<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>

#define ll long long
#define sz(x) (ll)x.size()
#define INF 1e9

class Solution {
    string s;
    ll dp[100005][2];
public:

    int minimumDeletions(string ss) {
         s = ss;
         memset(dp, -1, sizeof(dp));
         return rec(0, 0);
    }
private:

ll rec(ll idx, bool f){
        if (idx == sz(s))  return 0;

        ll &ret = dp[idx][f];
        if (~ret) return ret;

        ret = INF;

        if (!f){
            ret = min(ret, abs('a' - s[idx]) + rec(idx + 1, f));
            ret = min(ret,  abs(('b') - s[idx]) + rec(idx + 1, 1));
        }else{
            ret = min(ret,  abs(('b') - s[idx]) + rec(idx + 1, f));
        } 

        return ret;
}

};

auto init = []() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    return 'c';
}();
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