# 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 {
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;
}
}

• 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 {
public:
int countSubMultisets(const vector<int>& nums, int l, int r) {
int cnt[20001] = {};
int memo[20001] = {};
const int mod = 1000000007;
for (int n : nums) {
++cnt[n];
}
fill_n(memo, cnt[1] + 1, 1);
for (int n = 2, total = cnt[1]; n <= r; ++n) {
if (!cnt[n]) {
continue;
}
int top = (cnt[n] + 1) * n;
total += n * cnt[n];
for (int i = n, ii = min(total, r); i <= ii; ++i) {
memo[i] = (memo[i] + memo[i - n]) % mod;
}
for (int i = min(total, r); i >= top; --i) {
memo[i] = (mod + memo[i] - memo[i - top]) % mod;
}
}
return accumulate(memo + l, memo + r + 1, 0LL) * (cnt[0] + 1) % mod;
}
};


• func countSubMultisets(nums []int, l int, r int) int {
multiset := make(map[int]int)
for _, num := range nums {
multiset[num]++
}
mem := make([]int, r+1)
mem[0] = 1
prefix := make([]int, len(mem))
for num, occ := range multiset {
copy(prefix, mem)
for sum := num; sum <= r; sum++ {
prefix[sum] = (prefix[sum] + prefix[sum-num]) % mod
}
for sum := r; sum >= 0; sum-- {
if num > 0 {
mem[sum] = prefix[sum]
if sum >= num*(occ+1) {
mem[sum] = (mem[sum] - prefix[sum-num*(occ+1)] + mod) % mod
}
} else {
mem[sum] = (mem[sum] * (occ + 1)) % mod
}
}
}
var result int
for sum := l; sum <= r; sum++ {
result = (result + mem[sum]) % mod
}
return result
}
var mod int = 1e9 + 7


• function countSubMultisets(nums: number[], l: number, r: number): number {
const cnt: number[] = Array(20001).fill(0);
const memo: number[] = Array(20001).fill(0);
const mod: number = 1000000007;
for (const n of nums) {
cnt[n]++;
}
memo.fill(1, 0, cnt[1] + 1);
let total: number = cnt[1];
for (let n = 2; n <= r; ++n) {
if (!cnt[n]) {
continue;
}
const top: number = (cnt[n] + 1) * n;
total += n * cnt[n];
for (let i = n, ii = Math.min(total, r); i <= ii; ++i) {
memo[i] = (memo[i] + memo[i - n]) % mod;
}
for (let i = Math.min(total, r); i >= top; --i) {
memo[i] = (mod + memo[i] - memo[i - top]) % mod;
}
}
let result: number = 0;
for (let i = l; i <= r; i++) {
result = (result + memo[i]) % mod;
}
return (result * (cnt[0] + 1)) % mod;
}