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2902. Count of Sub-Multisets With Bounded Sum

Description

You are given a 0-indexed array nums of non-negative integers, and two integers l and r.

Return the count of sub-multisets within nums where the sum of elements in each subset falls within the inclusive range of [l, r].

Since the answer may be large, return it modulo 109 + 7.

A sub-multiset is an unordered collection of elements of the array in which a given value x can occur 0, 1, ..., occ[x] times, where occ[x] is the number of occurrences of x in the array.

Note that:

  • Two sub-multisets are the same if sorting both sub-multisets results in identical multisets.
  • The sum of an empty multiset is 0.

 

Example 1:

Input: nums = [1,2,2,3], l = 6, r = 6
Output: 1
Explanation: The only subset of nums that has a sum of 6 is {1, 2, 3}.

Example 2:

Input: nums = [2,1,4,2,7], l = 1, r = 5
Output: 7
Explanation: The subsets of nums that have a sum within the range [1, 5] are {1}, {2}, {4}, {2, 2}, {1, 2}, {1, 4}, and {1, 2, 2}.

Example 3:

Input: nums = [1,2,1,3,5,2], l = 3, r = 5
Output: 9
Explanation: The subsets of nums that have a sum within the range [3, 5] are {3}, {5}, {1, 2}, {1, 3}, {2, 2}, {2, 3}, {1, 1, 2}, {1, 1, 3}, and {1, 2, 2}.

 

Constraints:

  • 1 <= nums.length <= 2 * 104
  • 0 <= nums[i] <= 2 * 104
  • Sum of nums does not exceed 2 * 104.
  • 0 <= l <= r <= 2 * 104

Solutions

  • class Solution:
    def countSubMultisets(self, nums: List[int], l: int, r: int) -> int:
        kMod = 1_000_000_007
        # dp[i] := # of submultisets of nums with sum i
        dp = [1] + [0] * r
        count = collections.Counter(nums)
        zeros = count.pop(0, 0)

        for num, freq in count.items():
            # stride[i] := dp[i] + dp[i - num] + dp[i - 2 * num] + ...
            stride = dp.copy()
            for i in range(num, r + 1):
                stride[i] += stride[i - num]
            for i in range(r, 0, -1):
                if i >= num * (freq + 1):
                    # dp[i] + dp[i - num] + dp[i - freq * num]
                    dp[i] = stride[i] - stride[i - num * (freq + 1)]
                else:
                    dp[i] = stride[i]

        return (zeros + 1) * sum(dp[l : r + 1]) % kMod


  • 
class Solution {
    static final int MOD = 1_000_000_007;
    public int countSubMultisets(List<Integer> nums, int l, int r) {
        Map<Integer, Integer> count = new HashMap<>();
        int total = 0;
        for (int num : nums) {
            total += num;
            if (num <= r) {
                count.merge(num, 1, Integer::sum);
            }
        }
        if (total < l) {
            return 0;
        }
        r = Math.min(r, total);
        int[] dp = new int[r + 1];
        dp[0] = count.getOrDefault(0, 0) + 1;
        count.remove(Integer.valueOf(0));
        int sum = 0;
        for (Map.Entry<Integer, Integer> e : count.entrySet()) {
            int num = e.getKey();
            int c = e.getValue();
            sum = Math.min(sum + c * num, r);
            // prefix part
            // dp[i] = dp[i] + dp[i - num] + ... + dp[i - c*num] + dp[i-(c+1)*num] + ... + dp[i %
            // num]
            for (int i = num; i <= sum; i++) {
                dp[i] = (dp[i] + dp[i - num]) % MOD;
            }
            int temp = (c + 1) * num;
            // correction part
            // subtract dp[i - (freq + 1) * num] to the end part.
            // leves dp[i] = dp[i] + dp[i-num] +...+ dp[i - c*num];
            for (int i = sum; i >= temp; i--) {
                dp[i] = (dp[i] - dp[i - temp] + MOD) % MOD;
            }
        }
        int ans = 0;
        for (int i = l; i <= r; i++) {
            ans += dp[i];
            ans %= MOD;
        }
        return ans;
    }
}

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