<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:
   struct ItemsStepInfo {
        int left  = 0;
        int right = 0;
        int value = 0;
    };

    int findLatestStep(std::vector<int>& arr, int m) {
        std::vector<ItemsStepInfo> group;
        group.resize(arr.size());

        int step = 0;
        auto merge = [&group, m, &step](int i) {
            int next  = i + 1;
            int prev  = i - 1;
            int value = 1;
            if (i > 0 && next < group.size() && group[prev].value > 0 && group[next].value > 0) {
                if (group[prev].value == m)
                    --step;
                if (group[next].value == m)
                    --step;
                int left           = group[prev].left;
                int right          = group[next].right;
                group[left].right  = right;
                value              = group[prev].value + group[next].value + 1;
                group[left].value  = value;
                group[right].left  = left;
                group[right].value = value;
            }
            else if ((i == 0 || group[prev].value == 0) && next < group.size() && group[next].value > 0) {
                if (group[next].value == m)
                    --step;
                int right          = group[next].right;
                group[right].left  = i;
                group[i].left      = i;
                group[i].right     = right;
                value              = group[next].value + 1;
                group[right].value = value;
                group[i].value     = value;
            }
            else if ((next == group.size() || group[next].value == 0) && i > 0 && group[prev].value > 0) {
                if (group[prev].value == m)
                    --step;
                int left          = group[prev].left;
                group[left].right = i;
                group[i].right    = i;
                group[i].left     = left;
                value             = group[prev].value + 1;
                group[left].value = value;
                group[i].value    = value;
            }
            else {
                group[i].right = i;
                group[i].left  = i;
                group[i].value = 1;
            }
            if (value == m)
                ++step;
        };

        int pos = -1;
        for (int i = 0; i < arr.size(); ++i) {
            merge(arr[i] - 1);
            if (step > 0)
                pos = i + 1;
        }
        return pos;
    }
};
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