Problem Description

This problem asks you to count the number of sub-multisets from a given array nums where the sum of elements falls within the range [l, r].

A sub-multiset is a collection of elements from the array where:

• Each element can appear 0 or more times (up to its frequency in the original array)
• Order doesn't matter (it's unordered)
• Two sub-multisets are considered the same if they contain the same elements with the same frequencies

For example, if nums = [1,2,2,3]:

• You can use 0, 1, or 2 copies of the element 2 (since it appears twice)
• You can use 0 or 1 copy of elements 1 and 3 (since they appear once each)
• The empty set is valid and has a sum of 0

The solution uses dynamic programming with an optimization for handling multiple occurrences of the same element:

1. Initial Setup: Create a DP array where dp[i] represents the count of sub-multisets with sum i. Handle zeros separately since they don't affect the sum but multiply the number of ways (you can include 0, 1, 2, ... up to all zeros).

2. Processing Each Unique Element: For each unique number and its frequency:

   • Create a stride array that accumulates sums: stride[i] = dp[i] + dp[i-num] + dp[i-2*num] + ...
   • This represents all possible ways to form sum i using any number of copies of num
   • Then adjust for the frequency limit: if we can use at most freq copies, we subtract the contributions from using more than freq copies

3. Final Count: Sum up all dp[i] for i in range [l, r] and multiply by (zeros + 1) to account for all possible ways to include zeros.

The modulo operation 10^9 + 7 is applied to keep the result within bounds since the count can be very large.

Intuition

The key insight is recognizing this as a variant of the classic "subset sum" problem, but with a twist: elements can be used multiple times based on their frequency in the original array.

Let's think about how to build sub-multisets incrementally. If we have a DP solution for some elements, how do we incorporate a new element that appears freq times?

The naive approach would be to iterate through each possible count (0 to freq) of the new element and update our DP. For an element num appearing freq times, we'd need to consider adding 0*num, 1*num, 2*num, ..., freq*num to existing sums. This would be inefficient with nested loops.

The clever optimization comes from recognizing a pattern. When we want to know how many ways to form sum i using up to freq copies of num, we can think of it as:

• Ways to form i without using num (which is dp[i])
• Plus ways to form i-num (then add one num)
• Plus ways to form i-2*num (then add two num)
• And so on...

This forms a cumulative sum pattern! The stride array efficiently captures this: stride[i] contains the sum dp[i] + dp[i-num] + dp[i-2*num] + ..., representing all ways to form sum i using any number of copies of num.

But wait - we can only use up to freq copies. So if stride[i] includes contributions from using more than freq copies, we need to subtract those out. That's why we compute stride[i] - stride[i - num*(freq+1)] when possible - this removes the contribution from using freq+1 or more copies.

Zeros are handled separately because they're special: adding zeros doesn't change the sum, but each zero doubles our options (include it or not). So if we have zeros zeros, we multiply our final count by (zeros + 1) to account for all 2^zeros ways to include them.

This approach transforms an exponential problem into a polynomial one by cleverly reusing computations through the stride technique.

Learn more about Dynamic Programming and Sliding Window patterns.

Solution Approach

The implementation uses dynamic programming with a clever optimization for handling multiple occurrences of elements:

1. Initialization and Setup

dp = [1] + [0] * r
count = collections.Counter(nums)
zeros = count.pop(0, 0)

• Create a DP array where dp[i] tracks the number of sub-multisets with sum exactly i
• Initialize dp[0] = 1 (one way to make sum 0: empty set)
• Count frequencies of each element using Counter
• Extract and handle zeros separately since they don't affect sums

2. Processing Each Unique Element

For each unique number and its frequency:

for num, freq in count.items():
    stride = dp.copy()
    for i in range(num, r + 1):
        stride[i] += stride[i - num]

• Create a stride array as a copy of current dp
• Build cumulative sums: stride[i] = dp[i] + dp[i-num] + dp[i-2*num] + ...
• This represents using 0, 1, 2, ... copies of num to form sum i

3. Applying Frequency Constraints

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

• Iterate backwards through sums to avoid interference
• If i >= num * (freq + 1), we can use the difference formula:
  • stride[i] includes all ways using unlimited copies of num
  • stride[i - num * (freq + 1)] represents ways using at least freq + 1 copies
  • The difference gives us ways using at most freq copies
• Otherwise, stride[i] already respects the frequency limit (can't use more than i/num copies anyway)

4. Computing Final Result

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

• Sum all valid sub-multiset counts in range [l, r]
• Multiply by (zeros + 1) to account for all ways to include/exclude zeros
• Apply modulo to keep result within bounds

Key Insights:

• The stride technique transforms O(freq) operations per sum into O(1) by precomputing cumulative sums
• Processing in reverse order prevents using updated values prematurely
• Separating zeros simplifies logic since they're multiplicative factors rather than additive
• Total time complexity: O(n * r) where n is the number of unique elements and r is the upper bound

Example Walkthrough

Let's walk through a small example with nums = [1, 2, 2, 3], l = 1, r = 5.

