How do you solve Count of Sub-Multisets With Bounded Sum on LeetCode?

Count of Sub-Multisets With Bounded Sum (LeetCode #2902) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(m) space complexity. Key insight: Dynamic programming can significantly reduce the complexity of counting problems.

What is the brute force solution for Count of Sub-Multisets With Bounded Sum?

The brute force approach for Count of Sub-Multisets With Bounded Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible sub-multisets and check their sums. This approach is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n * m).

What algorithm or data structure is used to solve Count of Sub-Multisets With Bounded Sum?

Count of Sub-Multisets With Bounded Sum uses the following concepts: Array, Hash Table, Dynamic Programming, Sliding Window. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count of Sub-Multisets With Bounded Sum?

Explain your thought process clearly before jumping into coding. Consider edge cases, such as empty arrays or all elements being the same. Practice writing clean and efficient code.

What are common mistakes when solving Count of Sub-Multisets With Bounded Sum?

Failing to account for the frequency of elements in the array. Not using modulo operations correctly to avoid overflow.

#2902

Count of Sub-Multisets With Bounded Sum

Hard

ArrayHash TableDynamic ProgrammingSliding WindowHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m)
Space	O(1)	O(m)

💡

Intuition

Time O(n * m)Space O(m)

We can use dynamic programming to count the number of ways to form sums using the elements in the array. This approach is efficient and avoids the overhead of generating all subsets.

⚙️

Algorithm

3 steps

1Step 1: Create a dp array where dp[x] represents the number of ways to form the sum x using elements from nums.
2Step 2: Iterate through each unique number in nums and update the dp array for all possible sums that can be formed with that number.
3Step 3: Finally, sum the values in dp from index l to r to get the total count of valid sub-multisets.

solution.py15 lines

1# Full working Python code
2
3def count_subsets(nums, l, r):
4    MOD = 10**9 + 7
5    dp = [0] * (20001)
6    dp[0] = 1  # Base case: one way to form sum 0 (empty set)
7
8    for num in set(nums):
9        count = nums.count(num)
10        for j in range(20000, num - 1, -1):
11            for k in range(1, count + 1):
12                if j >= num * k:
13                    dp[j] = (dp[j] + dp[j - num * k]) % MOD
14
15    return sum(dp[l:r + 1]) % MOD

ℹ

Complexity note: Here, n is the number of unique elements in nums, and m is the maximum sum (20000). The complexity is efficient because we only iterate through the unique elements and their counts.

1Dynamic programming can significantly reduce the complexity of counting problems.
2Understanding how to manage counts of elements is crucial in combinatorial problems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.