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# Program to find minimum number of characters to be deleted to make A's before B's in Python

Suppose we have a string s consisting only two letters A and B, we have to find the minimum number of letters that need to be deleted from s to get all occurrences of As before all occurrences of Bs.

So, if the input is like S = "AABAABB", then the output will be 1, as We can remove the last A to get AABBB

To solve this, we will follow these steps:

a_right := number of occurrences of "A" in s

b_left := 0

ans := a_right

for each index i and character c in s, do

if c is same as "A", then

a_right := a_right - 1

otherwise,

b_left := b_left + 1

ans := minimum of ans and a_right + b_left

return ans

Let us see the following implementation to get better understanding:

## Example

class Solution: def solve(self, s): a_right = s.count("A") b_left = 0 ans = a_right for i, c in enumerate(s): if c == "A": a_right -= 1 else: b_left += 1 ans = min(ans, a_right + b_left) return ans ob = Solution() S = "AABAABB" print(ob.solve(S))

## Input

"AABAABB"

## Output

1

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