1888. Minimum Number of Flips to Make the Binary String Alternating

Problem Link

Description

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You are given a binary string s. You are allowed to perform two types of operations on the string in any sequence:

• Type-1: Remove the character at the start of the string s and append it to the end of the string.
• Type-2: Pick any character in s and flip its value, i.e., if its value is '0' it becomes '1' and vice-versa.

Return the minimum number of type-2 operations you need to perform such that s becomes alternating.

The string is called alternating if no two adjacent characters are equal.

• For example, the strings "010" and "1010" are alternating, while the string "0100" is not.

 

Example 1:

Input: s = "111000"
Output: 2
Explanation: Use the first operation two times to make s = "100011".
Then, use the second operation on the third and sixth elements to make s = "101010".

Example 2:

Input: s = "010"
Output: 0
Explanation: The string is already alternating.

Example 3:

Input: s = "1110"
Output: 1
Explanation: Use the second operation on the second element to make s = "1010".

 

Constraints:

• 1 <= s.length <= 105
• s[i] is either '0' or '1'.

Solution

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Python3

class Solution:
    def minFlips(self, s: str) -> int:
        n = len(s)
        s += s
        
        A, B = '', ''
        for i in range(len(s)):
            A += '1' if i % 2 == 0 else '0'
            B += '0' if i % 2 == 0 else '1'
        
        res = float('inf')
        count1 = count2 = 0
        
        for i in range(len(s)):
            if A[i] != s[i]: count1 += 1
            if B[i] != s[i]: count2 += 1
            
            if i >= n:
                if A[i - n] != s[i - n]: count1 -= 1
                if B[i - n] != s[i - n]: count2 -= 1
            
            if i >= n - 1:
                res = min(res, count1, count2)
            
        return res