<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 DSU:
    def __init__(self, size: int, occ_size: int):
        self.parent = list(range(size))
        self.cc_size = [0] * size
        self.size_occ = [0] * (occ_size + 1)
    
    def is_in(self, x: int) -> bool:
        return self.cc_size[x] > 0
    
    def add(self, p: int):
        self.cc_size[p] = 1
        self.size_occ[1] += 1
        
    def find(self, p: int) -> int:
        if self.parent[p] != p:
            self.parent[p] = self.find(self.parent[p])
        return self.parent[p]
    
    def join(self, p: int, q: int):
        pp, pq = self.find(p), self.find(q)
        if pp != pq:
            if self.cc_size[pp] <  len(self.size_occ):
                self.size_occ[self.cc_size[pp]] -= 1
            if self.cc_size[pq] <  len(self.size_occ):
                self.size_occ[self.cc_size[pq]] -= 1
            if self.cc_size[pq] + self.cc_size[pp] < len(self.size_occ):
                self.size_occ[self.cc_size[pq] + self.cc_size[pp]] += 1
            
            if self.cc_size[pq] > self.cc_size[pp]:
                pp, pq = pq, pp
            self.parent[pq] = pp
            self.cc_size[pp] += self.cc_size[pq]
        

class Solution:
    def findLatestStep(self, arr: List[int], m: int) -> int:
        n = len(arr)
        dsu = DSU(n, m)
        last = -1
        for i, x in enumerate(arr, start=1):
            x -= 1
            dsu.add(x)
            if x - 1 >= 0 and dsu.is_in(x - 1):
                dsu.join(x, x - 1)
                
            if x + 1 < n and dsu.is_in(x + 1):
                dsu.join(x, x + 1)
            
            if dsu.size_occ[-1]:
                last = i
        return last
            
            
                
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