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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:

Related Topics:
Brainteaser

## Solution 1.

• 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')
else if (c == 'R')
}
return indices;
}
}

############

class Solution {
public boolean canTransform(String start, String end) {
int n = start.length();
int i = 0, j = 0;
while (true) {
while (i < n && start.charAt(i) == 'X') {
++i;
}
while (j < n && end.charAt(j) == 'X') {
++j;
}
if (i == n && j == n) {
return true;
}
if (i == n || j == n || start.charAt(i) != end.charAt(j)) {
return false;
}
if (start.charAt(i) == 'L' && i < j || start.charAt(i) == 'R' && i > j) {
return false;
}
++i;
++j;
}
}
}

// 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

• func canTransform(start string, end string) bool {
n := len(start)
i, j := 0, 0
for {
for i < n && start[i] == 'X' {
i++
}
for j < n && end[j] == 'X' {
j++
}
if i == n && j == n {
return true
}
if i == n || j == n || start[i] != end[j] {
return false
}
if start[i] == 'L' && i < j {
return false
}
if start[i] == 'R' && i > j {
return false
}
i, j = i+1, j+1
}
}