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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
Updated on: 10-Nov-2020
620 Views
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