<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:
        counts = {key: 0 for key in ["Q", "W", "E", "R"]}
        for c in s:
            counts[c] += 1
        
        needs = {key: 0 for key in ["Q", "W", "E", "R"]}
        ret_early = True
        for key in counts.keys():
            needs[key] = max(counts[key] - len(s) // 4, 0)
            if needs[key] != 0:
                ret_early = False
        if ret_early:
            return 0
        
        def still_meets_needs(left_ind, curr_counts):
            for key in ["Q", "W", "E", "R"]:
                if key == s[left_ind]:
                    if curr_counts[key] - 1 < needs[key]:
                        return False
                else:
                    if curr_counts[key] < needs[key]:
                        return False
            return True
        def meets_needs(curr_counts):
            for key in ["Q", "W", "E", "R"]:
                if curr_counts[key] < needs[key]:
                    return False
            return True

        left_ind = 0
        curr_counts = {key: 0 for key in ["Q", "W", "E", "R"]}
        ret = len(s)

        # print(needs)
        for i in range(len(s)):
            # print(left_ind, curr_counts)
            curr_counts[s[i]] += 1
            while still_meets_needs(left_ind, curr_counts):
                curr_counts[s[left_ind]] -= 1
                left_ind += 1
            if meets_needs(curr_counts):
                ret = min(i - left_ind + 1, ret)
        return ret

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