1756 Minimum Deletions To Make String Balanced

Minimum Deletions to Make String Balanced

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*.

 

Example 1:

**Input:** s = "aababbab"
**Output:** 2
**Explanation:** You can either:
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").

Example 2:

**Input:** s = "bbaaaaabb"
**Output:** 2
**Explanation:** The only solution is to delete the first two characters.

 

Constraints:

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

class Solution:
    def minimumDeletions(self, s: str) -> int:
        la = [0] * len(s)
        lb = [0] * len(s)
        i = 0
        temp = 0
        for k in s:
            la[i] = temp
            if k == 'b':
                temp += 1
            
            i += 1
        j = len(s) - 1
        temp = 0
        for k in range(len(s)):
            lb[j] = temp
            if s[len(s) -k-1] == 'a':
                temp += 1
            
            j -= 1
        
        
        result = 100000
        for k in range(len(s)):
            result = min(result, la[k] + lb[k])
        return result


        # for every a at i , find number of bs to left
        # for every b at i , find number of as to right. 
        # navigate each array one by one, 
        # we can remove any element and then all numbers to left or right reduce by 1,
        # we can remove the number altogather
        #0,0,b,1,b,b,3,0
        #a,a,2,a,1,1,a,0