<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 findWays(vector<int> &num, int tar){
     int n = num.size();

    vector<vector<int>> dp(n,vector<int>(tar+1,0));
    
    if(num[0] == 0) dp[0][0] =2;  // 2 cases -pick and not pick
    else dp[0][0] = 1;  // 1 case - not pick
    
    if(num[0]!=0 && num[0]<=tar) dp[0][num[0]] = 1;  // 1 case -pick
    
    for(int ind = 1; ind<n; ind++){
        for(int target= 0; target<=tar; target++){
            
            int notTaken = dp[ind-1][target];
    
            int taken = 0;
                if(num[ind]<=target)
                    taken = dp[ind-1][target-num[ind]];
        
            dp[ind][target]= (notTaken + taken);
        }
    }
    return dp[n-1][tar];
}
int countPartitions(int n, int d, vector<int>& arr){
    int totSum = 0;
    for(int i=0; i<n;i++){
        totSum += arr[i];
    }
    
    //Checking for edge cases
    if(totSum-d <0 || (totSum-d)%2 ) return 0;
    
    return findWays(arr,(totSum-d)/2);
}
    int ways(vector<int>& nums, int target,int index,vector<vector<int>>&dp){
         if (index < 0) return target == 0 ? 1 : 0;
        if(dp[index][target+1000]!=-1)return dp[index][target+1000];
        int minus_take=ways(nums,target+nums[index],index-1,dp);
        int plus_take=ways(nums,target-nums[index],index-1,dp);
        return dp[index][target+1000]=minus_take+plus_take;
    }
    int findTargetSumWays(vector<int>& nums, int target) {
        int n=nums.size();
        vector<vector<int>>dp(n,vector<int>(2001,-1));
       return countPartitions(n,target,nums);
    }
};
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