<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();
        map<char,int>mp;
        for(auto it: s){
            mp[it]++;
        }
        map<char,int>mp1;
        map<char,int>mp2;
        map<char,int>mp3;
        for(auto it: mp){
            if(it.second>n/4){
                mp3[it.first]=it.second-n/4;
            }
        }
        cout<<n<<endl;
        for(auto it: mp3){
            cout<<it.first<<" "<<it.second<<endl;
        }
        if(mp3.size()==0) return 0;
        mp1['Q']=0;
        mp1['W']=1;
        mp1['E']=2;
        mp1['R']=3;
        vector<map<int,int>>temp(4);
        temp[0][0]=-1;
        temp[1][0]=-1;
        temp[2][0]=-1;
        temp[3][0]=-1;
        int res=n;
        for(int i=0; i<n; i++){
            mp2[s[i]]++;
            int ans=n;
            int f=0;
            for(auto it: mp3){
                if(temp[mp1[it.first]].find(mp2[it.first]-it.second)!=temp[mp1[it.first]].end()){
                    ans=min(ans,temp[mp1[it.first]][mp2[it.first]-it.second]);
                }
                else{
                    f=1;
                }
                temp[mp1[it.first]][mp2[it.first]]=i;
            }
            if(f==0){
                res=min(res,i-ans);
            }
        }
        return res;
    }
};
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