<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 {
public:
    int balancedString(string s) {
        int n = s.length();
        int required = n / 4;
        unordered_map<char, int> count = {{'Q', 0}, {'W', 0}, {'E', 0}, {'R', 0}}; 
        for (const char& c: s) {
            count[c] += 1;
        }

        auto is_balanced = [&count, &required]() -> bool {
            bool ans = count['Q'] <= required && count['W'] <= required && count['E'] <= required && count['R'] <= required;
            return ans;
        };

        if (is_balanced()) {
            return 0;
        } 

        // sliding window
        int min_length = n;
        int left = 0;
        for (int right = 0; right < n; ++right) {
            // Reduce the count of the character in the current window
            count[s[right]]--;
            
            // While the remaining characters outside the window are balanced
            while (is_balanced()) {
                // Update the minimum length of the window
                min_length = min(min_length, right - left + 1);
                
                // Move the left pointer of the window
                count[s[left]]++;
                left++;
            }
        }
        return min_length;
    }
};
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