<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:
    int findLatestStep(vector<int>& arr, int m) {
        int n = arr.size();
        vector<int> ranks(n, 1);
        vector<int> parents(n, 0);
        vector<int> bin(n, 0);
        unordered_map<int,int> M;
        for (int i = 0; i < n; i++) {
            parents[i] = i;
        }
        int latest = -1;
        for (int i = 0; i < n; i++) {
            int idx = arr[i]-1;
            bin[idx] = 1;
            M[1]++;
            if (idx-1>=0 && bin[idx-1]==1) {
                Union(idx, idx-1, parents, ranks, m, M);
            }
            if (idx+1<n && bin[idx+1]==1) {
                Union(idx, idx+1, parents, ranks, m, M);
            }
            if (M[m]>0) {
                latest = i+1;
            }
        }
        return latest;
    }
    int find(vector<int> & parents, int p) {
        while(p != parents[p]) {
            parents[p] = parents[parents[p]];
            p = parents[p];
        }
        return p;
    }
    void Union(int i, int j, vector<int> & parents, vector<int> & ranks, int m, unordered_map<int,int>& M) {
        int pi = find(parents, i);
        int pj = find(parents, j);
        if (ranks[pi]>ranks[pj]) {
            M[ranks[pi]]--;
            M[ranks[pj]]--;
            ranks[pi]+=ranks[pj];
            M[ranks[pi]]++;
            parents[pj] = pi;
        }
        else {
            M[ranks[pi]]--;
            M[ranks[pj]]--;
            ranks[pj]+=ranks[pi];
            M[ranks[pj]]++;
            parents[pi] = pj;
        }
        return;
    }
};
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