<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 dp[20][10000];
    Solution(){
        memset(dp,-1,sizeof(dp));
    }
    int solve(vector<int>& nums, int target,int pos){
        if(dp[pos][target+1000] != -1) return dp[pos][target+1000];
      if(pos == nums.size()-1){
        if(nums[pos] == target || nums[pos] == (-1*target)){
            dp[pos][target+1000] = 1;
             return dp[pos][target+1000];
            }
        else{
            dp[pos][target+1000] = 0;
            return dp[pos][target+1000];
        }
      }
      dp[pos+1][target+nums[pos]+1000] = solve(nums,target+nums[pos],pos+1);
      dp[pos+1][target-nums[pos]+1000] = solve(nums,target-nums[pos],pos+1);
      return dp[pos+1][target+nums[pos]+1000] +dp[pos+1][target-nums[pos]+1000]; 
    }
    int findTargetSumWays(vector<int>& nums, int target) {
      return (nums.back() == 0)?solve(nums,target,0)*2:solve(nums,target,0); 
    }
};
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