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:


 Live Demo

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
            b_left += 1
         ans = min(ans, a_right + b_left)
      return ans

ob = Solution()





Updated on: 10-Nov-2020


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