Count Subset With Target Sum II

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Intuition

Count subsets with given sum where array may have duplicates. DP with proper handling of duplicate elements.

Algorithm

1Sort array. DP similar to 0-1 knapsack: dp[j] = count of subsets with sum j.
2For each num: update dp right-to-left (0-1 knapsack style).

Common Pitfalls

•Same as LC 416 count variant. Handle duplicates: 0-1 knapsack treats each element independently.

Count Subset With Target Sum II.java

Java

// Approach: DP (unbounded). dp[j] = ways to achieve sum j using array elements with repetition allowed.
// Time: O(n * sum) Space: O(sum)
import java.util.*;

class Solution {
    private HashMap<String, Integer> dp;

    public int countSubset(int[] arr, int k) {
        int psum = 0, nsum = 0;
        for (int num : arr) {
            if (num > 0)
                psum += num;
            else
                nsum += num;
        }
        if (k > psum || k < nsum)
            return 0;

        dp = new HashMap<>();
        return solve(0, arr, k);
    }

    private int solve(int i, int[] arr, int k) {
        if (i == arr.length)
            return k == 0 ? 1 : 0;

        String key = i + "_" + k;
        if (dp.containsKey(key))
            return dp.get(key);

        int inc = solve(i + 1, arr, k - arr[i]);
        int exc = solve(i + 1, arr, k);

        dp.put(key, inc + exc);
        return inc + exc;
    }
}

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