#2027

Easy

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Minimum Moves to Convert String

Minimize moves to convert a string of 'X' and 'O' to all 'O' by converting three consecutive characters to 'O'.

Problem Statement

You are given a string s consisting of 'X' and 'O' characters. A move is defined as selecting three consecutive characters of s and converting them to 'O'. If a 'O' is selected, it remains unchanged.

The goal is to find the minimum number of moves required to convert all characters of s to 'O'.

Examples

Example 1

Input: s = "XXX"

Output: 1

XXX -> OOO We select all the 3 characters and convert them in one move.

Example 2

Input: s = "XXOX"

Output: 2

XXOX -> OOOX -> OOOO We select the first 3 characters in the first move, and convert them to 'O'. Then we select the last 3 characters and convert them so that the final string contains all 'O's.

Example 3

Input: s = "OOOO"

Output: 0

There are no 'X's in s to convert.

Constraints

3 <= s.length <= 1000
s[i] is either 'X' or 'O'.

Solution Approach

Greedy Approach

The greedy approach focuses on selecting the leftmost 'X's. In each step, try to change a group of three consecutive characters starting from the leftmost 'X'. This minimizes the number of operations needed.

Invariant Validation

The approach guarantees that each move affects three characters, so after every move, the string gets closer to having no 'X's. At the end, all characters are guaranteed to become 'O'.

Edge Case Handling

Ensure that if there are no 'X's, the result is zero, and handle strings of length three, as they require one move if 'X's are present.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on how many moves are required to convert all 'X's. Since each move works on a fixed group of three characters, the number of moves is proportional to the number of 'X's in the string. Space complexity is O(1), as the algorithm only uses a few extra variables to track the number of moves.

What Interviewers Usually Probe

Checks if the candidate can apply a greedy strategy for string manipulation.
Evaluates understanding of invariant validation in greedy algorithms.
Assesses the ability to handle edge cases and different string lengths.

Common Pitfalls or Variants

Common pitfalls

Forgetting to handle strings with no 'X's properly (result should be zero).
Overcomplicating the solution by trying to track extra state or avoid greedy selection.
Missing the fact that you only need to select three consecutive characters at a time.

Follow-up variants

Allowing variable-length groups of consecutive characters to be selected.
Modifying the problem so that you can convert one or two characters at a time instead of three.
Requiring a condition where some 'O's can be converted back into 'X's.

FAQ

What is the greedy strategy for the Minimum Moves to Convert String problem?

The greedy strategy focuses on converting the leftmost 'X' to 'O' by selecting three consecutive characters, ensuring minimum moves.

How do I handle edge cases for the Minimum Moves to Convert String problem?

Edge cases include strings with no 'X's (result should be 0) or strings with exactly three characters. Ensure you handle these correctly.

What pattern is used in the Minimum Moves to Convert String problem?

The problem uses a greedy choice plus invariant validation, where each move gradually reduces the number of 'X's in the string.

Why does the solution focus on three consecutive characters at a time?

The problem specifies that each move changes exactly three consecutive characters, which aligns with the greedy approach to minimize operations.

Can the solution be optimized further for the Minimum Moves to Convert String problem?

The greedy approach is optimal in this problem. Any extra optimization might unnecessarily complicate the solution without improving performance.

terminal

Solution

Solution 1: Greedy Algorithm

Traverse the string $s$. Whenever you encounter `'X'`, move the pointer $i$ three steps forward and add $1$ to the answer; otherwise, move the pointer $i$ one step forward.

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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 ans

Continue Practicing

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