<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>

#include <vector>
using namespace std;

class DisjointSet {
public:
    vector<int> gp;
    vector<int> parent, size;

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

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

    void UnionBySize(int u, int v) {
        int ulu = findPar(u);
        int ulv = findPar(v);
        gp[size[ulu]]--;
        gp[size[ulv]]--;
        if (ulu == ulv) return;
        if (size[ulu] > size[ulv]) {
            parent[ulv] = ulu;
            size[ulu] += size[ulv];
            gp[size[ulu]]++;
        } else {
            parent[ulu] = ulv;
            size[ulv] += size[ulu];
            gp[size[ulv]]++;
        }
    }
};

class Solution {
public:
    int findLatestStep(vector<int>& arr, int m) {
        int n = arr.size();
        vector<int> vis(n + 1, 0);
        DisjointSet ds(n);
        int ans = -1;

        for (int i = 0; i < n; i++) {
            ds.gp[1]++;
            vis[arr[i]] = 1;

            if (arr[i] - 1 > 0 && vis[arr[i] - 1]) {
                ds.UnionBySize(arr[i], arr[i] - 1);
            }
            if (arr[i] + 1 <= n && vis[arr[i] + 1]) {
                ds.UnionBySize(arr[i], arr[i] + 1);
            }
            if (ds.gp[m] > 0) ans = i + 1;
        }
        return ans;
    }
};

91700