<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 n = nums.size();
        
        int tsum = accumulate(nums.begin(), nums.end(), 0);

        if(tsum %2) {
            return false;
        }

        vector<bool> dp(20000);
        dp[0] = dp[nums[0]] = true;

        // dp[i][j] represent if sum 'j' is possible at ith index or not
        for(int i=1; i<n; i++) {
            vector<bool> temp = dp;
            for(int j=0; j<=20000; j++) {
                bool pick = j >= nums[i] ? dp[j-nums[i]] : false;
                bool notPick = dp[j];
                temp[j] = pick || notPick;
            }
            dp = temp;
        }

        return dp[tsum/2];
    }
};
33578