<p>You are given an integer array <code>nums</code> and an integer <code>target</code>.</p>

<p>You want to build an <strong>expression</strong> out of nums by adding one of the symbols <code>&#39;+&#39;</code> and <code>&#39;-&#39;</code> before each integer in nums and then concatenate all the integers.</p>

<ul>
	<li>For example, if <code>nums = [2, 1]</code>, you can add a <code>&#39;+&#39;</code> before <code>2</code> and a <code>&#39;-&#39;</code> before <code>1</code> and concatenate them to build the expression <code>&quot;+2-1&quot;</code>.</li>
</ul>

<p>Return the number of different <strong>expressions</strong> that you can build, which evaluates to <code>target</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,1,1,1], target = 3
<strong>Output:</strong> 5
<strong>Explanation:</strong> 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
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1], target = 1
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 20</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 1000</code></li>
	<li><code>0 &lt;= sum(nums[i]) &lt;= 1000</code></li>
	<li><code>-1000 &lt;= target &lt;= 1000</code></li>
</ul>

class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int target) {
        int sum = 0;
        for (int i : nums) 
            sum += i;

        if (target > sum || target < -sum) return 0;
        if ((sum - target) % 2 != 0) return 0;

        // ways to reach curr_sum after +-index i
        vector<int> dp(2 * sum + 1, 0);
        dp[sum - nums[0]] += 1;
        dp[sum + nums[0]] += 1;

        for (int i = 1; i < nums.size(); ++i) {
            vector<int> next(2 * sum + 1, 0);
            for (int k = 0; k <= 2 * sum; ++k) {
                if (k - nums[i] >= 0)
                    next[k] += dp[k - nums[i]];

                if (k + nums[i] <= 2 * sum)
                    next[k] += dp[k + nums[i]];
            }

            dp = next;
        }

        return dp[target + sum];
    }
};
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