<p>You are given a string s of length <code>n</code> containing only four kinds of characters: <code>&#39;Q&#39;</code>, <code>&#39;W&#39;</code>, <code>&#39;E&#39;</code>, and <code>&#39;R&#39;</code>.</p>

<p>A string is said to be <strong>balanced</strong><em> </em>if each of its characters appears <code>n / 4</code> times where <code>n</code> is the length of the string.</p>

<p>Return <em>the minimum length of the substring that can be replaced with <strong>any</strong> other string of the same length to make </em><code>s</code><em> <strong>balanced</strong></em>. If s is already <strong>balanced</strong>, return <code>0</code>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;QWER&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong> s is already balanced.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;QQWE&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> We need to replace a &#39;Q&#39; to &#39;R&#39;, so that &quot;RQWE&quot; (or &quot;QRWE&quot;) is balanced.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;QQQW&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> We can replace the first &quot;QQ&quot; to &quot;ER&quot;. 
</pre>

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

<ul>
	<li><code>n == s.length</code></li>
	<li><code>4 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>n</code> is a multiple of <code>4</code>.</li>
	<li><code>s</code> contains only <code>&#39;Q&#39;</code>, <code>&#39;W&#39;</code>, <code>&#39;E&#39;</code>, and <code>&#39;R&#39;</code>.</li>
</ul>

class Solution:
    def balancedString(self, s: str) -> int:
        """
        s contains only 'Q', 'W', 'E', and 'R'
        s.length is a multiple of 4

        Return the minimum length of the substring
        that can be replaced with any other string of the same length to make s balanced.
            (s is already balanced => 0)

        idea:
            outside the wdw:
            if all(counter.value <= n //4 )
                then it can be balanced by replacing wdw
        """
        ctr = Counter(s)
        n = len(s)
        min_len = n 
        i = 0
        for j, c in enumerate(s):
            ctr[c] -= 1
            if all(v <= n / 4 for v in ctr.values()):
                while i < n and all(v <= n / 4 for v in ctr.values()):
                    ctr[s[i]] += 1
                    i += 1
                # [i, j] can't , [i-1, j] can
                min_len = min(min_len, j - i + 2)
                
        return min_len
329