Count of Subsets With Sum K

📋

Problem

Imagine you have a collection of coins and want to find out how many different ways you can pick some coins so that their total value equals exactly a target amount.

Given an array of integers and a target sum K, find the number of subsets of the array whose elements sum exactly to K. Return the count of such subsets.

1 ≤ n ≤ 10^5 (for optimized solutions)Array elements can be positive integers0 ≤ K ≤ sum of all elements

Edge cases: Empty array with K=0 → 1 (empty subset counts)Array with all elements greater than K → 0Array with multiple zeros and K=0 → multiple subsets including empty and zeros

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IDE

def count_subsets(arr: list[int], K: int) -> int:public int countSubsets(int[] arr, int K)int countSubsets(vector<int> &arr, int K)function countSubsets(arr, K)

def count_subsets(arr, K):
    # Write your solution here
    pass

class Solution {
    public int countSubsets(int[] arr, int K) {
        // Write your solution here
        return 0;
    }
}

#include <vector>
using namespace std;

int countSubsets(vector<int> &arr, int K) {
    // Write your solution here
    return 0;
}

function countSubsets(arr, K) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: 0Returning 0 for empty array with K=0 instead of 1 (missing empty subset).✅ Return 1 when index == len(arr) and current_sum == K.

Wrong: Overcounted subsetsCounting subsets with repeated use of elements (unbounded) instead of 0/1 usage.✅ Ensure each element is included at most once by advancing index in recursion.

Wrong: Undercount due to ignoring duplicatesTreating duplicate values as one element, missing subsets with different instances.✅ Treat each element as unique in recursion/DP even if values repeat.

Wrong: Non-zero count when no subsets sum to KNot pruning subsets exceeding K or incorrect base case handling.✅ Return 0 when no valid subsets found and prune sums exceeding K.

Wrong: TLEUsing brute force recursion for large n causing exponential time.✅ Implement O(n*K) DP solution with memoization or tabulation.

✓

Test Cases

Focus on handling base cases like empty arrays and zeros correctly.

Consider how duplicates and usage constraints affect counting subsets.

Optimize your solution to use DP instead of brute force to handle large inputs.

t1_01basic

Input{"arr":[1,2,3,3],"K":6}

Expected3

⏱ Performance - must finish in 2000ms

The subsets are [1,2,3] (using first 3) and [3,3] (using both 3s). The subset [1,2,3] is counted once per unique element combination, so total 2 subsets.

💡 Consider each element's inclusion or exclusion to form subsets.

💡 Use recursion or dynamic programming to count subsets summing to K.

💡 Count subsets by exploring both choices at each index, treating duplicates as distinct elements.

Why it failed: Incorrect count likely due to double counting or missing subsets with duplicate elements; fix by considering each element's inclusion/exclusion separately.

✓ Correctly counts all subsets summing to K including duplicates.

t1_02basic

Input{"arr":[2,4,6,10],"K":16}

Expected2

⏱ Performance - must finish in 2000ms

Subsets [6,10] and [2,4,10] sum to 16.

💡 Try enumerating subsets and summing their elements.

💡 Check combinations that exactly match K.

💡 Use DP or recursion to count subsets efficiently.

Why it failed: Missed subsets that sum to K due to incorrect recursion or DP state; fix by properly tracking sums at each step.

✓ Correctly identifies all subsets summing to K.

t2_01edge

Input{"arr":[],"K":0}

Expected1

⏱ Performance - must finish in 2000ms

Empty subset sums to 0, so count is 1.

💡 Consider the empty subset as a valid subset.

💡 Base case: if no elements and sum is 0, count is 1.

💡 Return 1 when array is empty and K is 0.

Why it failed: Returned 0 instead of 1 for empty array with K=0; fix by returning 1 when index reaches end and current sum equals K.

✓ Correctly counts empty subset for K=0.

t2_02edge

Input{"arr":[5,6,7],"K":4}

Expected0

⏱ Performance - must finish in 2000ms

All elements are greater than K, no subsets sum to 4.

💡 Check if elements exceed K and cannot be included.

💡 No subset can sum to K if all elements > K.

💡 Return 0 when no valid subsets found.

Why it failed: Returned non-zero count despite no subsets summing to K; fix by pruning subsets exceeding K.

✓ Correctly returns 0 when no subsets sum to K.

t2_03edge

Input{"arr":[0,0,0],"K":0}

Expected8

⏱ Performance - must finish in 2000ms

All subsets of zeros sum to 0; total subsets = 2^3 = 8.

💡 Zeros can be included or excluded without changing sum.

💡 Count all subsets when K=0 and zeros present.

💡 Total subsets = 2^(number of zeros) when K=0.

Why it failed: Undercounted subsets with zeros for K=0; fix by counting all subsets including/excluding zeros.

✓ Correctly counts all subsets of zeros summing to 0.

t2_04edge

Input{"arr":[7],"K":7}

Expected1

⏱ Performance - must finish in 2000ms

Single element equal to K forms one valid subset.

💡 Check single element equal to K.

💡 Include or exclude the single element.

💡 Return 1 if element equals K.

Why it failed: Returned 0 for single element equal to K; fix by checking if single element equals K and returning 1.

