How do you solve Swap Adjacent in LR String on LeetCode?

Swap Adjacent in LR String (LeetCode #777) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The order of 'L' and 'R' matters significantly; 'L' must always be to the left of 'R' in the final configuration.

What is the time complexity of Swap Adjacent in LR String?

The optimal solution for Swap Adjacent in LR String runs in O(n) time and uses O(1) space. The time complexity is O(n) because we traverse both strings linearly, ensuring we only check each character once.

What is the brute force solution for Swap Adjacent in LR String?

The brute force approach for Swap Adjacent in LR String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible configurations of the string by performing moves and checking if we can reach the target string. This method is straightforward but inefficient due to the large number of possible configurations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Swap Adjacent in LR String?

Swap Adjacent in LR String uses the following concepts: Two Pointers, String. The recommended patterns to study are: Two Pointers, Sliding Window.

What are the interview tips for Swap Adjacent in LR String?

Always clarify the problem constraints and edge cases before jumping into coding. Think about the implications of each character's position and how they can interact with each other. Practice explaining your thought process as you code; it helps in interviews to articulate your reasoning.

What are common mistakes when solving Swap Adjacent in LR String?

Failing to check if the counts of 'L' and 'R' match in both strings before proceeding. Not considering the constraints imposed by the positions of 'L' and 'R' during traversal.

#777

Swap Adjacent in LR String

Medium

Two PointersStringTwo PointersSliding Window

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution involves checking if we can transform the string by ensuring that the positions of 'L' and 'R' are valid relative to each other, considering the 'X' characters as placeholders. This approach is efficient and avoids unnecessary state exploration.

⚙️

Algorithm

3 steps

1Step 1: Check if the counts of 'L' and 'R' in both strings are the same (ignoring 'X').
2Step 2: Use two pointers to traverse both strings simultaneously, ensuring that 'L' in start can only move left and 'R' can only move right.
3Step 3: If at any point the conditions are violated (i.e., an 'L' tries to move right of an 'R'), return False.

solution.py19 lines

1def canTransform(start, end):
2    if start.replace('X', '') != end.replace('X', ''):
3        return False
4    p1, p2 = 0, 0
5    while p1 < len(start) and p2 < len(end):
6        while p1 < len(start) and start[p1] == 'X':
7            p1 += 1
8        while p2 < len(end) and end[p2] == 'X':
9            p2 += 1
10        if p1 < len(start) and p2 < len(end):
11            if start[p1] != end[p2]:
12                return False
13            if start[p1] == 'L' and p1 < p2:
14                return False
15            p1 += 1
16            p2 += 1
17        else:
18            break
19    return True

ℹ

Complexity note: The time complexity is O(n) because we traverse both strings linearly, ensuring we only check each character once.

1The order of 'L' and 'R' matters significantly; 'L' must always be to the left of 'R' in the final configuration.
2The 'X' characters can be treated as flexible spaces that allow 'L' and 'R' to move, but they cannot cross each other.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.