<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) {
        vector<unordered_map<int,int>> memo(nums.size()+1); // memo[i][j] returns how many ways to get to sum j using first i nums
        vector<int> upper(nums.size() + 1);
        vector<int> lower(nums.size() + 1);
        for(int i = nums.size()-1;i >= 0;--i){
            upper[i] = upper[i+1] + nums[i];
            lower[i] = lower[i+1] - nums[i];
        }
        if((upper[0]-target) % 2 ==1){
            return 0;
        }
        memo[0][0] = 1;
        for(int i = 0; i<nums.size();++i){
            for(pair<int,int> p : memo[i]){
                int ways = p.second;
                int sum = p.first;
                if( sum + nums[i] + lower[i] <= target ){
                    memo[i+1][sum + nums[i]] += ways;
                }
                if( sum - nums[i] + upper[i] >= target){
                    memo[i+1][sum - nums[i]] += ways;
                }
            }
        }
        return memo.back()[target];
    }
};
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