# 1653. Minimum Deletions to Make String Balanced

## Problem Statement

<br>

You are given a string `s` consisting only of characters `'a'` and `'b'`​​​​.

You can delete any number of characters in `s` to make `s` **balanced**. `s` is **balanced** if there is no pair of indices `(i,j)` such that `i < j` and `s[i] = 'b'` and `s[j]= 'a'`.

Return *the **minimum** number of deletions needed to make* `s` ***balanced***.

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**Example 1:**

<pre><code><strong>Input: s = "aababbab"
</strong><strong>Output: 2
</strong><strong>Explanation: You can either:
</strong>Delete the characters at 0-indexed positions 2 and 6 ("aababbab" -> "aaabbb"), or
Delete the characters at 0-indexed positions 3 and 6 ("aababbab" -> "aabbbb").
</code></pre>

**Example 2:**

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

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**Constraints:**

* `1 <= s.length <= 105`
* `s[i]` is `'a'` or `'b'`​​.

## Intuition

```
Take count of all Those before B's and All after a's for all index

Now the addition for each index tells me the deletion I want to do 
To eliminiate the inversions
```

### Links

<https://leetcode.com/problems/minimum-deletions-to-make-string-balanced/description/?envType=daily-question&envId=2024-07-30>

### Video Links

### Approach 1:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
class Solution {
public:
    int minimumDeletions(string s) {
        int n = s.size();
        vector<int> b(n, 0);

        for(int i=1; i<n; i++){
            if(s[i-1] == 'b'){
                b[i] = b[i-1] + 1;
            }else{
                b[i] = b[i-1];
            }
        }


        int ans = INT_MAX;
        for(int i=0; i<n; i++){
            ans = min(ans, a[i]+b[i]);
        }

        return ans;
    }
};
```

{% endcode %}

### Approach 2:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 3:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 4:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

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