<p>Given two strings: <code>s1</code> and <code>s2</code> with the same&nbsp;size, check if some&nbsp;permutation of string <code>s1</code> can break&nbsp;some&nbsp;permutation of string <code>s2</code> or vice-versa. In other words <code>s2</code> can break <code>s1</code>&nbsp;or vice-versa.</p>

<p>A string <code>x</code>&nbsp;can break&nbsp;string <code>y</code>&nbsp;(both of size <code>n</code>) if <code>x[i] &gt;= y[i]</code>&nbsp;(in alphabetical order)&nbsp;for all <code>i</code>&nbsp;between <code>0</code> and <code>n-1</code>.</p>

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

<pre>
<strong>Input:</strong> s1 = &quot;abc&quot;, s2 = &quot;xya&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> &quot;ayx&quot; is a permutation of s2=&quot;xya&quot; which can break to string &quot;abc&quot; which is a permutation of s1=&quot;abc&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s1 = &quot;abe&quot;, s2 = &quot;acd&quot;
<strong>Output:</strong> false 
<strong>Explanation:</strong> All permutations for s1=&quot;abe&quot; are: &quot;abe&quot;, &quot;aeb&quot;, &quot;bae&quot;, &quot;bea&quot;, &quot;eab&quot; and &quot;eba&quot; and all permutation for s2=&quot;acd&quot; are: &quot;acd&quot;, &quot;adc&quot;, &quot;cad&quot;, &quot;cda&quot;, &quot;dac&quot; and &quot;dca&quot;. However, there is not any permutation from s1 which can break some permutation from s2 and vice-versa.
</pre>

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

<pre>
<strong>Input:</strong> s1 = &quot;leetcodee&quot;, s2 = &quot;interview&quot;
<strong>Output:</strong> true
</pre>

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

<ul>
	<li><code>s1.length == n</code></li>
	<li><code>s2.length == n</code></li>
	<li><code>1 &lt;= n &lt;= 10^5</code></li>
	<li>All strings consist of lowercase English letters.</li>
</ul>

class Solution {
private:
    bool solve(vector<char>& set1, vector<char>& set2){
        for(int i=0;i<set1.size();i++){
            if(set1[i]<set2[i]){
                return false;
            }
        }
        
    return true;    
    }
public:
    bool checkIfCanBreak(string s1, string s2) {
        vector<char> set1(s1.begin(),s1.end());
        vector<char> set2(s2.begin(),s2.end());

        sort(set1.begin(),set1.end());
        sort(set2.begin(),set2.end());

        return (solve(set1,set2) || solve(set2,set1));
        
    }
};
14300