EasyNo. 2027
Minimum Moves to Convert String — LeetCode 2027 Python Solution
- Pattern
- Greedy
- Topics
- Greedy, String
- Reading time
- 2 min
- Source
- leetcode.com
The problem
You are given a string s consisting of n characters which are either 'X' or 'O'. A move is defined as selecting three consecutive characters of s and converting them to 'O'.
Example
- Input
- s = "XXX"
- Output
- 1
- Explanation
- XXX -> OOO
Python solution
class Solution:
def minimumMoves(self, s: str) -> int:
ans = i = 0
while i < len(s):
if s[i] == "X":
ans += 1
i += 3
else:
i += 1
return ansComplexity
- Time
- O(n)where n represents the length of the string s
- Space · auxiliary
- O(1) to O(n)
Pattern: Greedy
Take the locally best option every time — when you can prove that never costs you later. LeetCode 2027. Minimum Moves to Convert String is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Greedy.
The greedy guide has the Python template for the pattern and the 346 LeetCode problems that use it.
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|---|---|---|
| 0179 | Largest Number | Medium |
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| 0649 | Dota2 Senate | Medium |
| 0680 | Valid Palindrome II | Easy |
Frequently asked questions
- How hard is LeetCode 2027. Minimum Moves to Convert String?
LeetCode 2027. Minimum Moves to Convert String is rated Easy on LeetCode.
- What topics does LeetCode 2027. Minimum Moves to Convert String cover?
LeetCode 2027. Minimum Moves to Convert String is tagged Greedy and String on LeetCode.