Is Minimum Moves to Convert String easy or hard?

Minimum Moves to Convert String is classified as an Easy problem on LeetCode with an acceptance rate around 57%. The main challenge is recognizing the greedy observation that converting the earliest 'X' immediately always leads to the optimal solution.

Minimum Moves to Convert String Python/Java solution

Most implementations follow the same greedy pattern. Loop through the string, increment the answer when encountering 'X', and advance the pointer by three positions. The logic is identical across Python, Java, C++, C#, and JavaScript because it relies only on basic string indexing.

How to solve Minimum Moves to Convert String in O(n)?

Iterate through the string with an index pointer. When the current character is 'X', increment the move counter and jump the index forward by 3 because a single operation converts positions i, i+1, and i+2 to 'O'. If the character is 'O', simply move to the next index. This single pass guarantees O(n) time complexity.

What is the best approach for Minimum Moves to Convert String?

The greedy approach is the best solution. Scan the string from left to right, and whenever you encounter an 'X', perform one move that converts that character and the next two positions to 'O'. Then skip the next two indices because they are already handled. This produces the minimum number of moves in O(n) time and O(1) space.

Is Minimum Moves to Convert String asked at Google/Amazon/Meta?

Greedy string scanning problems like this frequently appear in interviews at companies such as Amazon, Google, and Meta because they test pattern recognition and the ability to derive optimal local decisions. Variants of window-based greedy problems are especially common in coding interviews.

What data structure is used in Minimum Moves to Convert String?

The problem mainly operates on a string and does not require additional data structures. The optimal solution uses a greedy pointer-based traversal of the string and a simple counter to track the number of moves.

What is the time complexity of Minimum Moves to Convert String?

The optimal greedy solution runs in O(n) time because each character in the string is processed at most once during the linear scan. The space complexity is O(1) since only a counter and an index pointer are used without extra data structures.

Description Hints 2 Solution Video 4:57

2027. Minimum Moves to Convert String

Easy58.4% acceptance3 companies

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'. Note that if a move is applied to the character 'O', it will stay the same.

Return the minimum number of moves required so that all the characters of s are converted to 'O'.

Example 1:

Input: s = "XXX"
Output: 1
Explanation: XXX -> OOO
We select all the 3 characters and convert them in one move.

Example 2:

Input: s = "XXOX"
Output: 2
Explanation: 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
Explanation: There are no 'X's in s to convert.

Constraints:

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

Asked at 3 companies

Jeavio

Google Bloomberg

Topics

Greedy String

Hints open one at a time

Hint 1

Find the smallest substring you need to consider at a time.

Hint 2

Try delaying a move as long as possible.

4:57

Still stuck?Read 3 approaches, or watch a walkthrough.

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