Replace the Substring for Balanced String
<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>
String; Sliding Window
class Solution {
public:
    int balancedString(string s) {
        int cnt[4]{};
        string t = "QWER";
        int n = s.size();
        for (char& c : s) {
            cnt[t.find(c)]++;
        }
        int m = n / 4;
        if (cnt[0] == m && cnt[1] == m && cnt[2] == m && cnt[3] == m) {
            return 0;
        }
        int ans = n;
        for (int i = 0, j = 0; i < n; ++i) {
            cnt[t.find(s[i])]--;
            while (j <= i && cnt[0] <= m && cnt[1] <= m && cnt[2] <= m && cnt[3] <= m) {
                ans = min(ans, i - j + 1);
                cnt[t.find(s[j++])]++;
            }
        }
        return ans;
    }
};