Move Pieces to Obtain a String

#826 Most AskedMedium

Move Pieces to Obtain a String #826 Most Asked

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ArraysTwo PointersStrings

You are given two strings start and target, both of length n. Each string consists only of the characters 'L', 'R', and '_' where:

The characters 'L' and 'R' represent pieces, where a piece 'L' can move to the left only if there is a blank space directly to its left, and a piece 'R' can move to the right only if there is a blank space directly to its right.
The character '_' represents a blank space that can be occupied by any of the 'L' or 'R' pieces.

Return true if it is possible to obtain the string target by moving the pieces of the string start any number of times. Otherwise, return false.

Example 1:

Input: start = "_L__R__R_", target = "L______RR"
Output: true
Explanation: We can obtain the string target from start by doing the following moves:
- Move the first piece one step to the left, start becomes equal to "L___R__R_".
- Move the last piece one step to the right, start becomes equal to "L___R___R".
- Move the second piece three steps to the right, start becomes equal to "L______RR".
Since it is possible to get the string target from start, we return true.

Example 2:

Input: start = "R_L_", target = "__LR"
Output: false
Explanation: The 'R' piece in the string start can move one step to the right to obtain "_RL_".
After that, no pieces can move anymore, so it is impossible to obtain the string target from start.

Example 3:

Input: start = "_R", target = "R_"
Output: false
Explanation: The piece in the string start can move only to the right, so it is impossible to obtain the string target from start.

Constraints:

n == start.length == target.length
1 <= n <= 10⁵
start and target consist of the characters 'L', 'R', and '_'.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the input strings 'start' and 'target' contain characters other than '_' and 'L' or 'R'?
If it's impossible to transform 'start' into 'target' by moving pieces, what should I return? Should I return false, throw an exception, or is there a specific error code I should use?
Are the lengths of the 'start' and 'target' strings always equal, and are they non-empty strings?
Do 'L' and 'R' pieces always have to remain in the same relative order between 'start' and 'target', even if their positions shift due to '_' movements?
Are there any constraints on the number of 'L' and 'R' pieces within the strings, or is it possible to have only '_' characters or an uneven distribution of 'L' and 'R' pieces?

Brute Force Solution

Approach

The brute force method for this puzzle is like trying every single possible move of the pieces until we find the exact arrangement that matches the target string. We essentially explore all pathways, no matter how silly or inefficient, until a match is found. This method guarantees an answer if one exists but is not concerned about efficiency.

Here's how the algorithm would work step-by-step:

Start by thinking about all the possible moves each piece could make at the beginning.
For each of those first moves, think about all the possible moves each piece could make next.
Keep going, step by step, considering every possible sequence of moves.
After each sequence of moves, compare the current arrangement of pieces with the target string.
If the arrangement matches the target string, you've found a solution! Stop.
If you've explored all possible sequences of moves and haven't found a match, there may be no solution.

Code Implementation

def can_reach_target_brute_force(start, target):
    # Use a set to store visited states to avoid infinite loops.
    visited_states = {start}

    def find_all_possible_moves(current_string):
        possible_next_states = []
        empty_indices = [i for i, char in enumerate(current_string) if char == '_']
        left_indices = [i for i, char in enumerate(current_string) if char == 'L']
        right_indices = [i for i, char in enumerate(current_string) if char == 'R']

        for empty_index in empty_indices:
            for left_index in left_indices:
                if left_index > empty_index and current_string[left_index] != '_' :
                    new_string = list(current_string)
                    new_string[empty_index], new_string[left_index] = new_string[left_index], new_string[empty_index]
                    possible_next_states.append("".join(new_string))

            for right_index in right_indices:
                if right_index < empty_index and current_string[right_index] != '_':
                    new_string = list(current_string)
                    new_string[empty_index], new_string[right_index] = new_string[right_index], new_string[empty_index]
                    possible_next_states.append("".join(new_string))
        return possible_next_states

    def explore_all_paths(current_string):
        if current_string == target:
            return True

        possible_next_moves = find_all_possible_moves(current_string)

        for next_move in possible_next_moves:
            # Prevent re-visiting already explored states
            if next_move not in visited_states:
                visited_states.add(next_move)

                if explore_all_paths(next_move):
                    return True

        return False

    # Kick off the recursive search
    return explore_all_paths(start)

Big(O) Analysis

Time Complexity

O(3^n) – The brute force approach explores all possible move sequences. In the worst-case scenario, each 'L' or 'R' piece could potentially move to any of the empty positions between it and another piece or the end of the string. If we have 'n' characters in the string, and assuming a significant number of movable pieces and empty spaces, each piece might have multiple possible moves at each step (let's approximate it as 3 options to account for left, right, and staying put). This branching leads to a tree of possibilities where the depth of the tree is related to the number of moves required, and in the worst case, we could be exploring a significant fraction of all characters in the string. Therefore, the number of branches to explore grows exponentially with 'n', resulting in a time complexity of approximately O(3^n).

