<p>You are given a binary string <code>s</code>. In one second, <strong>all</strong> occurrences of <code>&quot;01&quot;</code> are <strong>simultaneously</strong> replaced with <code>&quot;10&quot;</code>. This process <strong>repeats</strong> until no occurrences of <code>&quot;01&quot;</code> exist.</p>

<p>Return<em> the number of seconds needed to complete this process.</em></p>

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

<pre>
<strong>Input:</strong> s = &quot;0110101&quot;
<strong>Output:</strong> 4
<strong>Explanation:</strong> 
After one second, s becomes &quot;1011010&quot;.
After another second, s becomes &quot;1101100&quot;.
After the third second, s becomes &quot;1110100&quot;.
After the fourth second, s becomes &quot;1111000&quot;.
No occurrence of &quot;01&quot; exists any longer, and the process needed 4 seconds to complete,
so we return 4.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;11100&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong>
No occurrence of &quot;01&quot; exists in s, and the processes needed 0 seconds to complete,
so we return 0.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
	<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong></p>

<p>Can you solve this problem in O(n) time complexity?</p>

class Solution:
    def secondsToRemoveOccurrences(self, s: str) -> int:
        ans = prefix = prev = 0 
        for i, ch in enumerate(s): 
            if ch == '1': 
                ans = max(prev, i - prefix)
                prefix += 1
                if ans: prev = ans+1
        return ans 
        # n0=0
        # dp=[0]* (len(s))
        # for i, n in enumerate(s):
        #     if n=='1':
        #         dp[i]=1
        #     else:
        #         n0+=1
        # if not n0:
        #     return 0
        
        # k=0
        # while sum(dp[-n0:]):
        #     i=0
        #     while i< len(s)-1:
        #         if dp[i]==0 and dp[i+1]==1:
        #             dp[i+1],dp[i]=dp[i],dp[i+1]               
        #             i+=2
        #         else:
        #             i+=1
        #     k+=1
        # return k
        

16000