Count of Subsets With Sum K

▶

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. You are given an array of integers and need to find the number of subsets whose bitwise OR equals the maximum possible OR value among all subsets. Which algorithmic approach guarantees an optimal solution for this problem?

easy

A. Greedy approach selecting elements with highest bits set first

B. Iterative bitmask enumeration generating all subsets and computing their OR

C. Dynamic programming based on subset sums

D. Sorting the array and using two pointers to find max OR subsets

Solution

1. Step 1: Understand the problem requires checking all subsets

   Since the maximum bitwise OR depends on any combination of elements, all subsets must be considered.

2. Step 2: Identify approach that enumerates all subsets efficiently

   Iterative bitmask enumeration systematically generates all subsets and computes their OR, ensuring no subset is missed.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Bitmask enumeration covers all subsets -> optimal [OK]

Hint: Bitmask enumeration covers all subsets exhaustively [OK]

Common Mistakes:

• Assuming greedy selection suffices for max OR
• Confusing subset sum DP with bitwise OR problem

2. What is the time complexity of the optimal backtracking solution for letter case permutation on a string of length n with k letters (non-digit characters)?

medium

A. O(2^k * n) because only letters toggle case and each permutation requires joining n characters

B. O(2^k * k) because only letters are toggled and digits are skipped

C. O(n * 2^n) because each character doubles the permutations

D. O(n^2) because of nested recursion and string concatenation

Solution

1. Step 1: Identify branching factor

   Only letters (k of them) cause branching, each with 2 choices, so total permutations are 2^k.

2. Step 2: Calculate work per permutation

   Each permutation requires joining n characters to form a string, so O(n) per permutation.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Time is exponential in letters, linear in string length [OK]

Hint: Complexity depends on letters, not total length [OK]

Common Mistakes:

• Assuming all characters double permutations
• Ignoring join cost per permutation

3. Examine the following buggy code snippet for partitioning an array into k equal sum subsets using backtracking with bitmask memoization. Which line contains the subtle bug that can cause incorrect results or inefficiency?

medium

A. Line 2: missing check if total is divisible by k

B. Line 6: sorting nums in descending order

C. Line 14: memoizing result after recursive call

D. Line 18: checking if element is unused and fits in current subset

Solution

1. Step 1: Identify missing early pruning

   Line 2 calculates total but does not check if total % k != 0, which is critical to avoid futile search.

2. Step 2: Verify other lines

   Sorting descending (line 6) is correct for pruning; memoization (line 14) and usage check (line 18) are correct.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Without divisibility check, algorithm wastes time exploring impossible partitions [OK]

Hint: Always check divisibility before backtracking [OK]

Common Mistakes:

• Forgetting early divisibility check
• Misplacing memoization

4. 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

5. Suppose you want to generate all subsets of nums but now elements can be reused unlimited times in each subset (i.e., multisets). Which modification to the backtracking approach correctly handles this?

hard

A. At each recursion step, recurse twice: once including current element and staying at same index to allow reuse, once excluding and moving to next index

B. At each recursion step, recurse with index + 1 only, excluding current element to avoid duplicates

C. Use bit manipulation as before, but allow bits to be set multiple times to represent reuse

D. Sort the array and use two pointers to generate subsets with repeated elements

Solution

1. Step 1: Understand reuse requirement

   Allowing unlimited reuse means after including an element, we can include it again without advancing index.

2. Step 2: Modify recursion accordingly

   Recurse twice: include current element and recurse with same index, or exclude and recurse with next index.

3. Step 3: Why other options fail

   At each recursion step, recurse with index + 1 only, excluding current element to avoid duplicates excludes reuse by always advancing index; Use bit manipulation as before, but allow bits to be set multiple times to represent reuse is invalid as bitmask can't represent multiple counts; Sort the array and use two pointers to generate subsets with repeated elements is unrelated.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Staying at same index after include enables repeated elements [OK]

Hint: Reuse means recurse with same index after include [OK]

Common Mistakes:

• Advancing index after include prevents reuse
• Trying bitmask for multisets