Minimum Moves to Convert String

Minimum Moves to Convert String

1. What is this problem about?

The "Minimum Moves to Convert String" interview question involves a simple string transformation task. You are given a string s consisting of 'X' and 'O' characters. A "move" consists of selecting three consecutive characters and converting all of them to 'O'.

The objective is to find the minimum number of moves required to change all 'X' characters in the string to 'O'. This is a straightforward problem that focuses on efficient traversal and decision-making.

2. Why is this asked in interviews?

This is a common "Easy" difficulty question used by companies like Zoho and Google to check a candidate's basic Greedy Algorithm intuition. It tests:

• Linear Traversal: Can the candidate process the string in a single pass?
• Greedy Strategy: Does the candidate realize that they only need to start a move when they encounter an 'X'?
• Simplicity: Can the candidate write clean, bug-free code for a simple task without overengineering it?

3. Algorithmic pattern used

The pattern used here is the Greedy Strategy.

1. Iterate through the string from left to right.
2. If you encounter an 'O', just move to the next character.
3. If you encounter an 'X', this must be converted. Since a move covers three characters, the most efficient way to use that move is to start it exactly at this 'X'.
4. Increment the move count and skip the next two characters (because they are covered by this move, regardless of whether they were 'X' or 'O').
5. Continue until the end of the string.

4. Example explanation

String: XXOX

1. Index 0: Found 'X'.
   • Perform a move. Count = 1.
   • This move covers indices 0, 1, and 2.
   • Skip to index 3.
2. Index 3: Found 'X'.
   • Perform a move. Count = 2.
   • This move covers indices 3, 4, 5 (even if they don't exist).
   • End of string. Total moves = 2.

5. Common mistakes candidates make

• Overlapping Moves: Thinking they need to carefully choose the "best" three characters. In reality, starting at the leftmost 'X' is always optimal.
• Nested Loops: Trying to manually change 'X' to 'O' in the string and then re-scanning. A single loop with an index increment i += 3 is sufficient.
• Off-by-one: Not handling the end of the string correctly when a move extends past the last character.

6. Interview preparation tip

Greedy problems often have a "left-to-right" or "smallest-to-largest" intuition. If you find a mandatory requirement (like an 'X' that must be removed), take the action that satisfies that requirement while covering as much future ground as possible.