<p>Given an array <code>arr</code> that represents a permutation of numbers from <code>1</code> to <code>n</code>.</p>

<p>You have a binary string of size <code>n</code> that initially has all its bits set to zero. At each step <code>i</code> (assuming both the binary string and <code>arr</code> are 1-indexed) from <code>1</code> to <code>n</code>, the bit at position <code>arr[i]</code> is set to <code>1</code>.</p>

<p>You are also given an integer <code>m</code>. Find the latest step at which there exists a group of ones of length <code>m</code>. A group of ones is a contiguous substring of <code>1</code>&#39;s such that it cannot be extended in either direction.</p>

<p>Return <em>the latest step at which there exists a group of ones of length <strong>exactly</strong></em> <code>m</code>. <em>If no such group exists, return</em> <code>-1</code>.</p>

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

<pre>
<strong>Input:</strong> arr = [3,5,1,2,4], m = 1
<strong>Output:</strong> 4
<strong>Explanation:</strong> 
Step 1: &quot;00<u>1</u>00&quot;, groups: [&quot;1&quot;]
Step 2: &quot;0010<u>1</u>&quot;, groups: [&quot;1&quot;, &quot;1&quot;]
Step 3: &quot;<u>1</u>0101&quot;, groups: [&quot;1&quot;, &quot;1&quot;, &quot;1&quot;]
Step 4: &quot;1<u>1</u>101&quot;, groups: [&quot;111&quot;, &quot;1&quot;]
Step 5: &quot;111<u>1</u>1&quot;, groups: [&quot;11111&quot;]
The latest step at which there exists a group of size 1 is step 4.
</pre>

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

<pre>
<strong>Input:</strong> arr = [3,1,5,4,2], m = 2
<strong>Output:</strong> -1
<strong>Explanation:</strong> 
Step 1: &quot;00<u>1</u>00&quot;, groups: [&quot;1&quot;]
Step 2: &quot;<u>1</u>0100&quot;, groups: [&quot;1&quot;, &quot;1&quot;]
Step 3: &quot;1010<u>1</u>&quot;, groups: [&quot;1&quot;, &quot;1&quot;, &quot;1&quot;]
Step 4: &quot;101<u>1</u>1&quot;, groups: [&quot;1&quot;, &quot;111&quot;]
Step 5: &quot;1<u>1</u>111&quot;, groups: [&quot;11111&quot;]
No group of size 2 exists during any step.
</pre>

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

<ul>
	<li><code>n == arr.length</code></li>
	<li><code>1 &lt;= m &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= arr[i] &lt;= n</code></li>
	<li>All integers in <code>arr</code> are <strong>distinct</strong>.</li>
</ul>

class Node {
public:
    int p, rank, cnt;
    Node() : p(0), rank(0), cnt(1) {}
    Node(int p) {
        this->p = p;
        rank = 0;
        cnt = 0;
    }
};
class Solution {
public:
    int find(vector<Node>& p, int x) {
        if (x == p[x].p) {
            return x;
        }
        return p[x].p = find(p, p[x].p);
    }
    int _union(vector<Node>& p, int n1, int n2) {
        int a = find(p, n1), b = find(p, n2);
        if (a == b) {
            // same
            return p[a].cnt;
        }
        if (p[a].rank > p[b].rank) {
            p[b].p = a;
            p[a].cnt += p[b].cnt;
            return p[a].cnt;
        } else if (p[a].rank < p[b].rank) {
            p[a].p = b;
            p[b].cnt += p[a].cnt;
            return p[b].cnt;
        } else {
            p[a].p = b;
            p[b].cnt += p[a].cnt;
            p[b].rank += 1;
            return p[b].cnt;
        }
    }
    int findLatestStep(vector<int>& arr, int m) {
        int n = arr.size();
        vector<Node> parents(n+1);
        string s(n+1, '0');
        for (int i = 1; i <= n; ++i) {
            parents[i] = Node(i);
        }
        map<int, int> check;
        int ans = -1;
        int step = 1;
        for (auto a : arr) {
            s[a] = '1';
            parents[a].cnt = 1;
            int tmp = 1;
            if (a > 1 && s[a-1] == '1') {
                int parent = find(parents, a-1);
                check[parents[parent].cnt] -= 1;
                tmp = _union(parents, a, a-1);
            }
            if (a+1 <= n && s[a+1] == '1') {
                int parent = find(parents, a+1);
                check[parents[parent].cnt] -= 1;
                tmp = _union(parents, a, a+1);
            }
            check[tmp]++;
            if (check[m] > 0) ans = step;
            step++;
        }
        return ans;
    }
};
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