Space Complexity

O(N^M) – The brute force approach explores all possible move sequences. In the worst case, we might need to store all the intermediate arrangements in a call stack or some explicit data structure to track the explored paths. Assume each piece can move to 'N' positions and we are performing 'M' moves, then the number of possible states we might have to keep track of could grow up to N^M. The space complexity is therefore O(N^M) where N is the number of possible positions a piece can move to and M is the number of moves to reach the target.

Optimal Solution

Approach

The goal is to determine if we can transform one string into another by only moving specific characters ('L' and 'R') while 'X' characters remain fixed. The key insight is to focus on the relative order of the movable characters and their final positions, not on directly moving characters.

Here's how the algorithm would work step-by-step:

First, ignore all the 'X' characters in both strings and only consider the sequences of 'L' and 'R' characters.
Check if the resulting sequences of 'L' and 'R' are identical in both strings. If they are not, the transformation is impossible.
Now, go character by character through both strings together, keeping track of only the 'L' and 'R' characters.
For an 'L' character, the position in the starting string must be greater than or equal to its position in the target string. This is because 'L' can only move to the left.
For an 'R' character, the position in the starting string must be less than or equal to its position in the target string. This is because 'R' can only move to the right.
If these conditions hold true for every 'L' and 'R' character, the transformation is possible. Otherwise, it is not.

Code Implementation

def can_transform(start_string, end_string):
    start_without_x = start_string.replace('X', '')
    end_without_x = end_string.replace('X', '')

    if start_without_x != end_without_x:
        return False

    start_index = 0
    end_index = 0

    while start_index < len(start_string) and end_index < len(end_string):
        while start_index < len(start_string) and start_string[start_index] == 'X':
            start_index += 1
        while end_index < len(end_string) and end_string[end_index] == 'X':
            end_index += 1

        if start_index == len(start_string) and end_index == len(end_string):
            break

        if start_index == len(start_string) or end_index == len(end_string):
            return False

        # L can only move to the left
        if start_string[start_index] == 'L':
            if start_index < end_index:
                return False
        # R can only move to the right
        elif start_string[start_index] == 'R':
            if start_index > end_index:
                return False

        start_index += 1
        end_index += 1

    return True

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through both strings, `start` and `target`, at most once each to filter out 'X' characters and compare the remaining 'L' and 'R' sequences. It then iterates through the strings again, character by character, to validate the movement conditions for 'L' and 'R' characters. Each of these iterations depends on the length, n, of the input strings. Thus, the time complexity is directly proportional to the input size, making it O(n).

Space Complexity

O(1) – The algorithm operates primarily in-place, using only a few constant space variables for indexing. We are iterating through the input strings directly without creating any auxiliary data structures that scale with the input size N, where N is the length of the strings. The comparison of characters and positions is done using variables that take up constant space. Therefore, the space complexity is O(1).

Edge Cases

Case	How to Handle
s or target is null or empty string	Return true if both are empty, false otherwise to avoid null pointer exceptions or incorrect comparisons.
s and target have different lengths	Return false immediately as the number of characters must be the same to rearrange.
s and target have different numbers of 'L' or 'R' characters	Return false as 'L' and 'R' can't be created or destroyed, only moved.
An 'L' in s is to the right of its corresponding position in target	Return false because 'L' can only move left.
An 'R' in s is to the left of its corresponding position in target	Return false because 'R' can only move right.
s or target contains characters other than 'L', 'R', and '_'	Return false as other characters cannot be moved or matched.
Very long strings	Ensure the solution uses linear time and constant space to scale efficiently with string length.
All characters in s and target are '_'	Return true, as an empty arrangement satisfies the requirement.

s or target is null or empty string

How to Handle:

Return true if both are empty, false otherwise to avoid null pointer exceptions or incorrect comparisons.

s and target have different lengths

How to Handle:

Return false immediately as the number of characters must be the same to rearrange.

s and target have different numbers of 'L' or 'R' characters

How to Handle:

Return false as 'L' and 'R' can't be created or destroyed, only moved.

An 'L' in s is to the right of its corresponding position in target

How to Handle:

Return false because 'L' can only move left.

An 'R' in s is to the left of its corresponding position in target

How to Handle:

Return false because 'R' can only move right.

s or target contains characters other than 'L', 'R', and '_'

How to Handle:

Return false as other characters cannot be moved or matched.

Very long strings

How to Handle:

Ensure the solution uses linear time and constant space to scale efficiently with string length.

All characters in s and target are '_'

How to Handle:

Return true, as an empty arrangement satisfies the requirement.