<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) {
        // dp[i, k] -> when the elements from ith to n-1th index can be picked and current diff of sum is k
        // dp[i, k] = (dp[i+1, k-2*nums[i]] || dp[i+1, k])
        // dp[i, 0] = 1 and dp[i, k!=0] = 0
        // dp[0, total_sum]

        int n = nums.size();
        int diff = accumulate(nums.begin(), nums.end(), 0);
        if(diff%2) return false;
        vector<int>nextRow(diff+1, 0);
        nextRow[0] = 1;
        for(int i = n-2; i>=0; --i){
            vector<int>currRow(diff+1, 0);
            for(int k = 0; k<=diff; ++k){
                int first = nextRow[k];
                int second = 0;
                if(k - (2*nums[i]) >= 0) second = nextRow[k-2*nums[i]];
                currRow[k] = (first || second);
            }
            nextRow = currRow;
        }
        return nextRow[diff]; 
    }
};
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