<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) {
        // Calculate the total sum of the array
        int total = accumulate(nums.begin(), nums.end(), 0);

        // If the total sum is odd, it's not possible to split the array into two equal subsets
        if (total % 2 != 0) return false;

        // Target is half of the total sum
        int target = total / 2;
        
        // Memoization map to store intermediate results
        unordered_map<string, bool> memo;

        // Helper function for the DP solution
        function<bool(int, int)> dp = [&](int n, int k) -> bool {
            // If the target subset sum is 0, we found a valid partition
            if (k == 0) return true;
            // If we've gone through all elements or the target sum becomes negative, return false
            if (n == 0 || k < 0) return false;

            // Memoization key
            string key = to_string(n) + "," + to_string(k);

            // Check if the result is already memoized
            if (memo.find(key) != memo.end()) return memo[key];

            // Recurrence relation: either include nums[n-1] in the subset or exclude it
            bool result = dp(n - 1, k - nums[n - 1]) || dp(n - 1, k);

            // Store the result in the memo table and return it
            return memo[key] = result;
        };

        // Start the recursion with the full array and the target sum
        return dp(nums.size(), target);
    }
};
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