<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:
    def findLatestStep(self, l: List[int], m: int) -> int:

        if len(l)==1:
            return 1
        if len(l)==2:
            return m

        n = len(l)
        arr = list(0 for _ in range(n+1))
        dic = defaultdict(int)
        ans = -1
        t = 1
    
        for i in l:
            if i!=1 and i!=n and arr[i-1]==0 and arr[i+1]==0:
                arr[i] = 1
            
                dic[1]+=1
            
            elif i==1 and arr[i+1]==0:
                arr[i] = 1
             
                dic[1]+=1
         
            elif i==n and arr[i-1]==0:
                arr[i] = 1
                
                dic[1]+=1
              
            elif i!=1 and i!=n and arr[i-1]!=0 and arr[i+1]!=0:
                leftmost = i - arr[i-1]
                leftValue = arr[i-1] 
                rightValue = arr[i+1]
                rightmost = i+arr[i+1]
                arr[leftmost] = leftValue + 1 + rightValue
                arr[rightmost] = leftValue + 1 + rightValue

                dic[leftValue + 1 + rightValue]+=1
                dic[leftValue]-=1
                dic[rightValue]-=1
             
            elif i!=1 and arr[i-1]!=0:
                leftmost = i - arr[i-1]
                leftValue = arr[i-1] 
                arr[leftmost] = leftValue + 1 
                arr[i] = leftValue + 1 
                dic[leftValue + 1 ]+=1
                dic[leftValue]-=1
            elif i!=n and arr[i+1]!=0:
                rightmost = i+arr[i+1]
                rightValue = arr[i+1]
                arr[rightmost] = 1 + rightValue
                arr[i] = 1 + rightValue
                dic[rightValue + 1 ]+=1
                dic[rightValue]-=1

           
            if dic[m]!=0:
                ans = t
            t+=1
       

        return ans

        





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