Count of Subsets With Sum K

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Steps

setup

Initialize dp array

Create a dp array of size K+1 initialized to 0, then set dp[0] = 1 to represent the empty subset sum.

💡 dp[0] = 1 because there is exactly one subset (empty) that sums to zero.

Line:

dp = [0] * (K + 1)
dp[0] = 1

💡 This initialization sets the base case for counting subsets.

traverse

Start processing first element: 1

Begin iterating over the first element in arr, which is 1, to update dp sums from K down to 1.

💡 Processing elements one by one updates dp to count subsets including that element.

Line:

for num in arr:
    for j in range(K, num - 1, -1):

💡 We will update dp for sums that can include the current element.

fill_cells

Update dp for sum = 6 with num=1

Update dp[6] by adding dp[6 - 1] = dp[5]. Since dp[5] is 0, dp[6] remains 0.

💡 Checking if subsets that sum to 5 can be extended by 1 to sum to 6.

Line:

dp[j] += dp[j - num]  # j=6, num=1

💡 No subsets currently sum to 5, so no new subsets sum to 6 by adding 1.

fill_cells

Update dp for sum = 5 with num=1

Update dp[5] by adding dp[4]. dp[4] is 0, so dp[5] stays 0.

💡 Checking if subsets that sum to 4 can be extended by 1 to sum to 5.

Line:

dp[j] += dp[j - num]  # j=5, num=1

💡 No subsets sum to 4 yet, so no new subsets sum to 5 by adding 1.

fill_cells

Update dp for sum = 1 with num=1

Update dp[1] by adding dp[0]. dp[0] is 1, so dp[1] becomes 1.

💡 Since dp[0] = 1, adding 1 creates a new subset summing to 1.

Line:

dp[j] += dp[j - num]  # j=1, num=1

💡 We found one subset [1] that sums to 1.

traverse

Start processing second element: 2

Begin processing the second element 2, updating dp sums from K down to 2.

💡 Each element updates dp to count subsets including it.

Line:

for num in arr:
    for j in range(K, num - 1, -1):

💡 We will update dp for sums that can include 2.

fill_cells

Update dp for sum = 3 with num=2

Update dp[3] by adding dp[1]. dp[1] is 1, so dp[3] becomes 1.

💡 Subsets summing to 1 can be extended by 2 to sum to 3.

Line:

dp[j] += dp[j - num]  # j=3, num=2

💡 Found one subset [1,2] that sums to 3.

fill_cells

Update dp for sum = 2 with num=2

Update dp[2] by adding dp[0]. dp[0] is 1, so dp[2] becomes 1.

💡 Empty subset can be extended by 2 to form subset summing to 2.

Line:

dp[j] += dp[j - num]  # j=2, num=2

💡 Found one subset [2] that sums to 2.

traverse

Start processing third element: 3

Begin processing the third element 3, updating dp sums from K down to 3.

💡 Each element updates dp to count subsets including it.

Line:

for num in arr:
    for j in range(K, num - 1, -1):

💡 We will update dp for sums that can include 3.

fill_cells

Update dp for sum = 6 with num=3

Update dp[6] by adding dp[3]. dp[3] is 1, so dp[6] becomes 1.

💡 Subsets summing to 3 can be extended by 3 to sum to 6.

Line:

dp[j] += dp[j - num]  # j=6, num=3

💡 Found one subset [1,2,3] that sums to 6.

fill_cells

Update dp for sum = 5 with num=3

Update dp[5] by adding dp[2]. dp[2] is 1, so dp[5] becomes 1.

💡 Subsets summing to 2 can be extended by 3 to sum to 5.

Line:

dp[j] += dp[j - num]  # j=5, num=3

💡 Found one subset [2,3] that sums to 5.

fill_cells

Update dp for sum = 4 with num=3

Update dp[4] by adding dp[1]. dp[1] is 1, so dp[4] becomes 1.

💡 Subsets summing to 1 can be extended by 3 to sum to 4.

Line:

dp[j] += dp[j - num]  # j=4, num=3

💡 Found one subset [1,3] that sums to 4.

fill_cells

Update dp for sum = 3 with num=3

Update dp[3] by adding dp[0]. dp[0] is 1, so dp[3] becomes 2 (previously 1).

