Explained 38ms, beat 100% ; avoid tmp DP array for optimization
IntuitionGroup identical numbers. Process each distinct number separately since total distinct numbers ≤ sqrt(sum). For each number, process all possible counts
# Intuition
Group identical numbers. Process each distinct number separately since total distinct numbers ≤ sqrt(sum). For each number, process all possible counts (0 to max_count) using sliding window technique to group positions with same remainder efficiently.

# Approach
1. Group identical numbers and count occurrences
2. Handle zeros separately - multiply existing combinations by (count+1)
3. For each non-zero number:
   - Process each remainder sequence using sliding window
   - Window size = count+1 (all possible quantities of current number)
   - Update dp array sum2count[i] = ways to make sum i
4. Return sum of combinations in range [l,r]

# Complexity
- Time complexity: $$O(N\sqrt{S} + R)$$ where N = nums.size(), S = sum(nums), R = r
  - Sorting: O(N log N)
  - Per distinct number: O(num * r/num) = O(r) for window operations
  - Total distinct numbers ≤ sqrt(S)

- Space complexity: $$O(R)$$ for dp array sum2count

# Code
```cpp []
class Solution {
public:
    int countSubMultisets(vector<int>& nums, int l, int r) {
        vector<pair<int, int>> num_counts;
        sort(nums.begin(), nums.end());
        for (int i = 1, beg = 0; i <= nums.size(); ++i) {
            if (i == nums.size() || nums[i] != nums[beg]) {
                num_counts.emplace_back(nums[beg], i - beg);
                beg = i;
            }
        }

        vector<long long> sum2count(r + 1); // [i] number of ways to make sum i
        sum2count[0] = 1;
        int sum_lim = 0; // maximum possible sum achieved so far
        for (const auto [num, count] : num_counts) {
            if (num == 0) { // zero - handle separately, multoply count+1
               for (long long& c : sum2count)
                    c *= count + 1;
                continue;
            }
            // process each remainder mod num using sliding window
            sum_lim = min(r, sum_lim + num * count);
            for (int num_rem_offset = 0; num_rem_offset < num; ++num_rem_offset) {
                long long window_sum = 0;
                int q = sum_lim - num_rem_offset + num;
                for (int i = 0; i <= count && q >= 0; ++i) { // initial window of size count+1
                    q -= num;
                    if (q < 0) break;
                    window_sum += sum2count[q];
                }
                for (int p = sum_lim - num_rem_offset; p >= 0 ; p -= num) { // slide window
                    long long up_count = window_sum % kMod; // allows to avoid tmp update_sum2count
                    window_sum -= sum2count[p]; // remove leftmost
                    sum2count[p] = up_count;    // update current
                    q -= num;
                    if (q >= 0)
                        window_sum += sum2count[q]; // add rightmost
                }
            }
        }

        // sum range [l,r]
        return (accumulate(sum2count.begin() + l, sum2count.end(), 0LL) % kMod + kMod) % kMod;
    }

private:
    static constexpr long long kMod = 1e9 + 7;
};

```
['C++']