<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 {
    int find(int n, vector<int>& parent) {
        while (n != parent[n]) n = parent[n];
        return n;
    }

    void unite(int u, int v, vector<int>& parent, vector<int>& size) {
        int root_u = find(u, parent);
        int root_v = find(v, parent);

        if (root_u != root_v) {
            if (size[root_u] > size[root_v]) {
                parent[root_v] = root_u;
                size[root_u] += size[root_v];
            } else {
                parent[root_u] = root_v;
                size[root_v] += size[root_u];
            }
        }
    }

public:
    int findLatestStep(vector<int>& arr, int m) {
        int n = arr.size();

        vector<int> parent(n + 1, 0);
        vector<int> size(n + 1, 1);
        unordered_map<int, int> groupCount;
        int ans = -1;

        for (int i = 0; i < n; i++) {
            int flip = arr[i];
            parent[flip] = flip;

            if (flip > 1 && parent[flip - 1] != 0) {
                int leftGroup = find(flip - 1, parent);

                if (--groupCount[size[leftGroup]] == 0) {
                    groupCount.erase(size[leftGroup]);
                }

                unite(flip - 1, flip, parent, size);
            }

            if (flip < n && parent[flip + 1] != 0) {
                int rightGroup = find(flip + 1, parent);
                
                if (--groupCount[size[rightGroup]] == 0) {
                    groupCount.erase(size[rightGroup]);
                }

                unite(flip + 1, flip, parent, size);
            }

            int newGroup = find(flip, parent);
            groupCount[size[newGroup]]++;

            if (groupCount[m] > 0) {
                ans = i + 1;
            }
        }

        return ans;
    }
};

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