🚀Beeats 91.53%🚀✅Super Easy Two Pointer (C++/Java/Python) Solution With Detailed Explanation✅
Intuition\nThe intuition of the code is to employ a two-pointer approach to traverse start and target strings simultaneously, skipping \'\' characters and ensur
# Intuition\nThe intuition of the code is to employ a two-pointer approach to traverse `start` and `target` strings simultaneously, skipping \'_\' characters and ensuring \'L\' and \'R\' characters can only move within the constraints of their respective rules. The process checks for a one-to-one match between non-\'_\' characters in the correct order and direction, returning `true` if it\'s possible to transform `start` into `target` by adhering to movement rules, and `false` otherwise.\n\n\n![Screenshot 2024-02-18 090102.png](https://assets.leetcode.com/users/images/65234468-25e1-4585-9c4c-59ff5ffcc000_1708227074.9881327.png)\n\n# Approach\n\n1. **Initialization**: Two pointers, `i` and `j`, are initialized at the start of `start` and `target` strings respectively. These pointers will traverse the strings to compare the characters.\n\n2. **Skipping \'_\':** Both pointers skip the \'_\' characters in their respective strings since these represent empty spaces and do not affect the transformation process.\n\n3. **Comparison**:\n   - If both pointers point to the same character (\'L\' or \'R\'), further checks are performed to ensure the movement rules are not violated:\n     - If the character is \'L\', it can only move left. Thus, if `i < j`, the transformation is not possible because an \'L\' in `start` appearing to the right of its position in `target` cannot be achieved.\n     - If the character is \'R\', it can only move right. Therefore, if `i > j`, the transformation is not possible because an \'R\' in `start` appearing to the left of its position in `target` cannot be achieved.\n   - If the characters at `i` and `j` do not match, the transformation is not possible.\n\n4. **Advancement**: If none of the above conditions fail, both pointers are advanced to check the next set of non-\'_\' characters.\n\n5. **Conclusion**: After exiting the loop, if both pointers have reached the end of their respective strings, it indicates that the transformation is possible. If either pointer has not reached the end, it means there are unmatched characters, making the transformation impossible.\n\n# Complexity\n- **Time Complexity**: The algorithm has a time complexity of \\(O(n)\\), where \\(n\\) is the length of the strings. This is because each character in both strings is visited at most once by the pointers. The while loops for skipping \'_\' and the main while loop for comparison ensure that each character is involved in a constant amount of work.\n\n- **Space Complexity**: The space complexity is \\(O(1)\\). The algorithm uses a constant amount of extra space for the pointers and variables used for iteration and comparison, regardless of the input size.\n\n# Code\n```C++ []\nclass Solution {\npublic:\n    bool canChange(string start, string target) {\n\n        int n = start.size();\n\n        int i=0;\n        int j=0;\n        while (i<n || j<n)\n        {\n            while (i<n && start[i] == \'_\')\n                i++;\n\n            while (j<n && target[j] == \'_\')\n                j++;\n\n            if (i==n || j==n)\n                break;\n\n\n            if (start[i] == target[j])\n            {\n                if (start[i] == \'L\' && i < j)\n                    return false;\n\n                else if (start[i] == \'R\' && i > j)\n                    return false;\n            }\n            else \n                return false;\n\n            i++;\n            j++;\n        }\n\n        return i==n && j==n;\n        \n    }\n};\n```\n```Java []\npublic class Solution {\n\n    public boolean canChange(String start, String target) {\n        int n = start.length();\n\n        int i = 0;\n        int j = 0;\n        while (i < n || j < n) {\n            while (i < n && start.charAt(i) == \'_\')\n                i++;\n\n            while (j < n && target.charAt(j) == \'_\')\n                j++;\n\n            if (i == n || j == n)\n                break;\n\n            if (start.charAt(i) == target.charAt(j)) {\n                if (start.charAt(i) == \'L\' && i < j)\n                    return false;\n\n                else if (start.charAt(i) == \'R\' && i > j)\n                    return false;\n            } else\n                return false;\n\n            i++;\n            j++;\n        }\n\n        return i == n && j == n;\n    }\n}\n\n```\n```Python []\nclass Solution:\n    def canChange(self, start: str, target: str) -> bool:\n        n = len(start)\n\n        i, j = 0, 0\n        while i < n or j < n:\n            while i < n and start[i] == \'_\':\n                i += 1\n\n            while j < n and target[j] == \'_\':\n                j += 1\n\n            if i == n or j == n:\n                break\n\n            if start[i] == target[j]:\n                if start[i] == \'L\' and i < j:\n                    return False\n\n                elif start[i] == \'R\' and i > j:\n                    return False\n            else:\n                return False\n\n            i += 1\n            j += 1\n\n        return i == n and j == n\n\n```
['Two Pointers', 'Python', 'C++', 'Java']