💡 Empty subset can be extended by 3 to form subset summing to 3, adding to existing count.

Line:

dp[j] += dp[j - num]  # j=3, num=3

💡 Found a new subset [3] in addition to [1,2] that sums to 3.

traverse

Start processing fourth element: 3 (duplicate)

Process the last element 3, updating dp sums from K down to 3 again to include subsets with this duplicate.

💡 Duplicate elements are processed similarly, counting distinct subsets including this element.

Line:

for num in arr:
    for j in range(K, num - 1, -1):

💡 We will update dp for sums that can include this second 3.

fill_cells

Update dp for sum = 6 with second 3

Update dp[6] by adding dp[3]. dp[3] is 2, so dp[6] becomes 3 (previously 1).

💡 Subsets summing to 3 can be extended by this 3 to sum to 6, adding to existing count.

Line:

dp[j] += dp[j - num]  # j=6, num=3

💡 Found two additional subsets [3,3] and [1,2,3] (using second 3) that sum to 6.

fill_cells

Update dp for sum = 5 with second 3

Update dp[5] by adding dp[2]. dp[2] is 1, so dp[5] becomes 2 (previously 1).

💡 Subsets summing to 2 can be extended by this 3 to sum to 5, adding to existing count.

Line:

dp[j] += dp[j - num]  # j=5, num=3

💡 Found an additional subset [2,3] using second 3 that sums to 5.

fill_cells

Update dp for sum = 4 with second 3

Update dp[4] by adding dp[1]. dp[1] is 1, so dp[4] becomes 2 (previously 1).

💡 Subsets summing to 1 can be extended by this 3 to sum to 4, adding to existing count.

Line:

dp[j] += dp[j - num]  # j=4, num=3

💡 Found an additional subset [1,3] using second 3 that sums to 4.

fill_cells

Update dp for sum = 3 with second 3

Update dp[3] by adding dp[0]. dp[0] is 1, so dp[3] becomes 3 (previously 2).

💡 Empty subset can be extended by this 3 to form subset summing to 3, adding to existing count.

Line:

dp[j] += dp[j - num]  # j=3, num=3

💡 Found a new subset [3] using second 3, increasing count of subsets summing to 3.

reconstruct

Return final answer dp[K]

Return dp[6], which is 3, representing the count of subsets summing to K.

💡 dp[K] holds the total count of subsets that sum exactly to K.

Line:

return dp[K]

💡 The algorithm has counted all valid subsets without enumerating them explicitly.

def count_subsets_space_optimized(arr, K):
    dp = [0] * (K + 1)  # STEP 1
    dp[0] = 1  # STEP 1
    for num in arr:  # STEP 2, 6, 9, 14
        for j in range(K, num - 1, -1):  # STEP 3-5, 7-8, 10-13, 15-18
            dp[j] += dp[j - num]  # update dp[j]
    return dp[K]  # STEP 19

if __name__ == '__main__':
    arr = [1, 2, 3, 3]
    K = 6
    print(count_subsets_space_optimized(arr, K))

📊

Count of Subsets With Sum K - Watch the Algorithm Execute, Step by Step

Watching each update to the dp array reveals how subsets are counted incrementally without enumerating all subsets explicitly.

enteringarr=[1,2,3,3]K=6

Call Path (current → root)

count_subsets_space_optimized(arr=[1,2,3,3], K=6)

Remaining

initialize dp array

choosingarr=[1,2,3,3]K=6

Call Path (current → root)

count_subsets_space_optimized(arr=[1,2,3,3], K=6)

Tried

num=1

Remaining

num=2num=3num=3

choosingarr=[1,2,3,3]K=6

Call Path (current → root)

count_subsets_space_optimized(arr=[1,2,3,3], K=6)

Tried

j=6

Remaining

j=5j=4j=3j=2j=1

choosingarr=[1,2,3,3]K=6

Call Path (current → root)

count_subsets_space_optimized(arr=[1,2,3,3], K=6)

Tried

j=6j=5

Remaining

j=4j=3j=2j=1

choosingarr=[1,2,3,3]K=6

Call Path (current → root)

count_subsets_space_optimized(arr=[1,2,3,3], K=6)

