<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>

struct DisjointSetUnion{

    vector<int> rank, parent, size;

    DisjointSetUnion(int n){
        rank.resize(n+1,0);
        parent.resize(n+1);
        size.resize(n+1);
        for(int i=0; i<=n; i++){
            parent[i]=i;
            size[i]=1;
        }
    }

    int FindUltimateParent(int node){
        if(node==parent[node])
            return node;
        return parent[node]=FindUltimateParent(parent[node]);
    }

    void UnionByRank(int u, int v){

        int Ult_u = FindUltimateParent(u);
        int Ult_v = FindUltimateParent(v);

        if(Ult_u==Ult_v) return;

        if(rank[Ult_u]<rank[Ult_v]){
            parent[Ult_u]=Ult_v;
            return;
        }

        else if(rank[Ult_v]<rank[Ult_u]){
            parent[Ult_v]=Ult_u;
            return;
        }

        parent[Ult_u]=Ult_v;
        rank[Ult_v]++;
        return;
    }

    void UnionBySize(int u, int v){
        int Ult_u = FindUltimateParent(u);
        int Ult_v = FindUltimateParent(v);

        if(Ult_u==Ult_v) return;

        if(size[Ult_u]<size[Ult_v]){
            parent[Ult_u]=Ult_v;
            size[Ult_v]+=size[Ult_u];
            return;
        }else{
            parent[Ult_v]=Ult_u;
            size[Ult_u]+=size[Ult_v];
            return;            
        }        
    }
};

class Solution {
public:
    int findLatestStep(vector<int>& arr, int m) {
        int n=arr.size();
        vector<int> hash(n,0);
        DisjointSetUnion ds(n);
        multiset<int> st;
        int ans=-1;
        for(int i=0; i<n; i++){
            int h=arr[i]-1;
            hash[h]=1;
            int a=0, b=0;
            if(h-1>=0 && hash[h-1]==1){
                a=ds.size[ds.FindUltimateParent(h-1)];
                ds.UnionBySize(h,h-1);
            }
            if(h+1<n && hash[h+1]==1){
                b=ds.size[ds.FindUltimateParent(h+1)];
                ds.UnionBySize(h,h+1);
            }
            if(a>0) st.erase(st.find(a));
            if(b>0) st.erase(st.find(b));
            st.insert(a+b+1);
            if(st.contains(m)) ans=i+1;
        }
        return ans;
    }
};
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