<p>You are given a string <code>s</code> consisting only of characters <code>&#39;a&#39;</code> and <code>&#39;b&#39;</code>​​​​.</p>

<p>You can delete any number of characters in <code>s</code> to make <code>s</code> <strong>balanced</strong>. <code>s</code> is <strong>balanced</strong> if there is no pair of indices <code>(i,j)</code> such that <code>i &lt; j</code> and <code>s[i] = &#39;b&#39;</code> and <code>s[j]= &#39;a&#39;</code>.</p>

<p>Return <em>the <strong>minimum</strong> number of deletions needed to make </em><code>s</code><em> <strong>balanced</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aababbab&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> You can either:
Delete the characters at 0-indexed positions 2 and 6 (&quot;aa<u>b</u>abb<u>a</u>b&quot; -&gt; &quot;aaabbb&quot;), or
Delete the characters at 0-indexed positions 3 and 6 (&quot;aab<u>a</u>bb<u>a</u>b&quot; -&gt; &quot;aabbbb&quot;).
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;bbaaaaabb&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> The only solution is to delete the first two characters.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s[i]</code> is&nbsp;<code>&#39;a&#39;</code> or <code>&#39;b&#39;</code>​​.</li>
</ul>

class Solution {
public:
    int minimumDeletions(string s) {
        // count (aa..a...bb..b) set
        vector<int> num_a, num_b;
        vector<pair<int, int>> dp;
        int i = 0;
        int n = s.length();

        while(i < n){
            int a = 0;
            int b = 0;
            while(i < n && s[i] == 'a'){
                a++;
                i++;
            }
            while(i < n && s[i] == 'b'){
                b++;
                i++;
            }
            num_a.push_back(a);
            num_b.push_back(b);
        }

        dp.resize(num_a.size()+1, {0, 0});
        for(int j=0; j<dp.size()-1; j++){
            dp[j+1].second = dp[j].second + num_b[j];
            dp[dp.size()-2-j].first = dp[dp.size()-1-j].first + num_a[dp.size()-2-j];
        }
        int min_sum = INT_MAX;
        for(int i=0;i<dp.size()-1; i++){
            min_sum = min(min_sum, dp[i+1].first + dp[i].second);
        }
        return min_sum;
    }
};
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