<p>Given an integer array <code>nums</code>, return <code>true</code> <em>if you can partition the array into two subsets such that the sum of the elements in both subsets is equal or </em><code>false</code><em> otherwise</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,5,11,5]
<strong>Output:</strong> true
<strong>Explanation:</strong> The array can be partitioned as [1, 5, 5] and [11].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,5]
<strong>Output:</strong> false
<strong>Explanation:</strong> The array cannot be partitioned into equal sum subsets.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 200</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
</ul>

class Solution {
public:
    bool recTab(vector<int>& v, int n, int k, vector<int>& prev)
    {
        prev[0] = 1;
        if(v[0]<=k) prev[v[0]]=1;

        for(int idx=1; idx<n; idx++)
        {
            vector<int> curr(k+1, 0);
            curr[0]=1;
            for(int tar=1; tar<=k; tar++)
            {
                bool inc = false;
                if(v[idx]<=tar) inc = prev[tar-v[idx]];
                bool exc = prev[tar];

                curr[tar]= inc || exc;
            }
            prev = curr;
        }
        
        return prev[k];
    }

    bool canPartition(vector<int>& nums) {
        int sum=0;
        int n = nums.size();
        for(int i=0; i<n; i++) sum += nums[i];

        int tar = sum/2;
        vector<int> prev(tar+1, 0);
        if(sum%2 != 0) return false;
        else return recTab(nums, n, tar, prev);
    }
};
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