<p>You are given a string <code>text</code>. You can swap two of the characters in the <code>text</code>.</p>

<p>Return <em>the length of the longest substring with repeated characters</em>.</p>

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

<pre>
<strong>Input:</strong> text = &quot;ababa&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> We can swap the first &#39;b&#39; with the last &#39;a&#39;, or the last &#39;b&#39; with the first &#39;a&#39;. Then, the longest repeated character substring is &quot;aaa&quot; with length 3.
</pre>

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

<pre>
<strong>Input:</strong> text = &quot;aaabaaa&quot;
<strong>Output:</strong> 6
<strong>Explanation:</strong> Swap &#39;b&#39; with the last &#39;a&#39; (or the first &#39;a&#39;), and we get longest repeated character substring &quot;aaaaaa&quot; with length 6.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> text = &quot;aaaaa&quot;
<strong>Output:</strong> 5
<strong>Explanation:</strong> No need to swap, longest repeated character substring is &quot;aaaaa&quot; with length is 5.
</pre>

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

<ul>
	<li><code>1 &lt;= text.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>text</code> consist of lowercase English characters only.</li>
</ul>

class Solution {
public:
    int maxRepOpt1(string text) {
        unordered_map<char,vector<pair<int,int>>> mp;
        int n=text.length();
        int ma=0;
        vector<char> temp;
        for(int i=0;i<n;i++){
            int start=i;
            int end=i;
            while(i<n-1 and text[i]==text[i+1]){
                i++;end++;
            }
            mp[text[start]].push_back({start,end});
            if(end-start+1>ma){
                temp.clear();
                ma=max(ma,end-start+1);
            }
            if(end-start+1==ma)
                temp.push_back(text[start]);
        }
        // cout<<temp.size()<<endl;
        // for(int i=0;i<temp.size();i++) if(mp[temp[i]].size()>1) {ma=ma+1;break;}
        for(auto &[x,v]:mp){
            int si=v.size();
            for(int i=0;i<si-1;i++){
                if(v[i].second+2==v[i+1].first){
                    if(si==2){
                        ma=max(ma,v[i+1].second-v[i].first);
                    }else{
                        ma=max(ma,v[i+1].second-v[i].first+1);
                    }
                }else{
                    ma=max(ma,1+max(v[i].second-v[i].first+1,v[i+1].second-v[i+1].first+1));
                }
            }
        }
        return ma;
    }
};
10600