494. 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.

S1: 2 hash map O(2^n)?

用hash_map记录到目前为止所有的sum和达到sum的可行组合数

    int findTargetSumWays(vector<int>& nums, int S) {
        unordered_map<int, int> sumCount;
        sumCount[0] = 1;
        for (int i : nums) {
            unordered_map<int, int> nextCount;
            for (auto sum : sumCount) {
                nextCount[sum.first + i] += sum.second;
                nextCount[sum.first - i] += sum.second;
            }
            sumCount = nextCount;
        }
        return sumCount[S];
    }

S2: DP + math O(mn)

假设最终的sum 由正数部分sum[p]和负数部分sum[n]组成

sum[p] + sum[n] = S => sum[p] + sum[n] + sum[p] - sum[n] = S + sum[p] - sum[n] => 2* sum[p] = S + nums.sum

=> sum[p] = (S + nums.sum) / 2 可转化为求sum为(s + nums.sum)/2的subset的个数。

因为原数组没有负数，S 一定小于nums.sum，S+nums.sum 大于等于0且为偶数

class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int S) {
        int sum = 0;
        for (int i : nums) {
            sum += i;
        }
        if (sum + S < 0 || sum < S || (sum + S) & 1) {
            return 0;
        } else {
            return findSum(nums, (sum + S) / 2);
        }
    }

    int findSum(vector<int>& nums, int target) {
        vector<int> sumCount(target + 1, 0);
        sumCount[0] = 1;
        for (int i : nums) {
            for (int j = target; j >= i; --j) {
                sumCount[j] += sumCount[j - i];
            }
        }
        return sumCount[target];
    }
};