Step 1: Initialize

• Count frequencies: {1: 1, 2: 2, 3: 1} (no zeros)
• Create DP array: dp = [1, 0, 0, 0, 0, 0] (index represents sum 0-5)
• dp[0] = 1 means one way to make sum 0 (empty set)

Step 2: Process element 1 (frequency = 1)

• Create stride array: stride = [1, 0, 0, 0, 0, 0]
• Build cumulative sums for i = 1 to 5:
  • stride[1] = stride[1] + stride[0] = 0 + 1 = 1
  • stride[2] = stride[2] + stride[1] = 0 + 1 = 1
  • stride[3] = stride[3] + stride[2] = 0 + 1 = 1
  • stride[4] = stride[4] + stride[3] = 0 + 1 = 1
  • stride[5] = stride[5] + stride[4] = 0 + 1 = 1
• Apply frequency constraint (freq = 1, so max 1 copy):
  • For i = 5: 5 >= 1*(1+1) = 2, so dp[5] = stride[5] - stride[3] = 1 - 1 = 0
  • For i = 4: 4 >= 2, so dp[4] = stride[4] - stride[2] = 1 - 1 = 0
  • For i = 3: 3 >= 2, so dp[3] = stride[3] - stride[1] = 1 - 1 = 0
  • For i = 2: 2 >= 2, so dp[2] = stride[2] - stride[0] = 1 - 1 = 0
  • For i = 1: 1 < 2, so dp[1] = stride[1] = 1
• Result: dp = [1, 1, 0, 0, 0, 0] (can make sum 0 with {} and sum 1 with {1})

Step 3: Process element 2 (frequency = 2)

• Create stride array: stride = [1, 1, 0, 0, 0, 0]
• Build cumulative sums:
  • stride[2] = 0 + stride[0] = 0 + 1 = 1
  • stride[3] = 0 + stride[1] = 0 + 1 = 1
  • stride[4] = 0 + stride[2] = 0 + 1 = 1
  • stride[5] = 0 + stride[3] = 0 + 1 = 1
• Apply frequency constraint (freq = 2, so max 2 copies):
  • For i = 5: 5 < 2*(2+1) = 6, so dp[5] = stride[5] = 1
  • For i = 4: 4 < 6, so dp[4] = stride[4] = 1
  • For i = 3: 3 < 6, so dp[3] = stride[3] = 1
  • For i = 2: 2 < 6, so dp[2] = stride[2] = 1
  • For i = 1: 1 < 6, so dp[1] = stride[1] = 1
• Result: dp = [1, 1, 1, 1, 1, 1]
• Possible sets: {}, {1}, {2}, {1,2}, {2,2}, {1,2,2}

Step 4: Process element 3 (frequency = 1)

• Create stride array: stride = [1, 1, 1, 1, 1, 1]
• Build cumulative sums:
  • stride[3] = 1 + stride[0] = 1 + 1 = 2
  • stride[4] = 1 + stride[1] = 1 + 1 = 2
  • stride[5] = 1 + stride[2] = 1 + 1 = 2
• Apply frequency constraint (freq = 1, so max 1 copy):
  • For i = 5: 5 < 3*(1+1) = 6, so dp[5] = stride[5] = 2
  • For i = 4: 4 < 6, so dp[4] = stride[4] = 2
  • For i = 3: 3 < 6, so dp[3] = stride[3] = 2
  • For i = 2: 2 < 6, so dp[2] = stride[2] = 1
  • For i = 1: 1 < 6, so dp[1] = stride[1] = 1
• Final: dp = [1, 1, 1, 2, 2, 2]

Step 5: Calculate result

• Sum for range [1, 5]: dp[1] + dp[2] + dp[3] + dp[4] + dp[5] = 1 + 1 + 2 + 2 + 2 = 8
• No zeros to multiply, so answer = 8

The 8 valid sub-multisets with sums in [1, 5] are:

1. {1} → sum = 1
2. {2} → sum = 2
3. {3} → sum = 3
4. {1,2} → sum = 3
5. {2,2} → sum = 4
6. {1,3} → sum = 4
7. {2,3} → sum = 5
8. {1,2,2} → sum = 5

Solution Implementation

Time and Space Complexity

Time Complexity: O(n * r) where n is the number of unique elements in nums and r is the upper bound of the sum range.

