Biweekly contest 140 C - Solution
Intuition\n Describe your first thoughts on how to solve this problem. \n## Problem Breakdown\n\nWe are given two strings, word1 and word2. We need to find a le
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n## Problem Breakdown\n\nWe are given two strings, `word1` and `word2`. We need to find a lexicographically smallest sequence of indices from `word1` such that the string formed by these indices is **almost equal** to `word2`. "Almost equal" means we can change at most one character in the sequence to match `word2`.\n\n### Basic Greedy Approach (Fails)\n\nAt first glance, a **greedy approach** might seem appropriate. The idea would be to iteratively match characters in both strings, always picking the lexicographically smallest option. However, this strategy fails because we can\'t predict whether taking or skipping a mismatched character will allow us to complete the rest of the sequence while maintaining the "almost equal" condition. As such, a greedy approach doesn\'t account for the flexibility of making a single character change, leading to incorrect results.\n\n### Brute Force Dynamic Programming (2D DP - Not Optimal)\n\nThe next step in refining our approach is to consider **dynamic programming**. A standard DP approach would involve maintaining a 2D DP array where:\n\n- `i` represents the current position in `word1`.\n- `j` represents the current position in `word2`.\n\nFor each position `i` and `j`, we would check:\n- If `word1[i] == word2[j]`, take the index and move both `i` and `j`.\n- If they don\'t match, we can either:\n  - **Skip** the current character in `word1` and move `i` (not `j`).\n  - **Take** the current character and flag that we\'ve used our single mismatch option, moving both `i` and `j`.\n\nHowever, this approach involves exploring multiple branches (taking vs. skipping) and keeping track of the mismatch, leading to **O(n1 * n2)** time complexity. While correct, this solution is too slow for large inputs, and would likely result in a **TLE** (Time Limit Exceeded).\n\n### Optimal Approach: MaxSuff DP\n\nA more optimal approach involves creating a **MaxSuff DP** array. This array stores, for each index `i` in `word1`, the maximum suffix of `word2` that can be formed starting from that index in `word1`.\n\nThe idea is:\n1. **Build a DP array** (`dp[i]`) that holds the length of the longest suffix of `word2` that can be formed starting from `i` in `word1`.\n2. **Iterate** through `word1` and `word2`. For each mismatch between `word1[i]` and `word2[j]`, check whether it\'s possible to form the remainder of `word2` using the suffix from `dp[i + 1]`.\n3. Ensure the resulting sequence is lexicographically smallest by always taking the first valid option.\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n### `createmaxsuff` function:\n- This function iterates from the end of `word1` and `word2` to build the `dp` array.\n- The while loop runs from `i = n1 - 1` to `i = 0`, so it executes at most `n1` times.\n- Inside the loop, the second pointer `j` decreases when characters match, but this does not affect the overall time complexity as both pointers move in a linear fashion.\n- Therefore, the time complexity of this function is **O(n1)**.\n\n### `manage` function:\n- This function iterates through both `word1` and `word2` starting from index `i` and `j` and runs until one of them reaches the end.\n- The while loop will run at most `n1` times, but since it is called only when certain conditions are met (during the main loop in `validSequence`), the time complexity contribution will be within **O(n1)** overall.\n\n### `validSequence` function:\n- The initial part of the function sets up the lengths `n1` and `n2`, and calls `createmaxsuff` with time complexity **O(n1)**.\n- Then, the while loop (starting at `while (j < n2)`) goes through `word1` and `word2`, matching characters.\n- In the worst case, this loop can iterate through the entire `word1` (with a total of `n1` iterations).\n- Inside the loop, it checks conditions and may call `manage`, but this does not add extra complexity as it runs at most once per iteration.\n- Therefore, the overall complexity of this loop is **O(n1)**.\n\n### Final Time Complexity:\n- The `createmaxsuff` runs in **O(n1)**.\n- The main loop and the possible call to `manage` inside `validSequence` also run in **O(n1)**.\n\nHence, the overall time complexity of the solution is **O(n1)**, where `n1` is the length of `word1`. Since `n1 >= n2` is ensured in the function, this is the dominating term.\n\n\n- Space complexity:\nThe space complexity is **O(n1)** since we are using a dp array to store the maximum suffix of word1 in word2.\n\n# Code\n```cpp []\nclass Solution {\npublic:\n    void createmaxsuff(string& w1, string& w2, int n1, int n2,\n                       vector<int>& dp) {\n        int i = n1 - 1;\n        int j = n2 - 1;\n        int temp = 0;\n        while (i >= 0) {\n            if (j >= 0) {\n                if (w1[i] == w2[j]) {\n                    temp++;\n                    dp[i] = temp;\n                    j--;\n                } else {\n                    dp[i] = temp;\n                }\n\n            } else {\n                dp[i] = temp;\n            }\n            i--;\n        }\n    }\n    int manage(int& i, int& j, int n1, int n2, string& w1, string& w2,\n               vector<int>& dp, vector<int>& ans) {\n\n        while (j < n2 && i < n1) {\n            if (w1[i] == w2[j]) {\n                ans.push_back(i);\n                j++;\n            }\n            i++;\n        }\n        if (j >= n2) {\n            return 1;\n        }\n        return -1;\n    }\n    vector<int> validSequence(string word1, string word2) {\n        int n1 = word1.size();\n        int n2 = word2.size();\n        if (n1 < n2) {\n            return {};\n        }\n        vector<int> dp(n1);\n        createmaxsuff(word1, word2, n1, n2, dp);\n        \n        int i = 0;\n        int j = 0;\n        vector<int> ans;\n        while (j < n2) {\n            if (i >= n1) {\n                return {};\n            }\n            if (word1[i] == word2[j]) {\n                ans.push_back(i);\n                i++;\n                j++;\n            } else {\n                if (n2 - j - 1 == 0) {\n                    ans.push_back(i);\n                    j++;\n                } else {\n                    if (i + 1 >= n1) {\n                        return {};\n                    } else {\n                        if (dp[i + 1] >= n2 - j - 1) {\n                            ans.push_back(i);\n                            j++;\n                            i++;\n                            if (manage(i, j, n1, n2, word1, word2, dp, ans) ==\n                                1) {\n                                return ans;\n                            } else {\n                                return {};\n                            }\n                        }\n                    }\n                    i++;\n                }\n            }\n        }\n        return ans;\n    }\n};\n```
['C++']