<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 approach_1(string s){

        int n = s.size(); int count = 0;

        stack<int> st;
        for(int i = 0; i<n; i++){
            if(!st.empty() && s[i] == 'a' && st.top() == 'b'){
                st.pop();
                count++;
            }
            else{
                st.push(s[i]);
            }
        }

        return count;
    }
    int approach_2(string s){
        
        int n = s.size();
        vector<int> count_b(n, 0);
        for(int i = 1; i<n; i++){  // 1 -> n-1
            count_b[i] = count_b[i-1];
            if(s[i-1] == 'b'){
                count_b[i] += 1;
            }
        }

        vector<int> count_a(n, 0); // 0 <- n-2 
        for(int i = n-2; i >= 0; i--){

            count_a[i] = count_a[i+1];

            if(s[i+1] == 'a') {
                count_a[i] += 1;
            }
        }

        int res = INT_MAX;

        for(int i = 0; i<n; i++){
            res = min(res, count_a[i] + count_b[i]);
        }

        return res;
    }

    int optimized(string s){
        int n = s.size();

        int count_a = 0;

        for(int i = n-1; i >= 0; i--){
            if(s[i] == 'a'){
                count_a++;
            }
        }

        int count = INT_MAX;
        int count_b = 0;

        for(int i = 0; i< n; i++){
            if(s[i] == 'a'){
                count_a--;
            }

            count = min(count, count_b + count_a);

            if(s[i] == 'b'){
                count_b++;
            }
        }

        return count;
    }
    int minimumDeletions(string s) {
        // return optimized(s);
        return approach_1(s);
    }
};
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