494. Target Sum

/**
 * Dynamic Programming.
 *
 * As per the question, Sum(P) + Sum(N) = Sum(nums). Also, Sum(P) - Sum(N) =
 * target. Thus if we add these two equations, we will get:
 *
 * 2 * Sum(P) = target + Sum(nums)
 *
 * OR
 *
 * Sum(P) = (target + Sum(nums)) / 2
 *
 * One we have the target sum of positive numbers, the probelm reduces to
 * 416-Partition Equal Subset Sum question. Refer to this question's solution
 * for explaination on how to find the subset.
 * (https://leetcode.com/problems/partition-equal-subset-sum)
 *
 * Only change here, we have to return the number of ways we can create these
 * subsets. Thus DP transformation will modify to:
 *
 * DP[i][s] = DP[i-1][s] + DP[i-1][s - nums[i]]
 *
 * This problem has the same explaination for finding the subset
 */
class Solution {
    public int findTargetSumWays(int[] nums, int S) {
        if (nums == null || (nums.length == 0 && S > 0)) {
            return 0;
        }
        if (nums.length == 0 && S == 0) {
            return 1;
        }

        int sum = 0;
        for (int num : nums) {
            sum += num;
        }
        if (S > sum || (S + sum) % 2 != 0) {
            return 0;
        }
        int target = (S + sum) / 2;

        int[] dp = new int[target + 1];
        dp[0] = 1;
        for (int num : nums) {
            for (int s = target; s >= num; s--) {
                dp[s] += dp[s - num];
            }
        }

        return dp[target];
    }
}