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. Consider the following Python code implementing the optimal backtracking with bitmask and memoization approach for the Matchsticks to Square problem. Given the input [1,1,2,2,2], what is the final return value of makesquare([1,1,2,2,2])?

easy

A. True

B. Depends on the order of matchsticks

C. Raises an exception due to index error

D. False

Solution

1. Step 1: Calculate total and side length

   Total = 1+1+2+2+2 = 8, side = 8/4 = 2.

2. Step 2: Trace backtracking calls

   Sorted sticks: [2,2,2,1,1]. The algorithm tries to form sides of length 2. It can form three sides with three 2's, and the last side with two 1's. Hence, returns true.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Sum divisible by 4 and sticks fit sides exactly [OK]

Hint: Sum divisible by 4 and sticks fit sides exactly [OK]

Common Mistakes:

• Forgetting to sort and prune
• Assuming order affects result
• Miscounting sides formed

2. You are given an integer array that may contain duplicates. Your task is to find all possible subsets without duplicate subsets in the result. Which approach guarantees generating all unique subsets efficiently without redundant computations?

easy

A. Generate all subsets using brute force recursion and then filter duplicates by storing subsets in a hash set.

B. Use a greedy algorithm that picks elements only if they are not duplicates of the previous element.

C. Use dynamic programming with a 2D table tracking subset sums to avoid duplicates.

D. Sort the array and use backtracking with skipping duplicates at the same recursion depth to avoid repeated subsets.

Solution

1. Step 1: Understand the problem constraints

   The problem requires generating all unique subsets from an array that may contain duplicates, avoiding duplicate subsets in the output.

2. Step 2: Identify the approach that handles duplicates efficiently

   Sorting the array groups duplicates together, allowing the backtracking algorithm to skip duplicates at the same recursion level, preventing repeated subsets without extra filtering.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Backtracking with sorting and skipping duplicates is the standard pattern for Subsets II [OK]

Hint: Sort and skip duplicates at recursion level [OK]

Common Mistakes:

• Using greedy approach misses some subsets
• DP approach is unrelated to subset generation here
• Brute force with set filtering is inefficient

3. What is the worst-case time complexity of the optimal backtracking solution with sum bound pruning for Combination Sum III (k numbers from 1 to 9 summing to n)?

medium

A. O(k!) because permutations of k numbers are explored

B. O(k * n) because we explore all sums up to n for k elements

C. O(n^k) because we try all sequences of length k with sum n

D. O(2^9) because each number 1-9 can be either included or excluded, pruning reduces calls but worst case remains exponential

Solution

1. Step 1: Identify search space size

   Numbers 1 to 9 can be included or excluded, so 2^9 subsets possible.

2. Step 2: Consider pruning effect

   Pruning reduces calls but worst-case complexity remains O(2^9) since all subsets might be explored.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Backtracking worst case is exponential in number of candidates [OK]

Hint: Backtracking subsets of 9 elements is O(2^9) worst case [OK]

Common Mistakes:

• Confusing with DP complexity O(k*n)
• Thinking permutations cause factorial complexity

4. Suppose you want to generate all combinations of size k from numbers 1 to n, but now elements can be reused multiple times in a combination (combinations with replacement). Which modification to the backtracking approach correctly handles this?

hard

A. Use iterative lexicographic generation without any changes

B. Change the recursive call to backtrack(i + 1, path) to prevent reuse of elements

C. Change the recursive call to backtrack(i, path) to allow reuse of the current element

D. Sort the input array before backtracking to handle duplicates

Solution

1. Step 1: Understand combinations with replacement allow repeated elements

   To allow reuse, the recursive call must not increment start index beyond current element.

2. Step 2: Modify recursion to backtrack(i, path) to allow current element reuse

   This ensures the same element can be chosen multiple times in subsequent recursive calls.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Using backtrack(i, path) enables combinations with replacement [OK]

Hint: Reuse requires recursive calls with same start index [OK]

Common Mistakes:

• Incrementing start index prevents element reuse

5. Suppose the problem is modified so that elements can be reused multiple times in the subset (i.e., the subset can contain duplicates). Which modification to the algorithm is necessary to correctly find the largest divisible subset under this new constraint?

hard

A. Modify DP to consider all previous elements including i itself, allowing repeated use, and remove early break

B. Switch to a graph cycle detection algorithm to handle repeated elements

C. Use the same DP approach but allow dp[i] to consider dp[i] itself for reuse, enabling cycles

D. Use a greedy approach picking the largest element repeatedly until no more divisible

Solution

1. Step 1: Understand reuse impact

   Allowing reuse means elements can appear multiple times, so dp must consider extending from the same element multiple times.

2. Step 2: Modify DP logic

   DP inner loop must consider j ≤ i (including i itself) and remove early break because sorted order no longer guarantees divisibility breaks.

3. Step 3: Reject incorrect options

   Greedy fails to find longest chain; graph cycle detection is unnecessary; naive DP without modification misses reuse.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   DP adjusted for reuse and no early break handles duplicates [OK]

Hint: Reuse requires DP to consider self and no early break [OK]

Common Mistakes:

• Using original DP without changes leads to incorrect results
• Trying greedy approach which fails on complex divisibility
• Confusing graph cycles with DP chains