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 <= 104
start.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
}
}
|