README_EN.md

comments	difficulty	edit_url	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/0700-0799/0777.Swap%20Adjacent%20in%20LR%20String/README_EN.md	Two Pointers String

777. Swap Adjacent in LR String

中文文档

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

Python3

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

Java

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;
        }
    }
}

C++

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;
        }
    }
};

Go

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
	}
}