How do you solve Move Pieces to Obtain a String on LeetCode?

Move Pieces to Obtain a String (LeetCode #2337) 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: The order of 'L' and 'R' must remain the same in both strings.

What is the time complexity of Move Pieces to Obtain a String?

The optimal solution for Move Pieces to Obtain a String runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only traverse the strings once using two pointers, making it efficient.

What is the brute force solution for Move Pieces to Obtain a String?

The brute force approach for Move Pieces to Obtain a String is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will try to simulate every possible move of the pieces from the start string to see if we can reach the target string. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Move Pieces to Obtain a String?

Move Pieces to Obtain a String uses the following concepts: Two Pointers, String. The recommended patterns to study are: Two Pointers, Sliding Window.

What are the interview tips for Move Pieces to Obtain a String?

Always clarify the movement rules before jumping into coding. Consider edge cases, such as strings with only '_' or no movable pieces. Use systematic approaches like two pointers to optimize your solution.

What are common mistakes when solving Move Pieces to Obtain a String?

Failing to check if the characters 'L' and 'R' match after removing '_'. Not considering the movement constraints of 'L' and 'R' properly.

#2337

Move Pieces to Obtain a String

Medium

Two PointersStringTwo PointersSliding 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)

In the optimal approach, we will use a two-pointer technique to efficiently check if we can transform the start string into the target string by moving the pieces according to the rules.

⚙️

Algorithm

3 steps

1Step 1: Check if the characters 'L' and 'R' in both strings are the same after removing all '_' characters.
2Step 2: Use two pointers to traverse both strings simultaneously, skipping over '_' characters.
3Step 3: For each character, ensure the movement rules are not violated (i.e., 'L' should not move right and 'R' should not move left).

solution.py23 lines

1# Full working Python code
2
3def canTransform(start: str, target: str) -> bool:
4    if start.replace('_', '') != target.replace('_', ''):
5        return False
6    n = len(start)
7    i, j = 0, 0
8    while i < n and j < n:
9        while i < n and start[i] == '_':
10            i += 1
11        while j < n and target[j] == '_':
12            j += 1
13        if i < n and j < n:
14            if start[i] != target[j]:
15                return False
16            if start[i] == 'L' and i < j:
17                return False
18            if start[i] == 'R' and i > j:
19                return False
20            i += 1
21            j += 1
22    return True
23

ℹ

Complexity note: The time complexity is O(n) because we only traverse the strings once using two pointers, making it efficient.

1The order of 'L' and 'R' must remain the same in both strings.
2The number of '_' characters can affect the movement of 'L' and 'R'.

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