777. Swap Adjacent in LR String

Problem

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&nbsp;<= 10^4

• start.length == end.length

• Both start and end will only consist of characters in 'L', 'R', and 'X'.

Solution (Java)

class Solution {
    public boolean canTransform(String start, String end) {
        int i1 = 0;
        int i2 = 0;
        char[] s1 = start.toCharArray();
        char[] s2 = end.toCharArray();
        while (i1 < s1.length && i2 < s2.length) {
            while (i1 < s1.length && s1[i1] == 'X') {
                i1++;
            }
            while (i2 < s2.length && s2[i2] == 'X') {
                i2++;
            }
            if (i1 == s1.length && i2 == s2.length) {
                return true;
            }
            if (i1 == s1.length || i2 == s2.length) {
                return false;
            }
            char c1 = s1[i1];
            char c2 = s2[i2];
            if (c1 != c2) {
                return false;
            }
            if (c1 == 'L' && i1 < i2) {
                return false;
            }
            if (c1 == 'R' && i1 > i2) {
                return false;
            }
            i1++;
            i2++;
        }
        while (i1 < s1.length && start.charAt(i1) == 'X') {
            i1++;
        }
        while (i2 < s2.length && end.charAt(i2) == 'X') {
            i2++;
        }
        return i1 == s1.length && i2 == s2.length;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).