<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 totalSum = 0;

        for(auto it: nums) {
            totalSum += it;
        }

        if(totalSum % 2 != 0) {
            return false;
        }

        int target = totalSum / 2;
        int n = nums.size();

        // Use an unordered_set to track all possible subset sums
        unordered_set<int> st;
        st.insert(0);

        for(int i = 0; i < n; i++) {

            unordered_set<int> newSums = st;  // Make a copy of the current set
            
            // For each sum already in the set, add nums[i] to form new sums
            for(auto it : st) {
                int newSum = it + nums[i];

                // If the new sum is equal to the target, return true
                if(newSum == target) {
                    return true;
                }

                // Add the new sum to the temporary set
                newSums.insert(newSum);
            }

            // Update the original set with the new sums
            st = newSums;
        }

        return st.find(target) != st.end();
    }
};

202539