<p>In a string composed of <code>&#39;L&#39;</code>, <code>&#39;R&#39;</code>, and <code>&#39;X&#39;</code> characters, like <code>&quot;RXXLRXRXL&quot;</code>, a move consists of either replacing one occurrence of <code>&quot;XL&quot;</code> with <code>&quot;LX&quot;</code>, or replacing one occurrence of <code>&quot;RX&quot;</code> with <code>&quot;XR&quot;</code>. Given the starting string <code>start</code> and the ending string <code>end</code>, return <code>True</code> if and only if there exists a sequence of moves to transform <code>start</code> to <code>end</code>.</p>

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

<pre>
<strong>Input:</strong> start = &quot;RXXLRXRXL&quot;, end = &quot;XRLXXRRLX&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> We can transform start to end following these steps:
RXXLRXRXL -&gt;
XRXLRXRXL -&gt;
XRLXRXRXL -&gt;
XRLXXRRXL -&gt;
XRLXXRRLX
</pre>

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

<pre>
<strong>Input:</strong> start = &quot;X&quot;, end = &quot;L&quot;
<strong>Output:</strong> false
</pre>

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

<ul>
	<li><code>1 &lt;= start.length&nbsp;&lt;= 10<sup>4</sup></code></li>
	<li><code>start.length == end.length</code></li>
	<li>Both <code>start</code> and <code>end</code> will only consist of characters in <code>&#39;L&#39;</code>, <code>&#39;R&#39;</code>, and&nbsp;<code>&#39;X&#39;</code>.</li>
</ul>

class Solution {
public:
    bool canTransform(string s, string e) {
        vector<int> sr,er,sl,el;
        int n = s.size();
        for(int i=0;i<n;i++){
            if(s[i]=='R') sr.push_back(i);
            if(s[i]=='L') sl.push_back(i);
        }
        for(int i=0;i<n;i++){
            if(e[i]=='R') er.push_back(i);
            if(e[i]=='L') el.push_back(i);
        }
        if(sl.size()!=el.size() || sr.size()!=er.size()) return 0;
        // int ind=sl[0];
        for(int i=0;i<sr.size();i++){
            if(sr[i] > er[i]) return 0;
            for(int j = sr[i]+1;j<=er[i];j++){
                if(s[j]=='L') return 0;
            }
        }
        for(int i=sl.size()-1;i>=0;i--){
            if(sl[i] < el[i]) return 0;
            for(int j = sl[i]-1;j>=el[i];j--){
                if(s[j]=='R') return 0;
            }
        }
        
        return 1;
    }
};
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