<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 call(int tar,vector<int>& nums,vector<vector<int>> &dp,int i){
        if(tar==0)return 1;
        if(i<0)return 0;
        if(dp[i][1000+tar]!=-1)return dp[i][1000+tar];
        int ans=0;
        if(nums[i]<=tar){
            ans=call(tar-nums[i],nums,dp,i-1);
        }
        ans+=call(tar,nums,dp,i-1);
        return dp[i][1000+tar]=ans;

    }
    int findTargetSumWays(vector<int>& nums, int target) {
        int sum=0;int c=0;
        vector<int> v;
        for(int j=0;j<nums.size();j++){
            sum+=nums[j];
            if(nums[j]!=0){
                  v.push_back(nums[j]);
            }else c++;
        }
        if(sum<target||(sum+target)%2!=0)return 0;
        int tar=(sum+target)/2;
        vector<vector<int>> dp(v.size(),vector<int> (2001,-1));
        int ans= call(tar,v,dp,v.size()-1);

     
        return ans*pow(2,c);
    
    }
};
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