Documentation Index
Fetch the complete documentation index at: /llms.txt
Use this file to discover all available pages before exploring further.
Tested Python solution for LeetCode 1758 with 20 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 1758 # by problem number
lcpy gen -s minimum_changes_to_make_alternating_binary_string # by problem name
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.
**Input:** s = "0100"
**Output:** 1
**Explanation:** If you change the last character to '1', s will be "0101", which is alternating.
**Input:** s = "10"
**Output:** 0
**Explanation:** s is already alternating.
**Input:** s = "1111"
**Output:** 2
**Explanation:** You need two operations to reach "0101" or "1010".
s[i] is either '0' or '1'.class Solution:
# Time: O(n)
# Space: O(1)
def min_operations(self, s: str) -> int:
start_zero = 0 # changes to reach pattern 0101...
for i, char in enumerate(s):
if int(char) != i % 2:
start_zero += 1
return min(start_zero, len(s) - start_zero)
| Time | Space |
|---|---|
| O(n) | O(1) |