends, so they become direct friends.
Request 3: Person 3 and person 4 cannot be friends since person 0 and person 1 would be indirect friends (0--4--3--1).
</pre>

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

<ul>
	<li><code>2 &lt;= n &lt;= 1000</code></li>
	<li><code>0 &lt;= restrictions.length &lt;= 1000</code></li>
	<li><code>restrictions[i].length == 2</code></li>
	<li><code>0 &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= n - 1</code></li>
	<li><code>x<sub>i</sub> != y<sub>i</sub></code></li>
	<li><code>1 &lt;= requests.length &lt;= 1000</code></li>
	<li><code>requests[j].length == 2</code></li>
	<li><code>0 &lt;= u<sub>j</sub>, v<sub>j</sub> &lt;= n - 1</code></li>
	<li><code>u<sub>j</sub> != v<sub>j</sub></code></li>
</ul>
Union Find; Graph
func friendRequests(n int, restrictions [][]int, requests [][]int) (ans []bool) {
	p := make([]int, n)
	for i := range p {
		p[i] = i
	}
	var find func(int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	for _, req := range requests {
		pu, pv := find(req[0]), find(req[1])
		if pu == pv {
			ans = append(ans, true)
		} else {
			ok := true
			for _, r := range restrictions {
				px, py := find(r[0]), find(r[1])
				if px == pu && py == pv || px == pv && py == pu {
					ok = false
					break
				}
			}
			ans = append(ans, ok)
			if ok {
				p[pv] = pu
			}
		}
	}
	return
}