<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:
    vector<int> mIdxList;
    vector<int> mSizeList;
    int getRootIndex(int idx){
        while(idx != mIdxList[idx]){
            idx = mIdxList[idx];
        }
        return idx;
    }
    map<int,int> mSizeMap;
    void connect(int id1,int id2){
        int root1 = getRootIndex(id1);
        int root2 = getRootIndex(id2);
        if(root1 == root2){
            return;
        }
        int size1 = mSizeList[root1];
        int size2 = mSizeList[root2];
        int tot = size1+size2;
        mIdxList[root2] = root1;
        mSizeList[root1] += size2;

        mSizeMap[tot]++;
        mSizeMap[size1]--;
        if(mSizeMap[size1] == 0){
            mSizeMap.erase(size1);
        }
        mSizeMap[size2]--;
        if(mSizeMap[size2] == 0){
            mSizeMap.erase(size2);
        }

    }
    int findLatestStep(vector<int>& arr, int m) {
        int N = arr.size();
        mIdxList.resize(N+1,0);
        for(int i = 1;i <= N;++i){
            mIdxList[i] = i;
        }
        mSizeList.resize(N+1,0);
        mSizeMap[0] = N;

        vector<int> nList(N+1,0);
        int step = 0;
        int lastone = -1;
        for(auto idx:arr){
            nList[idx] = 1;
            mSizeList[idx] = 1;
            mSizeMap[1]++;
            if(idx>1 && nList[idx-1] == 1){
                connect(idx,idx-1);
            }
            if(idx < N && nList[idx+1] == 1){
                connect(idx,idx+1);
            }
            step++;
            if(mSizeMap.count(m)){
                lastone = step;
            }
        }
        return lastone;
    }
};
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