<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) {
        unordered_map<int, int> groups; // <length of substr, number of substr>
        unordered_map<int, int> index_to_length; // pos, length of substr

        int ans = -1;
        // string binray_string('0', arr.size() + 1); // 1 index

        for (int i = 0; i < arr.size(); i += 1) {
            int len = 1;
            int left_len = 0, right_len = 0, start = arr[i], end = arr[i];
            // use a vector with size = arr.size() + 2
            if (arr[i] > 1 &&
                index_to_length.find(arr[i] - 1) != index_to_length.end()) {
                left_len = index_to_length[arr[i] - 1];
                index_to_length.erase(arr[i] - 1);
                groups[left_len] -= 1;
                start = arr[i] - 1 - left_len + 1;
            }
            if (arr[i] < arr.size() &&
                index_to_length.find(arr[i] + 1) != index_to_length.end()) {
                right_len = index_to_length[arr[i] + 1];
                index_to_length.erase(arr[i] + 1);
                groups[right_len] -= 1;
                end = arr[i] + 1 + right_len - 1;
            }

            len += left_len + right_len;
            index_to_length[start] = len;
            index_to_length[end] = len;
            groups[len] += 1;
            if (groups[m] > 0) { // check substring with len m
                ans = i + 1;
            }
        }
        return ans;
    }
};
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