<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:
long long subset(int n, long long s, vector<int> &arr, map<long long, map<long long, long long>> &dp){
       
        if(n==0){
            if(s==0 and arr[0]==0)  return 2;
            if(s==0 || s==arr[0]) return 1;
            return 0;
        }
        
        if(dp.count(n) and dp[n].count(s)) return dp[n][s];
        long long take=0, nottake=0;
        if(arr[n]<=s){
            take = (subset(n-1, s-arr[n], arr, dp)+ 
             subset(n-1, s, arr, dp));
        }
        else
            nottake = subset(n-1, s, arr, dp);
            return dp[n][s]=(take+nottake);
    }
    long long countPartitions(int n, long long d, vector<int>& arr) {
        // Code here
        long long sum=0;
        for(long long i=0;i<n;i++){
             sum+=arr[i];
    
        }
        if((sum+d)%2 or (sum-d)%2) return 0;
        
        long long s=(sum+d)/2;
        long long s2=(sum-d)/2;
        if(sum-d<0) return 0;
        map<long long, map<long long, long long>> dp;
        return subset(n-1, s,arr, dp);
    }
    int findTargetSumWays(vector<int>& nums, int target) {
        return countPartitions(nums.size(), target, nums);
    }
};
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