[Python3] Bitmask DP - Detailed Explanation
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n1. Preprocessing:\n\n
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n**1. Preprocessing:**\n\n- `n` is calculated as the length of the input list nums.\n- The list `nums` is sorted in ascending order to prune later.\n- If any number occurs more than `k` times, it\'s impossible to partition into `k` subsets, so the function returns -1.\n- If `k` is equal to `n`, it means each subset will have only one element, and the incompatibility is 0.\n- If `k` is 1, it means there will be only one subset, and the incompatibility is the difference between the maximum and minimum elements.\n\n**2. Calculating Next Larger Elements:**\n\n- `next_larger` is an array to store the index of the next larger element for each element in `nums`.\n- The loop iterates backward through nums, updating next_larger values.\n- It helps us to skip the duplicate number for next iteration of `i`.\n\n**3. Dynamic Programming:**\n\n- `len_set` is calculated as the desired size of each subset.\n- The `dp` function is memoized to store calculated results.\n- The `dp` function takes three parameters:\n    - `i`: the current index in nums\n    - `mask`: a bitmask representing which elements have been chosen\n    - `size`: the current size of the subset being formed\n- A pruning trick: We always take the smallest available number as the beginning of a new group.\n- The base cases are handled:\n    - If all elements have been chosen, the incompatibility is 0.\n    - If the current index is out of bounds, the incompatibility is infinity.\n    - If the current element has already been chosen, the function recursively calls itself with the next index and the same mask and size.\n- The function calculates the minimum incompatibility by considering two options:\n    - Include the current element in the current subset:\n        - Calculate the new index, mask, and size.\n        - If the current subset is not yet full, recursively call `dp` with the new values.\n        - If the current subset is full, return the incompatibility of the current subset plus the current element\'s value.\n    - Exclude the current element from the current subset:\n        - Recursively call `dp` with the next index and the same mask and size.\n- The function returns the minimum of these two options.\n\n**4. Final Result:**\n\n- The `dp` function is called with initial values of 0 for all parameters.\n- The returned value is the minimum incompatibility.\n\nThe code effectively solves the problem of finding the minimum incompatibility by using dynamic programming to explore all possible combinations of subsets and selecting the one with the minimum incompatibility. The `next_larger` array helps optimize the calculation by avoiding unnecessary recursive calls. This is called **Pruning**.\n\n# Complexity\n- Time complexity: $$O(2^N * N * N / k)$$ ~ $$O(2^N * N^2)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(2^N * N * N / k)$$ ~ $$O(2^N * N^2)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```python3 []\nclass Solution:\n    def minimumIncompatibility(self, nums: List[int], k: int) -> int:\n        n = len(nums)\n        nums.sort()\n        if max(Counter(nums).values()) > k: return -1\n        if k == n: return 0\n        if k == 1: return nums[n - 1] - nums[0]\n\n        next_larger, cur, next_ptr = [0 for _ in range(n)], nums[-1], n\n        for i in range(n - 1, -1, -1):\n            if nums[i] != cur: cur, next_ptr = nums[i], i + 1\n            next_larger[i] = next_ptr\n        \n        len_set = n // k - 1\n        @cache\n        def dp(i: int, mask: int, size: int) -> int:\n            if mask == (1 << n) - 1: return 0\n            if i == n: return float("inf")\n            if mask & (1 << i): return dp(i + 1, mask, size)\n            new_i, new_mask, new_size = next_larger[i], mask | (1 << i), size + 1\n            if size == 0: return dp(new_i, new_mask, new_size) - nums[i]\n            elif size < len_set: return min(\n                dp(new_i, new_mask, new_size),\n                dp(new_i, mask, size)\n            )\n            else: return min(\n                dp(0, new_mask, 0) + nums[i], dp(new_i, mask, size)\n            )\n        return dp(0, 0, 0)\n```
['Array', 'Dynamic Programming', 'Bit Manipulation', 'Python3']