<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 Solution {
public:
    class Disjoint{
        public : 
        vector<int>sz, par ;
        Disjoint(int n ){
            sz.resize(n +1,1) ; 
            par.resize(n +1 , 0 ) ; 
            for(int i = 0 ; i <=n;i++){
                par[i] =   i ; 
            } 

        }
        int findPar(int u ){
            if(u== par[u]) return  u ;
            return par[u] =  findPar(par[u]) ; 
        }
        void merge(int u, int v){
            int vp =  findPar(u) ; 
            int up =  findPar(v) ; 

            if(sz[vp] >  sz[up]){
                sz[vp]+=  sz[up] ; 
                par[up]  = vp  ; 
               
            }
            else{
                sz[up]+=  sz[vp] ; 
                par[vp]  = up  ; 
             
            }
        }
        int give(int n  ) { 
            return sz[findPar(n)] ; 
        }
    }; 
    int findLatestStep(vector<int>& arr, int m) {
        int n  = arr.size()  ; 
        Disjoint ds(n) ; 
        int tot =  0 ; 
        vector<int>vis(n+1, 0) ; 
        int latest  = -1 ; 
        for(int i = 0 ; i <n;i++){
           vis[arr[i]] = 1 ;  
           if(arr[i]>=2 && vis[arr[i]-1] ==1 ){
                if(ds.give(arr[i]-1)==  m ) tot-- ; 
                 ds.merge(arr[i] , arr[i]-1) ; 
           }    
           if(arr[i]+1<=n && vis[arr[i]+1]==1){
            if(ds.give(arr[i]+1)==m) tot-- ; 
          ds.merge(arr[i] , arr[i]+1) ; 
           }
           int z  =  ds.give(arr[i]) ; 
           if(z==  m) tot++ ;
         
           if(tot){
            latest =  i+1; 
           }
           

        }
        
        return latest ; 
    }
};
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