Smallest Sufficient Team Finder: Optimize Your Skillful Squad!
This is a Java code implementation for finding the smallest sufficient team based on required skills. The code utilizes a dynamic programming approach to solve 
This is a Java code implementation for finding the smallest sufficient team based on required skills. The code utilizes a dynamic programming approach to solve the problem efficiently.\n\n## Intuition\nThe intuition behind the code is to determine the minimum number of people required to form a team that possesses all the skills specified in the `req_skills` array.\n\n## Approach\nThe approach taken in this code is as follows:\n1. Create a `HashMap` to map each skill from the `req_skills` array to its corresponding index.\n2. Convert the `people_skills` list into a binary representation where each bit represents the presence or absence of a specific skill.\n3. Implement a recursive function, `solve`, that takes the following parameters:\n   - `people`: The list of binary representations of people\'s skills.\n   - `i`: The current index of the person being considered.\n   - `mask`: The binary representation of the skills accumulated so far.\n4. In the `solve` function:\n   - Check if all the required skills are covered by checking if the `mask` is equal to the binary representation of all ones (`req_mask`).\n   - Check if all the people have been considered by checking if the index `i` is equal to the number of people (`m`).\n   - If the result for the current state `(mask, i)` is already calculated, return it from the `dp` array.\n   - Calculate the number of people needed by recursively calling `solve` for two cases: adding the current person (`ans1`) and not adding the current person (`ans2`).\n   - Update the choice array to indicate whether the current person was added or not.\n   - Store the minimum of `ans1` and `ans2` in the `dp` array and return it.\n5. Create the necessary data structures and variables and initialize them:\n   - `req_mask`: The binary representation of all ones (`1 << n - 1`) where `n` is the number of required skills.\n   - `dp`: A 2D array to store the results of the dynamic programming calculations.\n   - `choice`: A 2D array to store the choices made during the calculations.\n6. Call the `solve` function with the initial parameters (`people`, `0`, `0`) to calculate the minimum number of people required.\n7. Traverse the `choice` array to determine the selected people and their corresponding skills.\n8. Convert the list of selected people into an array and return it as the final result.\n\n## Complexity Analysis\nThe complexity of the code is as follows:\n\n- Time complexity: The time complexity of the code is determined by the recursive function `solve`. Since the function can be called multiple times for each person, the time complexity can be described as **O(m * 2^n)**, where `m` is the number of people and `n` is the number of required skills.\n- Space complexity: The space complexity is determined by the `dp` and `choice` arrays. Both arrays have dimensions of **(2^n) x m**, where `n` is the number of required skills and `m` is the number of people. Therefore, the space complexity is **O(2^n * m)**.\n\nNote: The code assumes that the `java.util` package is imported.\n\n## Code\n\n```java\nimport java.util.*;\n\nclass Solution {\n    int[][] dp;\n    boolean[][] choice;\n    int req_mask, m;\n\n    public int solve(List<Integer> people, int i, int mask) {\n        if (mask == req_mask)\n            return 0;\n        if (i == m)\n            return 10000;\n        if (dp[mask][i] != -1)\n            return dp[mask][i];\n        int ans1 = solve(people, i + 1, mask | people.get(i)) + 1;\n        int ans2 = solve(people, i + 1, mask);\n        if (ans1 < ans2)\n            choice[mask][i] = true;\n        return dp[mask][i] = Math.min(ans1, ans2);\n    }\n\n    public int[] smallestSufficientTeam(String[] req_skills, List<List<String>> people_skills) {\n        int n = req_skills.length;\n        m = people_skills.size();\n\n        HashMap<String, Integer> skills = new HashMap<>();\n        for (int i = 0; i < n; i++) {\n            skills.put(req_skills[i], i);\n        }\n\n        List<Integer> people = new ArrayList<>();\n        for (List<String> v : people_skills) {\n            int h = 0;\n            for (String skill : v)\n                h |= 1 << skills.get(skill);\n            people.add(h);\n        }\n\n        req_mask = (1 << n) - 1;\n        dp = new int[1 << n][m];\n        choice = new boolean[1 << n][m];\n        for (int[] row : dp)\n            Arrays.fill(row, -1);\n        solve(people, 0, 0);\n\n        List<Integer> ans = new ArrayList<>();\n        int mask = 0;\n        for (int i = 0; i < m; i++) {\n            if (choice[mask][i]) {\n                ans.add(i);\n                mask |= people.get(i);\n            }\n            if (mask == req_mask)\n                break;\n        }\n\n        int[] result = new int[ans.size()];\n        for (int i = 0; i < ans.size(); i++)\n            result[i] = ans.get(i);\n\n        return result;\n    }\n}\n```\n
['Dynamic Programming', 'Recursion', 'Ordered Map', 'Java']