<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:
//window has the excess characters and is valid
    bool windowHasExcess(unordered_map<char, int>& freq, 
        unordered_map<char, int>& window_freq){
            return freq['Q'] <= window_freq['Q'] &&
                freq['W'] <= window_freq['W'] &&
                freq['E'] <= window_freq['E'] &&
                freq['R'] <= window_freq['R'];
        }
    int balancedString(string s) {
        //min length of substring to replace so that
        //Q W E R appear an equal # of times
        //(s.size is always a multiple of 4)

        //position matters
        //need to find total freq counts first
        //find smallest window w/ the excess chars
        unordered_map<char, int> freq;
        for(char c:s)
            freq[c]++;
        freq['Q'] -= s.size()/4, freq['W'] -= s.size()/4;
        freq['E'] -= s.size()/4, freq['R'] -= s.size()/4;

        // for(pair<char, int> p : freq)
        //     cout << p.first << ", " << p.second;
        
        unordered_map<char, int> window_freq;
        if(windowHasExcess(freq, window_freq))
            return 0; //was already balanced
        int l = 0;
        int ans = INT_MAX;
        for(int r=0; r<s.size(); r++){
            window_freq[s[r]]++;
            while(windowHasExcess(freq, window_freq)){
                ans = min(ans, r-l+1);
                //try shrinking left
                window_freq[s[l]]--;
                l++;
            }
        }
        return ans;
    }
};

/*
eeerrqqq -> have to replace the middle r's, too
*/
11700