2337. Move Pieces to Obtain a String

Problem Description

The given problem is about transforming one string into another using specific movement rules for the characters within the strings. The strings start and target are of the same length n and consist only of characters 'L', 'R', and ''. An 'L' can move only left if there is a blank space ('') directly to its left, and an 'R' can move only right if there is a blank space directly to its right. The goal is to determine if the start string can be transformed into the target string by applying these movements any number of times. The output should be true if the transformation is possible, otherwise false.

Intuition

The intuition behind the solution involves two key insights:

1. The relative order of the pieces 'L' and 'R' cannot change because 'L' can only move left and 'R' can only move right. Therefore, if in the target string, an 'R' appears to the left of an 'L' which was to its right in the start string, the transformation is not possible.

2. Given the same relative order, an 'L' can only move to the left, so its position in the target string must not be to the right of its position in the start string. Similarly, an 'R' can only move to the right, so its position in the target string must not be to the left of its position in the start string.

With these insights in mind, the solution approach is straightforward:

• Ignore the blank spaces and compare the positions of the non-blank characters in both strings. If there are a different number of 'L' and 'R' characters or they are in a different relative order, the target cannot be reached.

• If the characters are in the same order, check if 'L' in target is never to the right of its position in start, and 'R' is never to the left. If this is true for all characters, then the transformation is possible and return true. Otherwise, return false.

Learn more about Two Pointers patterns.

Solution Approach

The Python code provided earlier implements the solution approach effectively. It uses the following steps and Python-specific data structures and functions:

1. Filtering and Pairing: The comprehension lists a and b filter out the blank spaces '_' and create lists of tuples that contain the character and its index (position) from the start and target strings, respectively. This is done using the enumerate function which gives the index along with each character as you iterate over the string:

   1a = [(v, i) for i, v in enumerate(start) if v != '_']
   2b = [(v, i) for i, v in enumerate(target) if v != '_']

2. Checking the Length: This step checks if the lengths of the filtered lists are equal. If they are not, it is not possible to get target from start since the number of movable pieces is different, so the function returns False.

   1if len(a) != len(b):
   2    return False

3. Comparing Corresponding Pairs: The use of the zip function takes pairs from a and b in a parallel manner to ensure we're comparing pieces from the start and target strings that are supposed to correspond to each other.

4. Piece Type and Position Validation: For each pair of tuples (c, i) from a, and (d, j) from b, the following checks are performed:

   • If the piece type is not the same (c != d), it cannot be a valid transformation since 'L' cannot become 'R' and vice versa, hence False is returned.
   • If the piece is 'L', it should not be to the right in target as compared to start (i < j case), since 'L' can only move left.
   • If the piece is 'R', it should not be to the left in target as compared to start (i > j case), since 'R' can only move right.

   If any of these conditions are violated, the function returns False.

   1for (c, i), (d, j) in zip(a, b):
   2    if c != d or (c == 'L' and i < j) or (c == 'R' and i > j):
   3        return False

5. Returning True: If none of the above checks fail, it means the transformation from start to target is possible, hence True is returned at the end of the function.

By using tuples for storing character and index pairs, and the zip function to iterate over them, we avoid the need for more complex data structures. This simplifies the algorithm and improves its readability and performance.

Example Walkthrough

Let's consider two strings as an example:

• start: "R__L"
• target: "__RL"

Following the steps of the solution approach:

1. Filtering and Pairing:

   For start:

   1a = [(v, i) for i, v in enumerate("R__L") if v != '_']
   2  = [('R', 0), ('L', 3)]

   For target:

   1b = [(v, i) for i, v in enumerate("__RL") if v != '_']
   2  = [('R', 2), ('L', 3)]

2. Checking the Length:

   Length of a: 2 Length of b: 2

   Since both lengths are equal, we proceed.

3. Comparing Corresponding Pairs:

   We utilize zip to compare each pair from a and b.

   Pairs to compare:

   1('R', 0) from `start` and ('R', 2) from `target`
   2('L', 3) from `start` and ('L', 3) from `target`

4. Piece Type and Position Validation:

   For the first pair:

   • Piece type is the same, both are 'R'.
   • 'R' is in position 0 in start and position 2 in target, which is a valid move for 'R', since it can move to the right.

   For the second pair:

   • Piece type is the same, both are 'L'.
   • 'L' is in position 3 in start and position 3 in target, indicating no movement, which is also valid since 'L' cannot move to the right.

   Since both pairs are valid according to the movement rules, we can continue.

5. Returning True:

   Because none of the validation checks failed, the function would return True, meaning it is possible to transform the start string "R__L" into the target string "__RL" by moving the 'R' two spaces to the right.

Solution Implementation

Time and Space Complexity

Time Complexity

The given Python function canChange checks whether it is possible to transform the start string into the target string under certain conditions. Analyzing the time complexity involves a few steps:

1. Creation of list a: The list comprehension iterates over each character in the start string and includes only non-underscore characters. Therefore, this operation has a time complexity of O(n) where n is the length of the start string.

2. Creation of list b: Similarly, the creation of list b iterates over each character in the target string and has a time complexity of O(n).

3. Comparison of lengths: Checking if len(a) is equal to len(b) takes constant time, O(1).

4. Zipping and iterating: Zipping the two lists a and b and iterating over them to compare elements has a time complexity of O(m), where m is the number of non-underscore characters in the strings, which is at most n.

Given these steps occur sequentially, the overall time complexity is dominated by the terms with O(n), leading to a total time complexity of O(n).

Space Complexity

1. List a and b store the non-underscore characters and their respective indices from start and target. These take space proportional to the number of non-underscore characters, which is O(m) where m is the number of such characters.

2. The space taken by the zip object and iteration is negligible since they don't store values but only reference the elements in a and b.

Hence, the overall space complexity of the function canChange is O(m), where m is the number of non-underscore characters and m <= n. If we consider that all characters could be non-underscore, the space complexity would also be O(n) in the worst case.

Learn more about how to find time and space complexity quickly using problem constraints.