<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>

# WQWRQQQW   size = 2  {Q = 4, W = 3} -> window {Q = 2, W = 1}
class Solution:
    def balancedString(self, s: str) -> int:
        n = len(s); size = n / 4
        actualCharToCount = { "Q": 0, "W": 0, "E": 0, "R": 0 }
        charToCount = {}        

        for c in s:
            actualCharToCount[c] += 1

        for x, y in actualCharToCount.items():
            if y > size:
                charToCount[x] = y - size
        
        if len(charToCount) == 0: return 0

        i = 0; substringLen = n; count = len(charToCount)
        for (j, c) in enumerate(s):
            if c in charToCount:
                charToCount[c] -= 1
                if charToCount[c] == 0: count -= 1
            
            while count == 0:
                substringLen = min(substringLen, j - i + 1)
                if s[i] in charToCount:
                    charToCount[s[i]] += 1
                    if charToCount[s[i]] > 0: count += 1
                i += 1

        return substringLen





        
98