<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{
    vector<int> parent, size;
public:
    DisjointSet(int n){
        for (int i=0; i<n; i++) parent.push_back(i);

        size.resize(n, 0);
    }

    int findParent(int u) {
        if (u == parent[u]) return u;
        return parent[u] = findParent(parent[u]);
    }

    void set(int i){
        size[i] = 1;
        return;
    }

    int getSize(int i) {
        return size[findParent(i)];
    }

    void unionBySize(int u, int v, int& count, int m) {        
        int pU = findParent(u);
        int pV = findParent(v);

        if (pU == pV) return;

        if (size[pU] == m) count--;
        if (size[pV] == m) count--;

        if (size[pU] <= size[pV]) {
            size[pV] += size[pU];
            parent[pU] = pV;
            if (size[pV] == m) count++;
        }
        else {
            size[pU] += size[pV];
            parent[pV] = pU;
            if (size[pU] == m) count++;
        }

        
        return;
    }
};

class Solution {
public:
    int findLatestStep(vector<int>& arr, int m) {
        int n = arr.size();
        DisjointSet* ds = new DisjointSet(n+1);

        int ans = -1, count = 0;

        for (int i=0; i<arr.size(); i++) {
            int prevIdx = arr[i]-1, nextIdx = arr[i]+1;
            ds->set(arr[i]); 

            if (m==1) count++;                                   

            if (prevIdx>=0 and ds->getSize(prevIdx)) {
                ds->unionBySize(arr[i], prevIdx, count, m);
            }

            if (nextIdx<=n and ds->getSize(nextIdx)) {
                ds->unionBySize(arr[i], nextIdx, count, m);
            }

            // for (int j=1; j<=n; j++) {
            //     cout << ds->getSize(j) << " ";                
            // }
            // cout << endl;                
            if (count!=0) ans = i+1;
        }

        return ans;
    }
};
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