<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:
    def checkIfCanBreak(self, s1: str, s2: str) -> bool:
        if len(s1) != len(s2): return False
        # alp = 'abcdefghijklmnopqrstuvwxyz'

        s1counts = [0] * 26
        s2counts = [0] * 26

        for c1, c2 in zip(s1, s2):
            s1counts[ord(c1) - ord('a')] += 1
            s2counts[ord(c2) - ord('a')] += 1
        

        def canBreak(s1counts, s2counts):
            s = 0
            for i in range(0, 26):
                s += s1counts[i] - s2counts[i]
                if s < 0:
                    return False
            return True
        
        return canBreak(s1counts, s2counts) or canBreak(s2counts, s1counts)
        
17800