Description#
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 1:
Input: start = "RXXLRXRXL", end = "XRLXXRRLX"
Output: true
Explanation: We can transform start to end following these steps:
RXXLRXRXL ->
XRXLRXRXL ->
XRLXRXRXL ->
XRLXXRRXL ->
XRLXXRRLX
Example 2:
Input: start = "X", end = "L"
Output: false
Constraints:
1 <= start.length <= 104start.length == end.length- Both
start and end will only consist of characters in 'L', 'R', and 'X'.
Solutions#
Solution 1#
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| 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
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| 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;
}
}
}
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| class Solution {
public:
bool canTransform(string start, string end) {
int n = start.size();
int i = 0, j = 0;
while (true) {
while (i < n && start[i] == 'X') ++i;
while (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;
}
}
};
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| 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
}
}
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