# 0494. Target Sum

https://leetcode.com/problems/target-sum

## Description

You are given an integer array `nums` and an integer `target`.

You want to build an **expression** out of nums by adding one of the symbols `'+'` and `'-'` before each integer in nums and then concatenate all the integers.

* For example, if `nums = [2, 1]`, you can add a `'+'` before `2` and a `'-'` before `1` and concatenate them to build the expression `"+2-1"`.

Return the number of different **expressions** that you can build, which evaluates to `target`.

**Example 1:**

```
**Input:** nums = [1,1,1,1,1], target = 3
**Output:** 5
**Explanation:** There are 5 ways to assign symbols to make the sum of nums be target 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
+1 + 1 + 1 + 1 - 1 = 3
```

**Example 2:**

```
**Input:** nums = [1], target = 1
**Output:** 1
```

**Constraints:**

* `1 <= nums.length <= 20`
* `0 <= nums[i] <= 1000`
* `0 <= sum(nums[i]) <= 1000`
* `-1000 <= target <= 1000`

## ac1: knapsack dp

```java
class Solution {
    public int findTargetSumWays(int[] nums, int S) {
        // edge cases
        if (nums == null || nums.length == 0) return 0;

        int sum = 0;
        for (int i : nums) sum += i;  // get sum, dp range [-sum, sum]
        if (sum < S || -sum > S) return 0; // can't reach S

        int[] dp = new int[2*sum + 1];
        dp[0+sum] = 1; // when sum == 0, dp is 1

        for (int j = 0; j < nums.length; j++) {
            int[] dpnew = new int[2*sum + 1];
            for (int d = 0; d < dpnew.length; d++) {
                if (d - nums[j] >= 0 && d + nums[j] < dp.length) {
                dpnew[d - nums[j]] += dp[d];
                dpnew[d + nums[j]] += dp[d];
                }
            }
            dp = dpnew;
        }

        return dp[sum+S];
    }
}

/*
knapsack, for each element, you can take it or leave it
for each num, plus/minus it get a result, then cnt++ at that result
*/
```
