LeetCode Problem Workspace
Count of Sub-Multisets With Bounded Sum
This problem asks you to count the number of sub-multisets within a given array that have a sum in a specified range.
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Code langs
3
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Practice Focus
Hard · Array scanning plus hash lookup
Answer-first summary
This problem asks you to count the number of sub-multisets within a given array that have a sum in a specified range.
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The task involves counting sub-multisets of an array where the sum of the elements falls within a specified inclusive range. The sum of elements in each subset must lie between two given integers, l and r. This problem requires efficient use of array scanning and hash table techniques to find the solution while handling large arrays and sums efficiently.
Problem Statement
You are given a 0-indexed array of non-negative integers, nums, and two integers l and r. You need to find the number of sub-multisets of nums such that the sum of the elements in the subset falls within the inclusive range [l, r]. A sub-multiset is defined as any subset of the array, where elements can be repeated based on their frequency in the original array.
Since the output could be large, return the result modulo 10^9 + 7. The problem tests your ability to efficiently count subsets that match a sum condition, and requires the use of array scanning and hash lookup strategies to optimize the solution for large inputs.
Examples
Example 1
Input: nums = [1,2,2,3], l = 6, r = 6
Output: 1
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
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
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
Solution Approach
Dynamic Programming with Hash Map
Use dynamic programming (DP) to keep track of the possible subset sums as we iterate through the array. A hash map can be used to store these sums, where the key is the sum and the value is the count of subsets that result in that sum. For each element in the array, update the DP table by considering adding it to the existing sums.
Sliding Window for Range Counting
Once all subset sums are computed using the DP approach, the next step is to efficiently count how many of these sums fall within the given range [l, r]. This can be achieved by sliding over the possible sums and applying range counting techniques to avoid iterating over all possibilities repeatedly.
Optimizing with Frequency Count
To optimize the approach, note that the number of distinct elements in the array is limited to 200, as the sum of nums does not exceed 20000. You can use this frequency count to reduce redundant calculations and speed up the process by focusing on distinct elements and their frequencies.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time complexity depends on the number of subsets formed and the range of sums considered. Using dynamic programming and hash maps, the time complexity can be approximately O(n * sum) where n is the length of the array and sum is the maximum possible sum. The space complexity is O(sum) due to the need to store subset sums in a hash map. Optimizations with frequency counts can help reduce the constant factors in practice.
What Interviewers Usually Probe
- Look for understanding of dynamic programming and hash map usage in counting subsets.
- Test for optimization techniques when handling large arrays and sums, such as using frequency counts.
- Assess the candidate's ability to handle modular arithmetic efficiently in large numbers.
Common Pitfalls or Variants
Common pitfalls
- Failing to properly count subsets with repeated elements can lead to incorrect results.
- Inefficient range counting methods may lead to time limit exceeded errors.
- Overlooking the modulo constraint may result in integer overflow or incorrect final answers.
Follow-up variants
- Modify the problem to count only subsets where all elements are distinct.
- Change the sum range to be exclusive, not inclusive.
- Introduce constraints on the number of elements in each subset and track those as well.
FAQ
What is the key approach for solving the Count of Sub-Multisets With Bounded Sum problem?
The key approach is using dynamic programming with hash maps to track possible subset sums and applying sliding window techniques to count sums within the specified range.
How do you handle large inputs in this problem?
Efficient use of dynamic programming and hash maps, along with optimizations such as frequency counting, helps manage large inputs within time limits.
Why is a hash map useful in this problem?
A hash map stores subset sums efficiently and allows quick lookups to count how many subsets sum to a specific value, reducing redundant calculations.
How does the sliding window technique apply here?
The sliding window technique is used to efficiently count how many subset sums fall within the given range [l, r] without repeatedly iterating over all subsets.
What is the time complexity of this problem?
The time complexity is approximately O(n * sum), where n is the array length and sum is the maximum possible sum, due to the dynamic programming approach.
Solution
Solution 1
#### Python3
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]) % kModContinue Topic
array
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