#777

Medium

auto_awesomeTwo-pointer scanning with invariant tracking

LeetCode Problem Workspace

Swap Adjacent in LR String

Transform one string into another by swapping adjacent 'L' and 'R' characters in a sequence of moves.

Problem Statement

Given two strings, start and result, composed only of 'L', 'R', and 'X', you need to check if it's possible to transform the start string into the result string by repeatedly swapping adjacent 'L' and 'R' characters in specific patterns. The allowed operations are swapping 'XL' to 'LX' and 'RX' to 'XR'. Return True if such a transformation is possible, otherwise return False.

For example, starting with the string "RXXLRXRXL" and transforming it into "XRLXXRRLX" involves a series of valid swaps that respect the constraints. However, a transformation from "X" to "L" is not possible, since there are no valid operations to make the change.

Examples

Example 1

Input: start = "RXXLRXRXL", result = "XRLXXRRLX"

Output: true

We can transform start to result following these steps: RXXLRXRXL -> XRXLRXRXL -> XRLXRXRXL -> XRLXXRRXL -> XRLXXRRLX

Example 2

Input: start = "X", result = "L"

Output: false

Example details omitted.

Constraints

1 <= start.length <= 104
start.length == result.length
Both start and result will only consist of characters in 'L', 'R', and 'X'.

Solution Approach

Two-Pointer Scanning

Use two pointers to traverse the strings from left to right, comparing corresponding characters at each step. Track the positions of 'L' and 'R' characters, ensuring that they never move in a way that violates the rules (i.e., 'L' can only move leftward and 'R' can only move rightward).

Invariant Tracking

As you scan the strings, maintain an invariant that the relative order of 'L' and 'R' must remain consistent. The number of 'L' and 'R' characters in both strings must match at each pointer's position, ensuring that swaps happen in a manner that respects the problem's rules.

Efficiency Consideration

Given that the strings can be as large as 10^4 characters, an O(n) solution is ideal. A direct comparison with two pointers offers an optimal solution for this problem, ensuring that the algorithm handles the input size efficiently.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n), where n is the length of the string, as the algorithm only requires a single pass through the string. The space complexity is O(1), as we only need to track the two pointers and a few variables to maintain the character order.

What Interviewers Usually Probe

Test for edge cases where the start and result strings are identical or very small.
Look for solutions that efficiently handle large inputs within the constraints.
Evaluate if the candidate can explain the importance of character invariants in the two-pointer approach.

Common Pitfalls or Variants

Common pitfalls

Misunderstanding the invariant constraint, leading to incorrect swapping logic.
Failing to handle edge cases like strings with only 'X' characters or strings with an unequal number of 'L' and 'R' characters.
Overcomplicating the solution by not recognizing that the problem can be solved with linear time complexity.

Follow-up variants

Consider a variant where the transformation allows more than two types of operations, introducing additional constraints.
Modify the problem by requiring that no two 'L' characters are allowed to swap positions with each other.
Introduce a constraint that limits the number of swaps that can be performed in each transformation, testing the efficiency of the algorithm.

FAQ

What is the two-pointer scanning approach in the 'Swap Adjacent in LR String' problem?

The two-pointer scanning approach involves using two pointers to traverse the start and result strings simultaneously, ensuring that 'L' and 'R' characters are swapped according to the allowed operations without violating their movement constraints.

How can I solve the 'Swap Adjacent in LR String' problem efficiently?

To solve the problem efficiently, use two-pointer scanning with invariant tracking. This ensures a linear time solution with O(n) time complexity, where n is the length of the string.

Can the solution to this problem be done in constant space?

Yes, the solution can be done in constant space, as you only need a few variables to maintain the positions of the pointers and track character movement without using extra data structures.

How does the invariant tracking work in this problem?

Invariant tracking involves ensuring that the relative order of 'L' and 'R' characters remains consistent as you traverse both strings. The character movements must not violate the problem's transformation rules.

What are some common pitfalls in solving the 'Swap Adjacent in LR String' problem?

Common pitfalls include misunderstanding the invariant constraint, failing to handle edge cases like unequal numbers of 'L' and 'R' characters, or overcomplicating the solution with unnecessary complexity.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def canTransform(self, start: str, end: str) -> bool:
        n = len(start)
        i = j = 0
        while 1:
            while i < n and start[i] == 'X':
                i += 1
            while j < n and end[j] == 'X':
                j += 1
            if i >= n and j >= n:
                return True
            if i >= n or j >= n or start[i] != end[j]:
                return False
            if start[i] == 'L' and i < j:
                return False
            if start[i] == 'R' and i > j:
                return False
            i, j = i + 1, j + 1

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