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.

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.

Tips:

用简单的DFS可以做，类似Expression Add Operators，可以进行优化，用一个sum数组记录后面数的和，如果这个和小于target - 当前和的绝对值，则可以return。复杂度是O(2^n).

也可以用DP，非常快。（没看）

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

Note that the above formula has proved that target + sum(nums) must be even

Codes:

DFS:

public class Solution {
    int res = 0;
    public int findTargetSumWays(int[] nums, int S) {
        if (nums == null || nums.length == 0) return 0;
        int[] sums = new int[nums.length];
        sums[nums.length - 1] = nums[nums.length - 1];
        for (int i = nums.length - 2; i >= 0; i--) {
            sums[i] = sums[i + 1] + nums[i];
        }
        helper(nums, S, 0, 0, sums);
        return res;
    }

    private void helper(int[] nums, int S, int pos, long eval, int[] sums) {
        if (pos == nums.length) {
            if (eval == S) res++;
            return;
        }
        if (sums[pos] < Math.abs(S - eval)) return;
        helper(nums, S, pos + 1, eval + nums[pos], sums);
        helper(nums, S, pos + 1, eval - nums[pos], sums);
    }
}   

DP:

public int findTargetSumWays(int[] nums, int s) {
    int sum = 0;
    for (int n : nums)
        sum += n;
    return sum < s || (s + sum) % 2 > 0 ? 0 : subsetSum(nums, (s + sum) >>> 1); 
}   

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