Tried

j=6j=5j=4j=3j=2j=1

choosingarr=[1,2,3,3]K=6

Call Path (current → root)

count_subsets_space_optimized(arr=[1,2,3,3], K=6)

Tried

num=1num=2

Remaining

num=3num=3

choosingarr=[1,2,3,3]K=6

Call Path (current → root)

count_subsets_space_optimized(arr=[1,2,3,3], K=6)

Tried

j=6j=5j=4j=3

Remaining

j=2

choosingarr=[1,2,3,3]K=6

Call Path (current → root)

count_subsets_space_optimized(arr=[1,2,3,3], K=6)

Tried

j=6j=5j=4j=3j=2

choosingarr=[1,2,3,3]K=6

Call Path (current → root)

count_subsets_space_optimized(arr=[1,2,3,3], K=6)

Tried

num=1num=2num=3

Remaining

num=3

choosingarr=[1,2,3,3]K=6

Call Path (current → root)

count_subsets_space_optimized(arr=[1,2,3,3], K=6)

Tried

j=6

Remaining

j=5j=4j=3

choosingarr=[1,2,3,3]K=6

Call Path (current → root)

count_subsets_space_optimized(arr=[1,2,3,3], K=6)

Tried

j=6j=5

Remaining

j=4j=3

choosingarr=[1,2,3,3]K=6

Call Path (current → root)

count_subsets_space_optimized(arr=[1,2,3,3], K=6)

Tried

j=6j=5j=4

Remaining

j=3

choosingarr=[1,2,3,3]K=6

Call Path (current → root)

count_subsets_space_optimized(arr=[1,2,3,3], K=6)

Tried

j=6j=5j=4j=3

choosingarr=[1,2,3,3]K=6

Call Path (current → root)

count_subsets_space_optimized(arr=[1,2,3,3], K=6)

Tried

num=1num=2num=3num=3

choosingarr=[1,2,3,3]K=6

Call Path (current → root)

count_subsets_space_optimized(arr=[1,2,3,3], K=6)

Tried

j=6

Remaining

j=5j=4j=3

choosingarr=[1,2,3,3]K=6

Call Path (current → root)

count_subsets_space_optimized(arr=[1,2,3,3], K=6)

Tried

j=6j=5

Remaining

j=4j=3

choosingarr=[1,2,3,3]K=6

Call Path (current → root)

count_subsets_space_optimized(arr=[1,2,3,3], K=6)

Tried

j=6j=5j=4

Remaining

j=3

choosingarr=[1,2,3,3]K=6

Call Path (current → root)

count_subsets_space_optimized(arr=[1,2,3,3], K=6)

Tried

j=6j=5j=4j=3

backtrackingarr=[1,2,3,3]K=6

Call Path (current → root)

count_subsets_space_optimized(arr=[1,2,3,3], K=6)

Tried

num=1num=2num=3num=3

Found 1: [3]

Key Takeaways

✓ The dp array accumulates counts of subsets for each sum incrementally without enumerating subsets explicitly.

This insight is hard to see from code alone because the dp updates look like simple additions, but they represent counting subsets.

✓ Processing sums from K down to the current element prevents double counting subsets that include the same element multiple times.

The reverse iteration order is crucial and often overlooked when reading code.

✓ Duplicate elements are handled naturally by the dp updates, counting distinct subsets that differ by element indices.

Seeing dp counts increase with duplicates clarifies how the algorithm counts subsets with repeated numbers.

Practice

1. Consider the following Python code implementing the optimal backtracking solution for Combination Sum III. What is the final value of res after calling combinationSum3(3, 7)?

easy

A. [[1, 2, 4]]

B. [[1, 3, 3]]

C. [[2, 2, 3]]

D. [[1, 2, 3]]

Solution

1. Step 1: Trace backtracking calls for k=3, n=7

   Start from 1, try combinations of size 3 summing to 7. Valid combos are [1,2,4] only because 1+2+4=7.

