Day 14 - Subset Sum Problem.md

Difficulty	Medium
Source	160 Days of Problem Solving
Tags	Dynamic Programming

🚀 Day 14. Subset Sum Problem 🧠

The problem can be found at the following link: Question Link

💡 Problem Description:

Given an array of positive integers arr[] and a target value sum, determine if there exists a subset of arr[] whose sum is exactly equal to the given sum.

🔍 Example Walkthrough:

Example 1:

Input:

arr[] = {3, 34, 4, 12, 5, 2}
sum = 9

Output:

true

Explanation:

A subset {4, 3, 2} exists with sum = 9.

Example 2:

Input:

arr[] = {3, 34, 4, 12, 5, 2}
sum = 30

Output:

false

Explanation:

There is no subset with sum = 30.

Example 3:

Input:

arr[] = {1, 2, 3}
sum = 6

Output:

true

Explanation:

The entire array {1, 2, 3} forms the subset with sum = 6.

Constraints:

• $1 \leq \text{size of arr} \leq 200$
• $1 \leq arr[i] \leq 200$
• $1 \leq sum \leq 10^4$

🎯 My Approach:

Dynamic Programming (Optimized 1D DP) – O(N × sum) Time, O(sum) Space

1. Define a boolean DP array dp[sum + 1], where dp[j] tells whether sum j is possible.
2. Base Case: dp[0] = true (sum 0 is always achievable).
3. Iterate through each number in the array and update dp[j] for all j from sum down to num.
4. Transition Formula:
   $[ \text{dp}[j] = \text{dp}[j] \ OR\ \text{dp}[j - \text{num}] $]
   • If dp[j - num] was true, then dp[j] can also become true by including num.
5. Return dp[sum] as the final answer.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(N × sum), since we iterate through all elements and update the DP table.
• Expected Auxiliary Space Complexity: O(sum), as we use a 1D array of size sum + 1.

📝 Solution Code

Code (C++)

class Solution {
  public:
    bool isSubsetSum(vector<int>& arr, int sum) {
        vector<bool> dp(sum + 1, false);
        dp[0] = true;
        for (int num : arr)
            for (int j = sum; j >= num; --j)
                dp[j] = dp[j] || dp[j - num];
        return dp[sum];
    }
};

⚡ Alternative Approaches

2️⃣ Dynamic Programming (O(N×sum) Time, O(N×sum) Space) — 2D DP

Algorithm Steps:

1. Use a 2D DP table where dp[i][j] represents whether it's possible to achieve sum j using the first i elements.
2. Base Case:
   • dp[i][0] = true for all i (sum 0 is always achievable).
   • dp[0][j] = false for j > 0 (no elements can't form non-zero sum).
3. Recurrence Relation:
   $[ \text{dp}[i][j] = \text{dp}[i-1][j] || \text{dp}[i-1][j - arr[i-1]] $]
   • Exclude the element (dp[i-1][j]).
   • Include the element (dp[i-1][j - arr[i-1]]).

class Solution {
  public:
    bool isSubsetSum(vector<int>& arr, int sum) {
        int n = arr.size();
        vector<vector<bool>> dp(n + 1, vector<bool>(sum + 1, false));
        for (int i = 0; i <= n; i++) dp[i][0] = true;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= sum; j++) {
                dp[i][j] = dp[i - 1][j];
                if (j >= arr[i - 1]) dp[i][j] |= dp[i - 1][j - arr[i - 1]];
            }
        }
        return dp[n][sum];
    }
};

✅ Time Complexity: O(N × sum)
✅ Space Complexity: O(N × sum)

3️⃣ Recursive + Memoization (O(N×sum) Time, O(N×sum) Space)

Algorithm Steps:

1. Define a recursive function solve(index, sum):
   • Base Case: If sum == 0, return true.
   • If index < 0 or sum < 0, return false.
   • Memoize results to avoid recomputation.
2. Recurrence Relation:
   $[ \text{solve(index, sum)} = \text{solve(index - 1, sum)} \quad \text{OR} \quad \text{solve(index - 1, sum - arr[index])} ]$
   • Exclude the element.
   • Include the element.
3. Use memoization (dp[index][sum]) to avoid redundant calculations.

class Solution {
  public:
    vector<vector<int>> dp;
    bool solve(vector<int>& arr, int i, int sum) {
        if (sum == 0) return true;
        if (i < 0 || sum < 0) return false;
        if (dp[i][sum] != -1) return dp[i][sum];
        return dp[i][sum] = solve(arr, i - 1, sum) || solve(arr, i - 1, sum - arr[i]);
    }

    bool isSubsetSum(vector<int>& arr, int sum) {
        int n = arr.size();
        dp.assign(n, vector<int>(sum + 1, -1));
        return solve(arr, n - 1, sum);
    }
};

✅ Time Complexity: O(N × sum)
✅ Space Complexity: O(N × sum)

Comparison of Approaches

Approach	⏱️ Time Complexity	🗂️ Space Complexity	✅ Pros	⚠️ Cons
1D Space Optimized DP	🟡 O(N × sum)	🟢 O(sum)	Most efficient space-wise	Requires careful indexing
2D DP (Tabulation)	🟡 O(N × sum)	🔴 O(N × sum)	Easy to implement, intuitive	High space usage
Recursive + Memoization	🟡 O(N × sum)	🔴 O(N × sum)	Natural recursion flow	Stack overhead

✅ Best Choice?

• If optimizing space: Use 1D DP (Space-Optimized).
• If space is not a concern: Use 2D DP (Tabulation) for easy understanding.
• For recursion lovers: Use Recursive + Memoization.

Code (Java)

class Solution {
    static boolean isSubsetSum(int[] arr, int sum) {
        boolean[] dp = new boolean[sum + 1];
        dp[0] = true;
        for (int num : arr)
            for (int j = sum; j >= num; --j)
                dp[j] |= dp[j - num];
        return dp[sum];
    }
}

Code (Python)

class Solution:
    def isSubsetSum(self, arr, sum):
        dp = [False] * (sum + 1)
        dp[0] = True
        for num in arr:
            for j in range(sum, num - 1, -1):
                dp[j] |= dp[j - num]
        return dp[sum]

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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