<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 DisjointSet {
public:
    DisjointSet(int n) 
        : parent(vector<int>(n, -1)),
          size(vector<int>(n, 0)),
          count(vector<int>(n, 0)) {
    }
    
    void add(int x) {
        parent[x] = x;
        size[x] = 1;
        ++count[1];
    }
    
    int find(int x) {
        if (x == parent[x]) {
            return x;
        }
        
        parent[x] = find(parent[x]);
        return parent[x];
    } 
    
    void unite(int x, int y) {
        int px = find(x);
        int py = find(y);
        
        if (px == py) {
            return;
        }
        
        if (size[px] >= size[py]) {
            --count[size[px]];
            --count[size[py]];
            
            size[px] += size[py];
            parent[py] = px;
            
            ++count[size[px]];
        } else {
            --count[size[px]];
            --count[size[py]];
            
            size[py] += size[px];
            parent[px] = py;
            
            ++count[size[py]];
        }
    }
    
    bool query(int m) {
        return count[m];
    }
    
private:
    vector<int> parent;
    vector<int> size;
    vector<int> count;
};

class Solution {
public:
    int findLatestStep(vector<int>& arr, int m) {
        int n = arr.size();
        
        DisjointSet set(n + 1);
        string str(n + 1, '0');
        
        int last = -1;
        int step = 1;
        for (int i : arr) {
            str[i] = '1';
            set.add(i);
            
            if (i >= 0 && str[i - 1] == '1') {
                set.unite(i - 1, i);
            }
            if (i < n && str[i + 1] == '1') {
                set.unite(i, i + 1);
            }
            
            if (set.query(m)) {
                last = step;
            }
            ++step;
        }
        
        return last;
    }
};
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