<p>Given two strings <code>s</code> and <code>goal</code>, return <code>true</code> <em>if and only if</em> <code>s</code> <em>can become</em> <code>goal</code> <em>after some number of <strong>shifts</strong> on</em> <code>s</code>.</p>

<p>A <strong>shift</strong> on <code>s</code> consists of moving the leftmost character of <code>s</code> to the rightmost position.</p>

<ul>
	<li>For example, if <code>s = &quot;abcde&quot;</code>, then it will be <code>&quot;bcdea&quot;</code> after one shift.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s = "abcde", goal = "cdeab"
<strong>Output:</strong> true
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s = "abcde", goal = "abced"
<strong>Output:</strong> false
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

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


    // bool rotateString(string s, string t) {
    //      int n=s.size();
    //      int x=t.size();
    //      if(n!=x) return false;
    //      int first=t.find(s[0]);
    //     for (int i=0; i<n;i++){
          
    //         if(s[i]!=t[first]){
    //             return false;
    //         }
    //        first = (first + 1) % n; 
    //     }
    //     return true;
    
    // }
    // tle mar rha hai kyu ki t.c O(2n) hai
    class Solution {
public:
    bool rotateString(string s, string t) {
        int n = s.size();
        int m = t.size();

        // Check if lengths are not equal
        if (n != m) return false;

        // Concatenate s with itself
        string concatenated = s + s;

        // Check if t is a substring of concatenated
        return concatenated.find(t) != string::npos;
    }


};
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