What is the brute force approach and complexity for Subset Sum Problem?

Recurse on each number: skip it, or take it and lower the sum still needed with time complexity O(2^N) and space complexity O(N).

What is the optimal approach and complexity for Subset Sum Problem?

Keep one row dp[s]; for each number, update s from sum down to that number with optimal time complexity O(N · S) and space complexity O(S).

Subset Sum Problem

Q: How do you identify the pattern for Subset Sum Problem?

Look for key signals in the prompt: "some subset adds up to exactly sum" indicates Pick items so the total hits a target → Subset Sum, which is 0/1 knapsack where each value equals its weight.; "each number can be used at most once" indicates Each number is in or out, so the sum loop must go high → low.; "return true or false" indicates Only reachability matters, so each cell stores a boolean, not a best value..

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Problem Statement

You are given a list of positive whole numbers arr and a target sum. Decide whether some subset of the numbers adds up to exactly sum. Each number can be used at most once. The empty subset adds up to 0.

Return true if some subset adds up to sum, otherwise false. The function is bool isSubsetSum(vector<int>& arr, int sum).

Input

arr, a list of positive numbers. sum, the target.

Output

true if some subset of arr adds up to sum, otherwise false.

Example 1

Input: arr = [3, 34, 4, 12, 5, 2], sum = 9

Output: true

The numbers 4 and 5 add up to 9.

Example 2

Input: arr = [3, 34, 4, 12, 5, 2], sum = 30

Output: false

34 is too big. The other numbers add up to only 3 + 4 + 12 + 5 + 2 = 26, so 30 cannot be reached.

Example 3

Input: arr = [3], sum = 6

Output: false

There is only one 3. Using it twice (3 + 3) is not allowed.

Constraints

• 1 ≤ arr.size(), arr[i] ≤ 200
• 1 ≤ sum ≤ 10^4

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“some subset adds up to exactly sum”

→ Pick items so the total hits a target → Subset Sum, which is 0/1 knapsack where each value equals its weight.

“each number can be used at most once”

→ Each number is in or out, so the sum loop must go high → low.

“return true or false”

→ Only reachability matters, so each cell stores a boolean, not a best value.

Approach Complexity Ladder

Brute Force — skip or take every numberO(2^N)O(N)

Recurse on each number: skip it, or take it and lower the sum still needed

Better — 2-D table of true / falseO(N · S)O(N · S)

Store dp[i][s] for every prefix of the numbers and every sum; each cell is O(1)

Optimal — one row of bools, sum high → lowO(N · S)O(S)

Keep one row dp[s]; for each number, update s from sum down to that number

Brute Force Intuition

Go through the numbers one by one. For each one, skip it, or take it when it fits and lower the sum still needed. The answer is true if either branch reaches a sum of 0. Every subset is tried, with no memory between calls.

Recognition rule for this sub-pattern:

Each item once. dp[c] = max(dp[c], dp[c − w] + v), capacity loop goes high → low.

Remember this · DP — Knapsack

Trigger

Pick a subset to hit a target sum / capacity, partition into two groups, count the ways to reach an amount.

Think

Is this "choose some items so their total is X"?

Key question

Can each item be used once (loop capacity backwards) or many times (loop forwards)?

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