<p>You are given a <strong>0-indexed</strong> binary string <code>s</code> and two integers <code>minJump</code> and <code>maxJump</code>. In the beginning, you are standing at index <code>0</code>, which is equal to <code>&#39;0&#39;</code>. You can move from index <code>i</code> to index <code>j</code> if the following conditions are fulfilled:</p>

<ul>
	<li><code>i + minJump &lt;= j &lt;= min(i + maxJump, s.length - 1)</code>, and</li>
	<li><code>s[j] == &#39;0&#39;</code>.</li>
</ul>

<p>Return <code>true</code><i> if you can reach index </i><code>s.length - 1</code><i> in </i><code>s</code><em>, or </em><code>false</code><em> otherwise.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;<u>0</u>11<u>0</u>1<u>0</u>&quot;, minJump = 2, maxJump = 3
<strong>Output:</strong> true
<strong>Explanation:</strong>
In the first step, move from index 0 to index 3. 
In the second step, move from index 3 to index 5.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;01101110&quot;, minJump = 2, maxJump = 3
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
	<li><code>s[0] == &#39;0&#39;</code></li>
	<li><code>1 &lt;= minJump &lt;= maxJump &lt; s.length</code></li>
</ul>

class Solution:
    def canReach(self, s: str, minJump: int, maxJump: int) -> bool:
        n = len(s)
        if s[n - 1] == '1':
            return False
        
        line = 0
        delta = Counter()
        for i in range(n):
            line += delta[i]
            if s[i] == '0' and (i == 0 or line > 0):
                a, b = i + minJump, i + maxJump + 1
                delta[a] += 1
                delta[b] -= 1
            # print("iii", i, line, dict(delta))
        return line > 0


        # @lru_cache(None)
        # def dfs(i: int) -> bool:
        #     if i == n - 1:
        #         return True
        #     for j in range(min(i + maxJump, n - 1), i + minJump - 1, -1):
        #         if s[j] == '0' and dfs(j):
        #             return True
        #     return False
        # return dfs(0)

        # reachable, visited = {0}, {0}
        # while reachable:
        #     newreachable = set()
        #     for i in reachable:
        #         for j in range(i + minJump, min(i + maxJump + 1, n)):
        #             if s[j] == '1' or j in visited:
        #                 continue
        #             visited.add(j)
        #             newreachable.add(j)
        #     reachable = newreachable
        # return n-1 in visited
        
422