<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 canPartition(vector<int>& nums) {  
        int sum=0;
        for(int n:nums)
        sum+=n;
        if(sum%2!=0)
            return 0;
        int target = sum/2;
        vector<vector<bool>>possible_sums(target+1,vector<bool>(2,0));
        int cur_col=0;
        if(nums[0] < target)
        possible_sums[nums[0]][cur_col] = 1; //only two sums possible till idx 0, itself and 0
        possible_sums[0][cur_col]=1;
        if(target==0 || target==nums[0])
        return 1;
        cur_col = 1-cur_col;;
        for(int i=1;i<nums.size();i++)
        {
            for(int j=0;j<=target;j++)
            {
                
               
                if(possible_sums[j][1-cur_col])
                {
                    possible_sums[j][cur_col] = 1;
                     if(j + nums[i] > target)
                    continue;
                    possible_sums[j + nums[i]][cur_col]=1;
                    if(j + nums[i] == target)
                    return 1;
                }
            }
            cur_col = 1 - cur_col;
        }

        return 0;
        
    }
};
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