Formatted question description: https://leetcode.ca/all/1758.html
1758. Minimum Changes To Make Alternating Binary String
Level
Easy
Description
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 <= 10^4
s[i]
is either'0'
or'1'
.
Solution
If the first character of an alternating string is determined, the whole string can be determined. The first character of an alternating string can be either '0'
or '1'
. Calculate the number of operations for both cases to make s
alternating, and return the minimum number of operations.

class Solution { public int minOperations(String s) { int op1 = 0, op2 = 0; int length = s.length(); for (int i = 0; i < length; i++) { char c = s.charAt(i); if (i % 2 == 0) { if (c == '1') op1++; else op2++; } else { if (c == '1') op2++; else op1++; } } return Math.min(op1, op2); } }

Todo

# 1758. Minimum Changes To Make Alternating Binary String # https://leetcode.com/problems/minimumchangestomakealternatingbinarystring class Solution: def minOperations(self, s: str) > int: n = len(s) first, second = ["1"], ["0"] for _ in range(n1): first.append(second[1]) second.append(first[2]) def diff(word): return sum(a != b for a,b in zip(word, s)) return min(diff(first), diff(second))