Python3   ||    10 lines, w/ explanation  ||    T/S: 82%/ 40%
Here\'s the intuition:\n\n  1. The number of Ls, number of Rs, and number of _s must be equal between the two strings.\n\n  1. The ordering of Ls and Rs in the 
Here\'s the intuition:\n\n  1. The number of *L*s, number of *R*s, and number of _s must be equal between the two strings.\n\n  1. The ordering of *L*s and *R*s in the two strings must be the same.\n  2. *L*s can only move left and *R*s can only move right, so each *L* in start cannot be to the left of its corresponding *L* in target, and each *R* cannot be to the right of its corresponding *R* in target.\n```python3 []\nclass Solution:\n    def canChange(self, start: str, target: str) -> bool:\n\n        # Check initial conditions\n        if (len(start) != len(target)\n            or start.count(\'_\') != target.count(\'_\')): return False\n\n        n, i, j = len(start), 0, 0\n\n        while i < n or j < n:\n            # Move i and j to the next non-underscore character\n            while i < n and start[i]  == \'_\': i += 1\n            while j < n and target[j] == \'_\': j += 1\n\n            # If both indices are out of bounds, we are done\n            if i == n and j == n: return True\n\n            # If one index is out of bounds or characters differ, return False\n            if i == n or j == n or start[i] != target[j]: return False\n\n            # Check if the transformation is valid\n            if start[i] == \'L\' and i < j: return False\n            if start[i] == \'R\' and i > j: return False\n\n            # Move to the next character\n            i += 1\n            j += 1\n\n        return True\n```\n```cpp []\nclass Solution {\npublic:\n    bool canChange(string start, string target) {\n        // Check initial conditions\n        if (count(start.begin(), start.end(), \'_\') != count(target.begin(), target.end(), \'_\')) {\n            return false;\n        }\n\n        int n = start.size();\n        int i = 0, j = 0;\n\n        while (i < n || j < n) {\n            // Move i and j to the next non-underscore character\n            while (i < n && start[i] == \'_\') i++;\n            while (j < n && target[j] == \'_\') j++;\n\n            // If both indices are out of bounds, we are done\n            if (i == n && j == n) return true;\n\n            // If one index is out of bounds or characters differ, return false\n            if (i == n || j == n || start[i] != target[j]) return false;\n\n            // Check if the transformation is valid\n            if (start[i] == \'L\' && i < j) return false;\n            if (start[i] == \'R\' && i > j) return false;\n\n            // Move to the next character\n            i++;\n            j++;\n        }\n\n        return true;\n    }\n};\n\n```\n```java []\nclass Solution {\n    public boolean canChange(String start, String target) {\n        // Check initial conditions\n        if (start.length() != target.length() || countChar(start, \'_\') != countChar(target, \'_\')) {\n            return false;\n        }\n\n        int n = start.length();\n        int i = 0, j = 0;\n\n        while (i < n || j < n) {\n            // Move i and j to the next non-underscore character\n            while (i < n && start.charAt(i) == \'_\') i++;\n            while (j < n && target.charAt(j) == \'_\') j++;\n\n            // If both indices are out of bounds, we are done\n            if (i == n && j == n) return true;\n\n            // If one index is out of bounds or characters differ, return false\n            if (i == n || j == n || start.charAt(i) != target.charAt(j)) return false;\n\n            // Check if the transformation is valid\n            if (start.charAt(i) == \'L\' && i < j) return false;\n            if (start.charAt(i) == \'R\' && i > j) return false;\n\n            // Move to the next character\n            i++;\n            j++;\n        }\n\n        return true;\n    }\n\n    // Helper method to count occurrences of a character\n    private int countChar(String s, char ch) {\n        int count = 0;\n        for (char c : s.toCharArray()) {\n            if (c == ch) count++;\n        }\n        return count;\n    }\n}\n\n```\n[https://leetcode.com/problems/move-pieces-to-obtain-a-string/submissions/1470587049/?envType=daily-question&envId=2024-12-05](https://leetcode.com/problems/move-pieces-to-obtain-a-string/submissions/1470587049/?envType=daily-question&envId=2024-12-05)\n\nI could be wrong, but I think that time complexity is *O*(*N*) and space complexity is ~*O*(*N*)~ *O*(1), in which *N* ~ `len(start)`.
['C++', 'Java', 'Python3']