777. Swap Adjacent in LR String LeetCode Solution

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Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Swap Adjacent in LR String solution in C++
4. Swap Adjacent in LR String solution in Java
5. Swap Adjacent in LR String solution in Python
6. Additional Resources

Problem Statement of Swap Adjacent in LR String

In a string composed of ‘L’, ‘R’, and ‘X’ characters, like “RXXLRXRXL”, a move consists of either replacing one occurrence of “XL” with “LX”, or replacing one occurrence of “RX” with “XR”. Given the starting string start and the ending string result, return True if and only if there exists a sequence of moves to transform start to result.

Example 1:

Input: start = “RXXLRXRXL”, result = “XRLXXRRLX”
Output: true
Explanation: We can transform start to result following these steps:
RXXLRXRXL ->
XRXLRXRXL ->
XRLXRXRXL ->
XRLXXRRXL ->
XRLXXRRLX

Example 2:

Input: start = “X”, result = “L”
Output: false

Constraints:

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

Complexity Analysis

• Time Complexity: O(n)
• Space Complexity: O(1)

777. Swap Adjacent in LR String LeetCode Solution in C++

class Solution {
 public:
  bool canTransform(string start, string end) {
    if (removeX(start) != removeX(end))
      return false;

    int i = 0;  // start's index
    int j = 0;  // end's index

    while (i < start.length() && j < end.length()) {
      while (i < start.length() && start[i] == 'X')
        ++i;
      while (j < end.length() && end[j] == 'X')
        ++j;
      if (i == start.length() && j == end.length())
        return true;
      if (i == start.length() || j == end.length())
        return false;
      // L can only move to left.
      if (start[i] == 'L' && i < j)
        return false;
      // R can only move to right.
      if (start[i] == 'R' && i > j)
        return false;
      ++i;
      ++j;
    }

    return true;
  }

 private:
  string removeX(const string& s) {
    string t = s;
    std::erase(t, 'X');
    return t;
  }
};
/* code provided by PROGIEZ */

777. Swap Adjacent in LR String LeetCode Solution in Java

class Solution {
  public boolean canTransform(String start, String end) {
    if (!start.replace("X", "").equals(end.replace("X", "")))
      return false;

    int i = 0; // start's index
    int j = 0; // end's index

    while (i < start.length() && j < end.length()) {
      while (i < start.length() && start.charAt(i) == 'X')
        ++i;
      while (j < end.length() && end.charAt(j) == 'X')
        ++j;
      if (i == start.length() && j == end.length())
        return true;
      if (i == start.length() || j == end.length())
        return false;
      // L can only move to left.
      if (start.charAt(i) == 'L' && i < j)
        return false;
      // R can only move to right.
      if (start.charAt(i) == 'R' && i > j)
        return false;
      ++i;
      ++j;
    }

    return true;
  }
}
// code provided by PROGIEZ

777. Swap Adjacent in LR String LeetCode Solution in Python

class Solution:
  def canTransform(self, start: str, end: str) -> bool:
    if start.replace('X', '') != end.replace('X', ''):
      return False

    i = 0  # start's index
    j = 0  # end's index

    while i < len(start) and j < len(end):
      while i < len(start) and start[i] == 'X':
        i += 1
      while j < len(end) and end[j] == 'X':
        j += 1
      if i == len(start) and j == len(end):
        return True
      if i == len(start) or j == len(end):
        return False
      # L can only move to left.
      if start[i] == 'L' and i < j:
        return False
      # R can only move to right.
      if start[i] == 'R' and i > j:
        return False
      i += 1
      j += 1

    return True
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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