Leetcode #2337: Move Pieces to Obtain a String
In this guide, we solve Leetcode #2337 Move Pieces to Obtain a String in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Two Pointers, String
Intuition
The constraints hint that we can reason about two ends of the data at once, which is perfect for a two-pointer scan.
Moving one pointer at a time keeps the invariant intact and avoids nested loops.
Approach
Place pointers at the left and right ends and move them based on the comparison or target condition.
This yields a clean linear pass after any required sorting.
Steps:
- Set left and right pointers.
- Move a pointer based on the condition.
- Update the best answer while scanning.
Example
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.
Python Solution
class Solution:
def canChange(self, start: str, target: str) -> bool:
a = [(v, i) for i, v in enumerate(start) if v != '_']
b = [(v, i) for i, v in enumerate(target) if v != '_']
if len(a) != len(b):
return False
for (c, i), (d, j) in zip(a, b):
if c != d:
return False
if c == 'L' and i < j:
return False
if c == 'R' and i > j:
return False
return True
Complexity
The time complexity is O(n) (after optional sort O(n log n)). The space complexity is O(1).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.