<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>par;
    vector<int>size;
    public:
    vector<int>groupCount;
    DisjointSet(int n)
    {
        par.resize(n,1);
        size.resize(n,1);
        groupCount.resize(n+1,0);

        for(int i=0;i<n;i++)
            par[i]=i;
    }

    int findUPar(int v)
    {
        if(par[v]==v)
            return v;
        
      return  par[v]=findUPar(par[v]);
    }

    void unionBySize(int u , int v)
    {
        int upar_u=findUPar(u);
        int upar_v=findUPar(v);
        if(upar_u==upar_v)
            return;
        groupCount[size[upar_u]]--;
        groupCount[size[upar_v]]--;
        if(size[upar_u]<size[upar_v])
        {    
            size[upar_u]+=size[upar_v];
            groupCount[size[upar_u]]++;
            par[upar_v]=upar_u;
        }
        else
        {
            size[upar_v]+=size[upar_u];
            groupCount[size[upar_v]]++;
            par[upar_u]=upar_v;
        }
    }

    int findSize(int v)
    {
        int upar_v=findUPar(v);
        return size[upar_v];
    }

};
class Solution {
public:
    int findLatestStep(vector<int>& arr, int m) {
        

        int n = arr.size();
        vector<bool>isSet(n,false);
       
        DisjointSet ds(n);
        int ans=-1;
        for(int i=0;i<n;i++)
        {
            ds.groupCount[1]++;
            int currIdx=arr[i]-1;
            if(currIdx-1>=0 and isSet[currIdx-1])
            {
                ds.unionBySize(currIdx,currIdx-1);
            }

            if(currIdx+1<n and isSet[currIdx+1])
                 ds.unionBySize(currIdx,currIdx+1);
            if(ds.groupCount[m])
                ans=i+1;
            isSet[currIdx]=true;
        }

        return ans;
    }
};
88800