DP + Sliding window [EXPLAINED]
IntuitionWe need to count the number of subsets of a given list of integers that can sum to values between l and r. We also have to handle duplicates in the lis
# Intuition
We need to count the number of subsets of a given list of integers that can sum to values between l and r. We also have to handle duplicates in the list and perform efficient calculations using DP.

# Approach
I used DP where I maintain an array (dp) to keep track of the possible sums, and for each unique value in the list, I update possible sums by sliding a window to efficiently calculate how each number contributes to the total subsets that match the sum condition.

# Complexity
- Time complexity:
O(n * sum), where n is the number of unique elements in the list and sum is the total sum of all numbers. This is because we loop through the list of numbers and for each value, we update possible sums.

- Space complexity:
O(sum), where sum is the total sum of all numbers because the dp array stores sums up to that total. The space is used to store the count of all subset sums up to sum.

# Code
```python3 []
class Solution:
    def countSubMultisets(self, nums, l, r):
        M = 1_000_000_007
        ma = max(nums)  # maximum element
        total_sum = sum(nums)
        nums.append(0)  # clever handle case of nums having no 0
        
        # Create the count array
        cnt = [0] * (ma + 1)
        for v in nums:
            cnt[v] += 1
        
        # Sort and remove duplicates
        nums = sorted(set(nums))
        
        # Initialize the dp array
        dp = [0] * (total_sum + 1)
        dp[0] = cnt[0]
        
        for v in nums:
            for j in range(v):  # every modulo j
                cur = 0  # sum of all the appropriate DP values for the window
                p = total_sum - j  # points to left index of sliding window when the window is shifted
                while p >= 0 and (total_sum - j - p) // v <= cnt[v]:
                    cur = (cur + dp[p]) % M
                    p -= v
                for k in range(total_sum - j, -1, -v):
                    cur = (cur - dp[k] + M) % M  # take away, because dp[k] means we didn't use any occurrence of the number
                    dp[k] = (dp[k] + cur) % M
                    if p >= 0:
                        # Include the next DP in the sliding window
                        cur = (cur + dp[p]) % M
                        p -= v
        
        ans = 0
        for i in range(l, min(r, total_sum) + 1):
            ans = (ans + dp[i]) % M
        
        return ans

```
['Math', 'Dynamic Programming', 'Memoization', 'Sliding Window', 'Python3']