<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 minwinstr(string &ma,string &fo){
        map<char,int> mp;
        for(auto f:fo){
            mp[f]++;
        }
        int n=fo.size(),l=0,ans=100000;
        for(int i=0;i<ma.size();i++){
            if(mp[ma[i]]>0){
                n--;
            }
            mp[ma[i]]--;
            if(n==0){
                while(l<i && mp[ma[l]]<0){
                    mp[ma[l]]++;
                    l++;
                }
                ans=min(ans,i-l+1);
                mp[ma[l++]]++;
                n++;
            }
        }
        return ans;
    }
    int balancedString(string s) {
        int n=s.size();
        map<char,int> mp;
        for(auto a:s){
            mp[a]++;
        }
        string str="";
        for(auto m:mp){
            if(m.second>(n/4)){
                str+=string(m.second-(n/4),m.first);
            }
        }
        if(str==""){
            return 0;
        }
        return minwinstr(s,str);
    }
};// 3
//
10800