<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>

class Solution {
public:
    bool rotateString(string s, string goal) {
        if(s.size() != goal.size()){return false;}

        for(int i = 0; i < s.size(); ++i){
            if(s[0] == goal[i]){
                if(s.substr(0, goal.size() - i) == goal.substr(i, goal.size() - i) 
                && s.substr(goal.size()-i, i) == goal.substr(0, i)){
                    return true;
                }
            }
        }
        return false;
    }
};
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