494. Target Sum (Medium)

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.

Note:

1. The length of the given array is positive and will not exceed 20.
2. The sum of elements in the given array will not exceed 1000.
3. Your output answer is guaranteed to be fitted in a 32-bit integer.

Solution 1: Math + DP O(sum * n); O(sum)

set = {a,b,c,d} 如果 a+b -c-d = S,那么split positive and negative 就有a+b = S + c + d, 就有 a+b+c+d = S + 2(c + d), if total = a+b+c+d, then c+d = (total - S)/2 where total - S should be even, then we just need to find the number of subsets with target sum = (total - S) / 2.

    /**
     * Math + DP O(sum * n); O(sum)
     * 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
     * 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
     */
    public int findTargetSumWays(int[] nums, int s) {
        if (nums == null || nums.length == 0) {
            return 0;
        }

        int sum = 0;
        for (int num : nums) {
            sum += num;
        }
        return s > sum || s < -sum || (s + sum) % 2 == 1 ? 0 : subsetSum(nums, (s + sum) / 2);
    }

    private int subsetSum(int[] nums, int s) {
        int[] dp = new int[s + 1];
        dp[0] = 1;
        for (int num : nums) {
//            for (int i = num; i <= s; i++) { //这么写就不对，不知道为什么，应该是一样的啊
                  //应该是这里的dp[i]只能从后往前算，因为重复元素也是有序的，从左到右排列，不能认为是简单的组合
//                dp[i] += dp[i - num]; 
//            }
            for (int i = s; i >= num; i--) {
                dp[i] += dp[i - num];
            }
        }
        return dp[s];
    }