<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>&a, int target) {
        int n=a.size();
        int sum=0;
        for(int i=0;i<n;i++){
            sum+=a[i];
        }
        if(abs(target)>sum){
            return 0;
        }
        vector<vector<int>>dp(n+1,vector<int>(2*sum+1,0));
        // auto val=[&](int x)->int{
        //     return x-sum;
        // };
        dp[0][sum]=1;
        for(int i=1;i<=n;i++){
            for(int j=0;j<=2*sum;j++){
                  if(j-a[i-1]>=0){
                    dp[i][j]+=dp[i-1][j-a[i-1]];
                  }
                  if(j+a[i-1]<=2*sum){
                    dp[i][j]+=dp[i-1][j+a[i-1]];
                  }
            }
        }
        return dp[n][target+sum];
    }
};
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