Target Sum

You are given a list of non-negative integers, a1, a2, ..., an, and a target, S. Now you have 2 symbols + and -. For each integer, you should choose one from + and - as its new symbol.

Find out how many ways to assign symbols to make sum of integers equal to target S.

Example 1:

Input: nums is [1, 1, 1, 1, 1], S is 3. 
Output: 5
Explanation: 

-1+1+1+1+1 = 3
+1-1+1+1+1 = 3
+1+1-1+1+1 = 3
+1+1+1-1+1 = 3
+1+1+1+1-1 = 3

There are 5 ways to assign symbols to make the sum of nums be target 3.

Constraints:

• The length of the given array is positive and will not exceed 20.

• The sum of elements in the given array will not exceed 1000.

• Your output answer is guaranteed to be fitted in a 32-bit integer.

Explaination:

The original problem statement is equivalent to: Find a subset of nums that need to be positive, and the rest of them negative, such that the sum is equal to target

Let P be the positive subset and N be the negative subset For example: Given nums = [1, 2, 3, 4, 5] and target = 3 then one possible solution is +1-2+3-4+5 = 3 Here positive subset is P = [1, 3, 5] and negative subset is N = [2, 4]

Then let's see how this can be converted to a subset sum problem:

                  sum(P) - sum(N) = target
sum(P) + sum(N) + sum(P) - sum(N) = target + sum(P) + sum(N)
                       2 * sum(P) = target + sum(nums)

So the original problem has been converted to a subset sum problem as follows: Find a subset P of nums such that sum(P) = (target + sum(nums)) / 2

class Solution {
    public int findTargetSumWays(int[] nums, int targetSum) {
        int sum = 0;
        for (int n : nums)
            sum += n;
        return ((sum < targetSum) || (targetSum + sum) % 2 != 0) ? 0 : subsetSum(nums, (targetSum + sum) / 2);
    }

    public int subsetSum(int[] nums, int targetSum) {
        int[] dp = new int[targetSum + 1];
        dp[0] = 1;
        for (int n : nums)
            for (int i = targetSum; i >= n; i--)
                dp[i] += dp[i - n];
        return dp[targetSum];
    }
}