How do you solve Minimum Moves to Convert String on LeetCode?

Minimum Moves to Convert String (LeetCode #2027) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Converting three consecutive 'X's in one move is optimal, as it maximizes the number of 'X's converted per move.

What is the brute force solution for Minimum Moves to Convert String?

The brute force approach for Minimum Moves to Convert String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible triplet of characters in the string and converting them to 'O'. This method is straightforward but inefficient, as it may require multiple passes through the string. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Moves to Convert String?

Minimum Moves to Convert String uses the following concepts: String, Greedy. The recommended patterns to study are: Greedy, Sliding Window.

What are the interview tips for Minimum Moves to Convert String?

Always start with a brute-force solution to understand the problem before optimizing. Think about how you can reduce the number of operations by skipping already processed characters. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Minimum Moves to Convert String?

Not accounting for overlapping triplets correctly, which can lead to incorrect move counts. Failing to optimize the iteration by skipping characters that have already been converted.

#2027

Minimum Moves to Convert String

Easy

StringGreedyGreedySliding Window

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses a greedy strategy to convert as many 'X's as possible in one move. By processing the string in a single pass, we can minimize the number of moves required.

⚙️

Algorithm

3 steps

1Step 1: Initialize a counter for moves to 0.
2Step 2: Iterate through the string with an index.
3Step 3: Whenever an 'X' is encountered, increment the move counter and skip the next two characters (since they will also be converted).

solution.py15 lines

1# Full working Python code
2
3def minimum_moves(s):
4    moves = 0
5    i = 0
6    while i < len(s):
7        if s[i] == 'X':
8            moves += 1
9            i += 3  # Skip the next two characters
10        else:
11            i += 1
12    return moves
13
14# Example usage
15print(minimum_moves('XXOX'))  # Output: 2

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the string. The space complexity is O(1) since we are using a constant amount of extra space.

1Converting three consecutive 'X's in one move is optimal, as it maximizes the number of 'X's converted per move.
2Skipping characters after a conversion prevents double counting and reduces unnecessary checks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.