3D DP Python Beats 100%
Intuition\nThe problem is about finding a subsequence of indices in word1 such that the characters at those indices form a string almost equal to word2. The seq
# Intuition\nThe problem is about finding a subsequence of indices in `word1` such that the characters at those indices form a string almost equal to `word2`. The sequence should be lexicographically smallest, meaning we need to pick indices in increasing order as soon as we find valid matches. We are also allowed to change at most one character from `word1` to make it match `word2`.\n\n# Approach\n1. We use a recursive dynamic programming approach with memoization.\n2. We define a function `dp(i, j, k)` where:\n   - `i` is the current index in `word1`.\n   - `j` is the current index in `word2`.\n   - `k` indicates whether we can still make one character change (`k = 1` if change is allowed, `k = 0` if no change is allowed).\n3. The base case is when `j` reaches the length of `word2` (i.e., we successfully matched all characters of `word2`).\n4. If the characters at the current indices of `word1` and `word2` match, we proceed to the next indices in both strings.\n5. If they don\'t match, we check if we can still change one character (`k = 1`). If yes, we try to match by changing the current character.\n6. After checking for matches or changes, we continue searching through `word1` to find the next valid index.\n7. Finally, we return the answer in lexicographical order by reversing the result list.\n\n# Complexity\n- **Time complexity**: The time complexity is approximately $$O(n \\times m)$$, where `n` is the length of `word1` and `m` is the length of `word2`. This is because for each index in `word1`, we may recursively compare with each index in `word2`, and memoization ensures that each state is computed at most once.\n- **Space complexity**: The space complexity is $$O(n \\times m)$$ due to the recursive stack and memoization table.\n\n# Code\n```python\nclass Solution:\n    def validSequence(self, word1: str, word2: str) -> List[int]:\n        ans = []\n\n        # Dynamic programming with memoization (using cache decorator)\n        @cache \n        def dp(i, j, k):\n            # Base cases\n            if j == len(word2): return True \n            if i == len(word1): return False \n\n            # Case 1: Characters at current indices match, proceed with next indices\n            if word1[i] == word2[j]:\n                res = dp(i + 1, j + 1, k)\n                if res:\n                    ans.append(i)  # Add index to answer if valid\n                return res \n\n            # Case 2: Characters don\'t match, but we can still make one change\n            if k:\n                res = dp(i + 1, j + 1, 0)  # Use up the one allowed change\n                if res:\n                    ans.append(i)\n                    return True \n            \n            # Case 3: Continue scanning word1 until we find a matching character\n            while i < len(word1) and word1[i] != word2[j]:\n                i += 1 \n            \n            return dp(i, j, k)  # Continue with the updated index\n        return ans[::-1] if dp(0, 0, 1) else []\n
['Python3']