✓ Correctly counts single element subset equal to K.

t3_01corner

Input{"arr":[1,2,2,3],"K":4}

Expected2

⏱ Performance - must finish in 2000ms

Subsets: [1,3] and [2,2] sum to 4. The subset [1,2,1] mentioned is invalid as no extra 1 exists; duplicates treated as distinct elements.

💡 Beware of counting duplicates correctly.

💡 Each element is distinct even if values repeat.

💡 Count subsets by including/excluding each element separately.

Why it failed: Under-counted subsets due to treating duplicates as one; fix by treating each element as unique in recursion/DP.

✓ Correctly counts subsets with duplicate elements.

t3_02corner

Input{"arr":[1,2,3],"K":6}

Expected1

⏱ Performance - must finish in 2000ms

Only subset [1,2,3] sums to 6; tests 0/1 knapsack vs unbounded confusion.

💡 Check that each element is used at most once.

💡 Avoid counting subsets multiple times due to reuse.

💡 Use 0/1 knapsack logic, not unbounded.

Why it failed: Overcounted subsets by reusing elements multiple times; fix by ensuring each element is included at most once.

✓ Correctly counts subsets with 0/1 usage constraint.

t3_03corner

Input{"arr":[1,3,5],"K":4}

Expected1

⏱ Performance - must finish in 2000ms

No subset sums to 4; [1,3] sums to 4 but 3+1=4 is correct, but 3+1=4 is valid so expected should be 1. Re-checking: 1+3=4 is valid subset, so expected should be 1. Original expected was 1, so correct is 1.

💡 Greedy approach may pick largest elements first incorrectly.

💡 Try all combinations rather than greedy selection.

💡 Use recursion or DP to explore all subsets.

Why it failed: Greedy approach failed by missing valid subsets; fix by exploring all subsets via recursion or DP.

✓ Correctly counts subsets avoiding greedy trap.

t4_01performance

Input{"arr":[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100],"K":2500}

Expectednull

⏱ Performance - must finish in 2000ms

n=100, K=2500; O(n*K) DP must complete within 2 seconds.

💡 Brute force O(2^n) will time out here.

💡 Use DP with memoization or tabulation for efficiency.

💡 Optimize space by using 1D DP array if possible.

Why it failed: TLE due to exponential brute force approach; fix by implementing O(n*K) DP solution with memoization or tabulation.

✓ Efficient DP solution passes within time limit.

Practice

1. What is the worst-case time complexity of the optimized dynamic programming solution for the Largest Divisible Subset problem, where n is the number of elements in the input array?

medium

A. O(n^2) due to nested loops checking divisibility pairs

B. O(2^n) because all subsets are considered

C. O(n^3) because of triple nested loops for subset checks

D. O(n log n) due to sorting and linear DP

Solution

1. Step 1: Identify main operations

   Sorting takes O(n log n). The DP uses two nested loops: outer loop over n elements, inner loop up to i elements.

2. Step 2: Analyze nested loops

   Each pair (i,j) is checked once, resulting in O(n^2) divisibility checks. Early break may speed up in practice but worst case remains O(n^2).

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   DP nested loops dominate complexity [OK]

Hint: DP nested loops -> O(n^2) worst case [OK]

Common Mistakes:

• Confusing sorting time as dominant (O(n log n))
• Assuming triple nested loops due to subset checks
• Thinking brute force complexity applies to DP

2. Examine the following buggy code snippet from the trie-based solution. Which line contains the subtle bug that causes incorrect counts for some puzzles?

medium

A. Line where node.count is incremented

B. Line missing filtering words with more than 7 unique letters before insertion

C. Line checking if bit is set in mask during insertion

D. Line checking if puzzle_mask contains child bit during dfs

Solution

1. Step 1: Identify filtering step

   The code inserts all words regardless of unique letter count, causing large trie and slow traversal.

2. Step 2: Understand impact

   Words with >7 unique letters cannot match puzzles (7 letters max), so filtering them out is necessary for correctness and efficiency.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Filtering missing leads to incorrect counts and performance issues [OK]

Hint: Filter words with >7 unique letters before insertion [OK]

Common Mistakes:

• Not filtering large words
• Incorrect bit checks
• Misplaced count increments

3. Suppose the problem changes: now candidates can include negative numbers, and you still want all unique combinations summing to the target with reuse allowed. Which modification to the backtracking approach is necessary to handle this variant correctly?

hard

A. Remove pruning on target < 0 and add a visited set to detect cycles to avoid infinite recursion.

B. No change needed; the existing backtracking with pruning on target < 0 still works correctly.

C. Sort candidates in descending order and prune when target is less than the smallest candidate.

D. Use dynamic programming instead of backtracking because negative numbers break recursion.

Solution

1. Step 1: Understand impact of negative candidates

   Negative numbers allow sums to decrease indefinitely, causing infinite recursion if pruning on target < 0 is used or missing.

2. Step 2: Modify pruning and cycle detection

   Pruning on target < 0 is invalid because target can oscillate; must remove it and add a visited set or memoization to detect repeated states and avoid infinite loops.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Cycle detection is essential when negative numbers allow infinite recursion [OK]

Hint: Negative numbers require cycle detection to prevent infinite recursion [OK]

Common Mistakes:

• Assuming pruning on target < 0 always applies
• Ignoring infinite recursion caused by negatives

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