The algorithm iterates through each unique element in the counter (at most n unique elements). For each unique element with value num:

• Creating the stride array takes O(r) time
• Computing the prefix sums in stride takes O(r) time (iterating from num to r)
• Updating the dp array takes O(r) time (iterating from r down to 1)

Therefore, for each unique element, we perform O(r) operations, resulting in a total time complexity of O(n * r).

Space Complexity: O(r)

The space complexity is dominated by:

• The dp array of size r + 1: O(r)
• The stride array of size r + 1: O(r)
• The count dictionary storing at most n unique elements: O(n)

Since typically r is expected to be larger than n in problems involving sum ranges, and we're using O(r) space for the arrays, the overall space complexity is O(r).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Update Order in DP Array

One of the most critical pitfalls is updating the DP array in the wrong order when applying frequency constraints. If you update from low to high indices instead of high to low, you'll use already-modified values in your calculations, leading to incorrect results.

Wrong approach:

# This will cause errors - using updated values prematurely
for sum_value in range(1, r + 1):  # Forward iteration - WRONG!
    if sum_value >= number * (frequency + 1):
        dp[sum_value] = prefix_sum[sum_value] - prefix_sum[sum_value - number * (frequency + 1)]
    else:
        dp[sum_value] = prefix_sum[sum_value]

Correct approach:

# Process in reverse to avoid using modified values
for sum_value in range(r, 0, -1):  # Backward iteration - CORRECT!
    if sum_value >= number * (frequency + 1):
        dp[sum_value] = prefix_sum[sum_value] - prefix_sum[sum_value - number * (frequency + 1)]
    else:
        dp[sum_value] = prefix_sum[sum_value]

2. Forgetting to Handle Zero Elements Separately

Zeros require special handling because they don't contribute to the sum but multiply the number of valid sub-multisets. Forgetting to extract and handle zeros separately will give incorrect counts.

Wrong approach:

# Processing zeros like regular numbers - WRONG!
for number, frequency in frequency_map.items():
    # This will incorrectly handle zeros
    prefix_sum = dp.copy()
    for sum_value in range(number, r + 1):
        prefix_sum[sum_value] += prefix_sum[sum_value - number]

Correct approach:

# Extract zeros first and handle them separately
zero_count = frequency_map.pop(0, 0)  # Remove zeros from processing
# ... process non-zero elements ...
# Then multiply final result by (zero_count + 1)
result = (zero_count + 1) * sum(dp[l:r + 1]) % MOD

3. Modulo Operation Timing

Applying modulo operation at the wrong time or forgetting it entirely can cause integer overflow or incorrect results.

Wrong approach:

# Forgetting modulo during intermediate calculations - can cause overflow
for sum_value in range(number, r + 1):
    prefix_sum[sum_value] += prefix_sum[sum_value - number]  # No modulo - RISKY!

# Or applying modulo incorrectly at the end
result = (zero_count + 1) * sum(dp[l:r + 1])
return result % MOD  # Overflow might have already occurred

Correct approach:

# Apply modulo during intermediate steps if values get large
for sum_value in range(number, r + 1):
    prefix_sum[sum_value] = (prefix_sum[sum_value] + prefix_sum[sum_value - number]) % MOD

# And ensure final calculation doesn't overflow
result = ((zero_count + 1) % MOD * sum(dp[l:r + 1]) % MOD) % MOD

4. Off-by-One Errors in Range Calculations

The problem asks for sums in range [l, r] inclusive. Common mistakes include using exclusive ranges or incorrect array slicing.

Wrong approaches:

# Exclusive upper bound - WRONG!
return (zero_count + 1) * sum(dp[l:r]) % MOD  # Missing dp[r]

# Or incorrect range check
for sum_value in range(r + 1):  # Should start from r, not r+1 in reverse
    # ...

Correct approach:

# Include both l and r in the final sum
return (zero_count + 1) * sum(dp[l:r + 1]) % MOD  # r+1 for inclusive range

5. Not Creating a Copy of DP Array for Prefix Sum

Using the same array for both DP and prefix sum calculations will corrupt your data.

Wrong approach:

# Modifying dp directly - WRONG!
for sum_value in range(number, r + 1):
    dp[sum_value] += dp[sum_value - number]  # Corrupts original dp values

Correct approach:

# Create a separate copy for prefix sum calculations
prefix_sum = dp.copy()
for sum_value in range(number, r + 1):
    prefix_sum[sum_value] += prefix_sum[sum_value - number]