2. Step 2: Verify no other combos meet criteria

   Other combos like [1,3,3] or [2,2,3] are invalid due to duplicates or sum mismatch.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Only [1,2,4] sums to 7 with 3 distinct numbers [OK]

Hint: Only distinct numbers summing to 7 with size 3 is [1,2,4] [OK]

Common Mistakes:

• Including duplicates like [1,3,3]
• Miscounting sum or size

2. Given the following code to generate subsets, what is the value of result after processing num=2 when nums = [1, 2]?

easy

A. [[], [1], [2]]

B. [[], [1], [1, 2], [2]]

C. [[], [1], [2], [1, 2], [2, 2]]

D. [[], [1], [2], [1, 2]]

Solution

1. Step 1: Trace result after processing num=1

   Initially result = [[]]. After num=1, new_subsets = [[1]], result = [[], [1]].

2. Step 2: Trace result after processing num=2

   For each subset in [[], [1]], add 2: new_subsets = [[2], [1, 2]]. Extend result: [[], [1], [2], [1, 2]].

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   All subsets of [1,2] are present exactly once [OK]

Hint: Iterative subsets double result size each iteration [OK]

Common Mistakes:

• Forgetting to add subsets with current num
• Appending duplicates
• Wrong order of subsets

3. What is the time complexity of generating all subsets of an array of size n using bitmask enumeration, and why is the common misconception that it is O(2^n) incorrect?

medium

A. O(n^2) because we iterate over n elements twice

B. O(2^n) because there are 2^n subsets

C. O(n * 2^n) because each subset can have up to n elements and we build each subset by checking n bits

D. O(n) because we process each element once

Solution

1. Step 1: Count subsets and subset size

   There are 2^n subsets, and each subset can have up to n elements.

2. Step 2: Analyze bitmask enumeration cost

   For each of the 2^n masks, we check n bits to build the subset, resulting in O(n * 2^n) time.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Time depends on both number of subsets and subset size [OK]

Hint: Each subset requires O(n) to build, total O(n*2^n) [OK]

Common Mistakes:

• Ignoring subset size in complexity
• Assuming O(2^n) only
• Confusing with DP complexities

4. Suppose the problem is modified so that elements can be chosen multiple times (unlimited reuse) to form subsets. Which approach correctly adapts to count the number of max bitwise-OR subsets under this new constraint?

hard

A. Use dynamic programming over OR values with state compression to count combinations

B. Use backtracking with memoization to handle repeated elements and prune search

C. Use the same bitmask enumeration approach since subsets remain finite

D. Sort elements and greedily pick those with highest bits until max OR is reached

Solution

1. Step 1: Recognize infinite subsets due to unlimited reuse

   Bitmask enumeration is infeasible as subsets are infinite with reuse allowed.

2. Step 2: Use DP over OR states to count combinations efficiently

   DP with state compression tracks counts of subsets achieving each OR value, handling reuse.

3. Step 3: Understand greedy or naive backtracking fail due to infinite or redundant subsets

   Greedy misses combinations; naive backtracking is exponential and inefficient.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   DP over OR states handles reuse -> correct and efficient [OK]

Hint: Reuse means infinite subsets -> DP over OR states needed [OK]

Common Mistakes:

• Trying bitmask enumeration despite infinite subsets
• Using greedy approach ignoring subset counts

5. Suppose the problem is modified so that puzzles can have repeated letters and words can be counted multiple times if they match multiple subsets of the puzzle letters. Which modification to the optimal bitmask + trie approach correctly handles this variant?

hard

A. Preprocess puzzles to remove duplicates and apply the original algorithm unchanged.

B. Modify dfs to not require the puzzle's first letter to be present in the word and allow counting multiple matches per word.

C. Enumerate all subsets of puzzle letters including duplicates and sum counts from trie traversal without filtering first letter.

D. Use a multiset trie that stores counts for repeated letters and traverse all subsets including duplicates.

Solution

1. Step 1: Understand variant requirements

   Repeated letters in puzzles and multiple counting per word require handling multisets, not just sets.

2. Step 2: Modify data structure and traversal

   A multiset trie that stores counts for repeated letters and traverses all subsets including duplicates correctly counts all matches.

3. Step 3: Why other options fail

   Ignoring first letter breaks mandatory condition; removing duplicates loses information; naive enumeration is inefficient.

4. Final Answer:

   Option D -> Option D

5. Quick Check:

   Multiset trie handles repeated letters and multiple counts correctly [OK]

Hint: Multiset trie needed for repeated letters and multiple counts [OK]

Common Mistakes:

• Ignoring repeated letters
• Removing duplicates incorrectly
• Dropping first letter condition