Beginner Friendly Beats 90%
\n\n# Code\npython3 []\nfrom collections import Counter\nimport numpy as np\n\nclass Solution:\n    def countSubMultisets(self, nums: List[int], l: int, r: int)
\n\n# Code\n```python3 []\nfrom collections import Counter\nimport numpy as np\n\nclass Solution:\n    def countSubMultisets(self, nums: List[int], l: int, r: int) -> int:\n        # Step 1: Define the modulo value for large numbers\n        MOD = 10**9 + 7\n        \n        # Step 2: Count the frequency of each number in the nums list\n        count = Counter(nums)\n        \n        # Step 3: Create a dynamic programming (dp) array to store intermediate results\n        # We need an array with a size of \'r + 1\' to store the sum of different combinations\n        dp = np.zeros(r + 1, dtype=np.int64)\n        dp[0] = 1  # There\'s one way to make sum 0 (by choosing no elements)\n        \n        # Step 4: Iterate over each unique number in the count dictionary\n        for num in count:\n            if num == 0:\n                continue  # Skip the number 0 because it doesn\'t change the sum\n            \n            # Create a new dp array to store updated results after adding current number\n            new_dp = np.array(dp)\n            \n            # Step 5: For each frequency of the current number, update the new dp array\n            for i in range(1, count[num] + 1):\n                # Update the new dp array for sums that can be made by adding the number \'num\' multiple times\n                new_dp[num * i:] += dp[:-num * i]\n            \n            # Update the original dp array with the new values and apply the modulo to keep numbers manageable\n            dp = new_dp % MOD\n        \n        # Step 6: Calculate the final result by summing up the valid combinations\n        # We need to sum the dp values between \'l\' and \'r\' and account for how many times \'0\' appears in nums\n        return (count[0] + 1) * sum(dp[l:r + 1]) % MOD\n\n```
['Python3']