Problem Description

You have two strings start and target, both with the same length n. These strings contain only three types of characters: 'L', 'R', and '_'.

The characters represent:

• 'L' - a piece that can only move left
• 'R' - a piece that can only move right
• '_' - an empty/blank space

Movement rules:

• An 'L' piece can move one position to the left only if there's a blank space ('_') immediately to its left
• An 'R' piece can move one position to the right only if there's a blank space ('_') immediately to its right
• Pieces cannot jump over other pieces - they can only move into adjacent blank spaces

Your task is to determine if you can transform the start string into the target string by moving the pieces according to these rules. You can make any number of moves.

For example:

• "_L" can become "L_" (L moves left into the blank)
• "R_" can become "_R" (R moves right into the blank)
• "RL" cannot change (pieces are blocked by each other)

Return true if it's possible to transform start into target, otherwise return false.

Intuition

The key insight is that while pieces can move, they cannot pass through each other. This means the relative order of 'L' and 'R' pieces must remain the same in both strings.

Think about it this way: if we have "R_L", the R can never get past the L because R moves right and L moves left - they would collide. Similarly, two Rs or two Ls maintain their relative order since they move in the same direction and can't jump over each other.

This leads us to the first condition: if we remove all blanks from both strings, the sequence of 'L' and 'R' characters must be identical. If start has "RLL" after removing blanks, then target must also have "RLL" after removing blanks.

The second insight involves the movement constraints:

• An 'L' can only move left, so if an 'L' is at position i in start and needs to be at position j in target, then i >= j must be true (it can stay in place or move left, but never right)
• An 'R' can only move right, so if an 'R' is at position i in start and needs to be at position j in target, then i <= j must be true (it can stay in place or move right, but never left)

By checking these two conditions - matching sequence of pieces and valid movement directions for each piece - we can determine if the transformation is possible without actually simulating the moves. We extract the non-blank characters with their positions from both strings and verify that each piece can legally move from its starting position to its target position.

Learn more about Two Pointers patterns.

Solution Approach

The solution implements the intuition by extracting and comparing the non-blank characters from both strings:

1. Extract pieces with positions: Create two lists a and b that store tuples of (character, index) for all non-blank characters in start and target respectively:

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

   This gives us the sequence of pieces and their positions, filtering out all blanks.

2. Check piece count: First verify that both strings have the same number of pieces:

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

   If the counts don't match, transformation is impossible.

3. Validate each piece: Use zip to pair up corresponding pieces from both lists and check:

   for (c, i), (d, j) in zip(a, b):

   For each paired piece:

   • Check piece type matches: if c != d: return False - The i-th piece in start must be the same type as the i-th piece in target

   • Check movement validity for 'L': if c == 'L' and i < j: return False - An 'L' at position i cannot move right to position j (where j > i)

   • Check movement validity for 'R': if c == 'R' and i > j: return False - An 'R' at position i cannot move left to position j (where j < i)

4. Return result: If all checks pass, return True.

The algorithm runs in O(n) time where n is the length of the strings, as we iterate through each string once to extract pieces and then iterate through the pieces once to validate them. The space complexity is also O(n) for storing the extracted pieces.

Example Walkthrough

Let's walk through the solution with start = "R__L" and target = "_RL_".

Step 1: Extract pieces with positions

• From start = "R__L":
  • 'R' at index 0, 'L' at index 3
  • a = [('R', 0), ('L', 3)]
• From target = "_RL_":
  • 'R' at index 1, 'L' at index 2
  • b = [('R', 1), ('L', 2)]

Step 2: Check piece count

• Both lists have 2 pieces ✓

Step 3: Validate each piece

• First pair: ('R', 0) from start and ('R', 1) from target

  • Piece types match: 'R' == 'R' ✓
  • Movement check for 'R': Can it move from index 0 to index 1?
    • 'R' moves right, so we need i <= j: 0 <= 1 ✓

• Second pair: ('L', 3) from start and ('L', 2) from target

  • Piece types match: 'L' == 'L' ✓
  • Movement check for 'L': Can it move from index 3 to index 2?
    • 'L' moves left, so we need i >= j: 3 >= 2 ✓

Step 4: Return result All checks passed, so return True.

To verify this makes sense, let's trace the actual moves:

• "R__L" → "_R_L" (R moves right)
• "_R_L" → "_RL_" (L moves left)

The transformation is indeed possible!

Solution Implementation

Time and Space Complexity

Time Complexity: O(n) where n is the length of the input strings.

• Creating list a requires iterating through all characters in start: O(n)
• Creating list b requires iterating through all characters in target: O(n)
• The zip operation with the subsequent loop iterates through the filtered lists, which in the worst case contains all n characters: O(n)
• Overall time complexity: O(n) + O(n) + O(n) = O(n)

Space Complexity: O(n) where n is the length of the input strings.

• List a stores tuples of characters and indices from start, excluding underscores. In the worst case where there are no underscores, this stores n tuples: O(n)
• List b stores tuples of characters and indices from target, excluding underscores. In the worst case, this also stores n tuples: O(n)
• The zip operation creates an iterator but doesn't create additional space proportional to input size
• Overall space complexity: O(n) + O(n) = O(n)

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

Common Pitfalls

Pitfall 1: Forgetting that pieces cannot jump over each other

A common mistake is only checking if L pieces move left and R pieces move right, without considering that pieces block each other's movement. For example:

# Incorrect approach - only checks direction constraints
def canChange_wrong(start: str, target: str):
    for i in range(len(start)):
        if start[i] == 'L' and target[i] == 'L':
            # Just check if L is not moving right
            continue
        elif start[i] == 'R' and target[i] == 'R':
            # Just check if R is not moving left
            continue
    return True

This would incorrectly return True for cases like:

• start = "R_L", target = "L_R"
• The R cannot jump over L to reach the right position, and L cannot jump over R to reach the left position

Solution: The correct approach extracts pieces in order and ensures the relative ordering of all pieces remains the same. By comparing pieces sequentially using their extraction order, we implicitly ensure no jumping occurs.

Pitfall 2: Trying to simulate actual movements

Another mistake is attempting to simulate the actual movement process step by step:

# Inefficient and error-prone approach
def canChange_simulation(start: str, target: str):
    current = list(start)
    visited = set()
    queue = [current]
  
    while queue:
        state = queue.pop(0)
        if ''.join(state) == target:
            return True
        # Try all possible moves...
        # This becomes very complex and inefficient

This approach is inefficient (potentially exponential time complexity) and prone to implementation errors.

Solution: The key insight is that we don't need to simulate movements. We only need to verify:

1. The sequence of pieces (ignoring blanks) is identical
2. Each piece can reach its target position given its movement constraints

Pitfall 3: Incorrect handling of multiple pieces of the same type

Consider checking positions without maintaining order:

# Wrong approach
def canChange_wrong2(start: str, target: str):
    l_positions_start = [i for i, c in enumerate(start) if c == 'L']
    l_positions_target = [i for i, c in enumerate(target) if c == 'L']
  
    # Just checking if each L can reach some target position
    # This ignores which specific L should go where

For example, with start = "L_L" and target = "LL_", this might incorrectly try to move the second L to the first position.

Solution: By using zip to pair pieces in order, we ensure each piece in start is matched with its corresponding piece in target, maintaining the correct mapping.