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Formatted question description: https://leetcode.ca/all/777.html

777. Swap Adjacent in LR String (Medium)

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 end, return True if and only if there exists a sequence of moves to transform one string to the other.

Example:

Input: start = "RXXLRXRXL", end = "XRLXXRRLX"
Output: True
Explanation:
We can transform start to end following these steps:
RXXLRXRXL ->
XRXLRXRXL ->
XRLXRXRXL ->
XRLXXRRXL ->
XRLXXRRLX

Note:

1. 1 <= len(start) = len(end) <= 10000.
2. Both start and end will only consist of characters in {'L', 'R', 'X'}.

Companies:
Google

Related Topics:
Brainteaser

Solution 1.

// OJ: https://leetcode.com/problems/swap-adjacent-in-lr-string/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    bool canTransform(string start, string end) {
        int N = start.size();
        string a = start, b = end;
        a.erase(remove(a.begin(), a.end(), 'X'), a.end());
        b.erase(remove(b.begin(), b.end(), 'X'), b.end());
        if (a != b) return false;
        for (int i = 0, j = 0; i < N; ++i) {
            if (start[i] == 'L') {
                while (end[j] != 'L') ++j;
                if (i < j++) return false;
            }
        }
        for (int i = 0, j = 0; i < N; ++i) {
            if (start[i] == 'R') {
                while (end[j] != 'R') ++j;
                if (i > j++) return false;
            }
        }
        return true;
    }
}; 

Solution 2.

// OJ: https://leetcode.com/problems/swap-adjacent-in-lr-string/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    bool canTransform(string start, string end) {
        int i = 0, j = 0, N = start.size();
        for (; i < N && j < N; ++i, ++j) {
            while (i < N && start[i] == 'X') ++i;
            while (j < N && end[j] == 'X') ++j;
            if ((i < N) ^ (j < N)) return false;
            if (i < N && j < N) {
                if (start[i] != end[j]
                   || (start[i] == 'L' && i < j)
                   || (start[i] == 'R' && i > j)) return false;
            }
        }
        return true;
    }
};

Java

• Java
• C++
• Python

• class Solution {
      public boolean canTransform(String start, String end) {
          if (start == null || end == null || start.length() != end.length())
              return false;
          List<Integer> indices1 = getLRIndices(start);
          List<Integer> indices2 = getLRIndices(end);
          if (indices1.size() != indices2.size())
              return false;
          int size = indices1.size();
          for (int i = 0; i < size; i++) {
              int index1 = indices1.get(i), index2 = indices2.get(i);
              if (index1 > 0 && index2 < 0 || index1 < 0 && index2 > 0 || index1 > index2)
                  return false;
          }
          return true;
      }
  
      public List<Integer> getLRIndices(String str) {
          List<Integer> indices = new ArrayList<Integer>();
          int length = str.length();
          for (int i = 0; i < length; i++) {
              char c = str.charAt(i);
              if (c == 'L')
                  indices.add(-i - 1);
              else if (c == 'R')
                  indices.add(i + 1);
          }
          return indices;
      }
  }
  

• // OJ: https://leetcode.com/problems/swap-adjacent-in-lr-string/
  // Time: O(N)
  // Space: O(N)
  class Solution {
  public:
      bool canTransform(string start, string end) {
          int N = start.size();
          string a = start, b = end;
          a.erase(remove(a.begin(), a.end(), 'X'), a.end());
          b.erase(remove(b.begin(), b.end(), 'X'), b.end());
          if (a != b) return false;
          for (int i = 0, j = 0; i < N; ++i) {
              if (start[i] == 'L') {
                  while (end[j] != 'L') ++j;
                  if (i < j++) return false;
              }
          }
          for (int i = 0, j = 0; i < N; ++i) {
              if (start[i] == 'R') {
                  while (end[j] != 'R') ++j;
                  if (i > j++) return false;
              }
          }
          return true;
      }
  }; 
  

• 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
  
  ############
  
  class Solution(object):
      def canTransform(self, start, end):
          """
          :type start: str
          :type end: str
          :rtype: bool
          """
          i, j = 0, 0
          N = len(start)
          while i < N and j < N:
              while i < N - 1 and start[i] == 'X':
                  i += 1
              while j < N - 1 and end[j] == 'X':
                  j += 1
              if 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 += 1
              j += 1
          return True
  

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