<p>Given a string <code>s</code> containing only lowercase English letters and the <code>&#39;?&#39;</code> character, convert <strong>all </strong>the <code>&#39;?&#39;</code> characters into lowercase letters such that the final string does not contain any <strong>consecutive repeating </strong>characters. You <strong>cannot </strong>modify the non <code>&#39;?&#39;</code> characters.</p>

<p>It is <strong>guaranteed </strong>that there are no consecutive repeating characters in the given string <strong>except </strong>for <code>&#39;?&#39;</code>.</p>

<p>Return <em>the final string after all the conversions (possibly zero) have been made</em>. If there is more than one solution, return <strong>any of them</strong>. It can be shown that an answer is always possible with the given constraints.</p>

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

<pre>
<strong>Input:</strong> s = &quot;?zs&quot;
<strong>Output:</strong> &quot;azs&quot;
<strong>Explanation:</strong> There are 25 solutions for this problem. From &quot;azs&quot; to &quot;yzs&quot;, all are valid. Only &quot;z&quot; is an invalid modification as the string will consist of consecutive repeating characters in &quot;zzs&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;ubv?w&quot;
<strong>Output:</strong> &quot;ubvaw&quot;
<strong>Explanation:</strong> There are 24 solutions for this problem. Only &quot;v&quot; and &quot;w&quot; are invalid modifications as the strings will consist of consecutive repeating characters in &quot;ubvvw&quot; and &quot;ubvww&quot;.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 100</code></li>
	<li><code>s</code> consist of lowercase English letters and <code>&#39;?&#39;</code>.</li>
</ul>

class Solution:
    def modifyString(self, s: str) -> str:
        pp=list(s)
        a=s.count("?")
        # print(a)
        u=0
        yt=[]
        for h in range(len(s)):
            if(s[h]=="?"):
                yt.append(h)
        # if(s=="j?qg??b"):
        #     print(s.index("?",4),"tttt")
        for i in range(a):
            ore=0
            # print(i)
            for k in range(26):
                pp[yt[u]]=chr(97+ore)
                try:
                    din=0
                    try:
                        if(ord(pp[yt[u]])!=ord(pp[yt[u]+1])):
                            din+=1
                    except:
                        din+=1
                    try:
                        if(ord(pp[yt[u]])!=ord(pp[yt[u]-1])):
                            din+=1
                    except:
                        din+=1
                    if(din==2):
                        break
                except:
                    break
                ore+=1
            u+=1
        # print(pp)
        ip=""
        for j in range(len(pp)):
            ip+=pp[j]
        return ip

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