Date: 2026-06-06
Time: 17:59
minimum-moves-to-convert-string/solution.pySolves LeetCode 2027: Minimum Moves to Convert String. Given a string of 'X' and 'O' characters, each move converts any 3 consecutive characters to 'O'. The function returns the minimum number of moves to eliminate all 'X's.
Solution.maximumRemovals(self, s: str) -> int — The method name is wrong. It should be something like minimumMoves to match the problem; maximumRemovals is the name of a different LeetCode problem (1898). The docstring is correct, though — it describes the actual behavior.
The algorithm is a greedy left-to-right scan:
1. Pointer i walks through the string.
2. On 'O': advance by 1 — nothing to do.
3. On 'X': spend one move, then skip ahead by 3. The move covers positions i, i+1, i+2, so any 'X's in those next two positions are handled for free.
4. Return total moves.
This greedy choice is optimal because any 'X' must be covered by some move, and covering the leftmost uncovered 'X' as early as possible maximizes the reach of each move (you never waste coverage on positions you've already passed).
Greedy skip — a common idiom for interval-covering problems. Instead of tracking which positions are covered, the pointer jump (i += 3) implicitly marks the next two positions as "handled." This avoids needing a separate data structure.
No imports. The class follows the standard LeetCode Solution convention for submission compatibility. The Imported By list in the prompt is misleading — that's an artifact of the test harness importing Solution from each problem's own solution.py, not cross-problem imports.
'X' in the string is covered by exactly one move's 3-character window.i += 3, the pointer may land past len(s), which is fine because the while condition catches it.None. The function assumes valid input (a string of only 'X' and 'O'). An empty string returns 0 correctly since the while loop body never executes.