Fully explained | C++ ✅ | Easy to Understand ✅
\n# Approach\n1. First, we need to check if both strings, excluding \'X\', are identical in character order.\n   This ensures that the transformation is at leas
\n# Approach\n1. First, we need to check if both strings, excluding \'X\', are identical in character order.\n   This ensures that the transformation is at least possible based on character content.\n 2. Next, we iterate through both strings using two pointers. We skip \'X\' characters \n and compare the positions of \'L\' and \'R\' in both strings:\n a. For \'L\', it should never move right (so it can\'t be further to the right in the end string).\n b. For \'R\', it should never move left (so it can\'t be further to the left in the end string).\n3. If we find any inconsistency in these positions, we return false. If we complete the checks\n without finding inconsistencies, we return true.\n\n\n# Complexity\n- Time complexity: O(n)\n We traverse each string only once, comparing and checking positions of characters.\n\n- Space complexity: O(n)\n   The extra space used is for the filtered strings without \'X\', which in the worst case could be of length n.\n\n# Code\n```\nclass Solution {\npublic:\n\n    bool canTransform(string start, string end) {\n        // If lengths are different, transformation is impossible\n        if (start.length() != end.length()) {\n            return false;\n        }\n        \n        // Create strings without \'X\' to check if character sequences match\n        string startChars, endChars;\n        for (char c : start) {\n            if (c != \'X\') startChars += c;\n        }\n        for (char c : end) {\n            if (c != \'X\') endChars += c;\n        }\n        \n        // If the sequences without \'X\' don\'t match, return false\n        if (startChars != endChars) {\n            return false;\n        }\n        \n        // Initialize pointers for both strings\n        int p1 = 0, p2 = 0;\n        int n = start.size();\n        \n        while (p1 < n && p2 < n) {\n            // Skip \'X\' in both strings\n            while (p1 < n && start[p1] == \'X\') p1++;\n            while (p2 < n && end[p2] == \'X\') p2++;\n            \n            // If both pointers reach the end, the strings are transformable\n            if (p1 == n && p2 == n) {\n                return true;\n            }\n            \n            // If only one pointer reaches the end, strings aren\'t transformable\n            if (p1 == n || p2 == n) {\n                return false;\n            }\n            \n            // If current characters don\'t match, transformation is impossible\n            if (start[p1] != end[p2]) {\n                return false;\n            }\n            \n            // \'L\' should not move right (should be left or same position)\n            if (start[p1] == \'L\' && p1 < p2) {\n                return false; \n            }\n            // \'R\' should not move left (should be right or same position)\n            if (start[p1] == \'R\' && p1 > p2) {\n                return false; \n            }\n            \n            // Move to the next character in both strings\n            p1++;\n            p2++;\n        }\n        \n        // If all checks pass, the transformation is possible\n        return true;\n    }\n};\n```
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