<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:
	// Minimum Window Substring
    int minwindowsubstr(string s, string text)
    {
        vector<int> ct(26,0);
        int n=text.length();
        for(int i=0;i<n;i++) ct[text[i]-'A']++;
        int r=0,l=0,mi=INT_MAX;
        while(r<s.length()){
            ct[s[r]-'A']--;
            if(ct[s[r]-'A']>=0) n--;
            r++;
            while(l<r && n==0){
                mi=min(mi,r-l);
                ct[s[l]-'A']++;
                if(ct[s[l]-'A']>0) n++;
                l++;
            }
        }
        return mi;
    }
    
    int balancedString(string s) {
        int n=s.length();
        unordered_map<char,int> m;
        for(int i=0;i<n;i++) m[s[i]]++;
        int len=n/4;
        //cout<<len<<endl;
        string str="";
        for(auto it:m){
            if(it.second>len) str+=string(it.second-len,it.first);
        }
        if(str.length()==0) return 0;
        return minwindowsubstr(s,str);
    }
};
11100