<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() + 1;
        vector<int> par(n), sz(n, 1);
        iota(par.begin(), par.end(), 0);
        function<int(int)> get = [&](int v) -> int {
            if (v == par[v]) return v;
            return par[v] = get(par[v]);
        };
        function<void(int, int)> unite = [&](int u, int v) {
            u = get(u), v = get(v);
            if (u != v) {
                if (sz[u] < sz[v]) {
                    swap(u, v);
                }
                par[v] = u;
                sz[u] += sz[v];
            }
        };
        int out = -1;
        vector<int> active(n);
        vector<int> sizes(n);
        for (int i = 0; i < n - 1; i++) {
            if (arr[i] < n - 1 && active[arr[i] + 1]) {
                sizes[sz[get(arr[i] + 1)]]--;
                unite(arr[i], arr[i] + 1);
            }
            if (arr[i] > 0 && active[arr[i] - 1]) {
                sizes[sz[get(arr[i] - 1)]]--;
                unite(arr[i], arr[i] - 1);
            }
            active[arr[i]]++;
            sizes[sz[get(arr[i])]]++;
            if (sizes[m]) {
                out = i + 1;
            }
        }
        return out;
    }
};
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