LeetCode 1758. Minimum Changes To Make Alternating Binary String

Description

https://leetcode.com/problems/minimum-changes-to-make-alternating-binary-string/

You are given a string s consisting only of the characters '0' and '1'. In one operation, you can change any '0' to '1' or vice versa.

The string is called alternating if no two adjacent characters are equal. For example, the string "010" is alternating, while the string "0100" is not.

Return the minimum number of operations needed to make s alternating.

Example 1:

Input: s = "0100"
Output: 1
Explanation: If you change the last character to '1', s will be "0101", which is alternating.

Example 2:

Input: s = "10"
Output: 0
Explanation: s is already alternating.

Example 3:

Input: s = "1111"
Output: 2
Explanation: You need two operations to reach "0101" or "1010".

Constraints:

  • 1 <= s.length <= 104
  • s[i] is either '0' or '1'.

Explanation

There are two Alternating Binary Strings: one starts with ‘0’, the other starts with ‘1’. We can check which one we can use minimum operations to convert the s to.

Python Solution

class Solution:
    def minOperations(self, s: str) -> int:
        
        s_arr = list(s)
        
        
        zero_start_arr = ['0' if i % 2 else '1' for i in range(len(s))]
        one_start_arr = ['1' if i % 2 else '0' for i in range(len(s))]

        min_count = sys.maxsize
        
        zero_start_count = 0
        for c, z_c in zip(s_arr, zero_start_arr):
            if c != z_c:
                zero_start_count += 1
        
        one_start_count = 0
        for c, o_c in zip(s_arr, one_start_arr):
            if c != o_c:
                one_start_count += 1        
        
        
        return min(one_start_count, zero_start_count)
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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