<p>Given an integer array <code>nums</code> of <code>2n</code> integers, group these integers into <code>n</code> pairs <code>(a<sub>1</sub>, b<sub>1</sub>), (a<sub>2</sub>, b<sub>2</sub>), ..., (a<sub>n</sub>, b<sub>n</sub>)</code> such that the sum of <code>min(a<sub>i</sub>, b<sub>i</sub>)</code> for all <code>i</code> is <strong>maximized</strong>. Return<em> the maximized sum</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,4,3,2]
<strong>Output:</strong> 4
<strong>Explanation:</strong> All possible pairings (ignoring the ordering of elements) are:
1. (1, 4), (2, 3) -&gt; min(1, 4) + min(2, 3) = 1 + 2 = 3
2. (1, 3), (2, 4) -&gt; min(1, 3) + min(2, 4) = 1 + 2 = 3
3. (1, 2), (3, 4) -&gt; min(1, 2) + min(3, 4) = 1 + 3 = 4
So the maximum possible sum is 4.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [6,2,6,5,1,2]
<strong>Output:</strong> 9
<strong>Explanation:</strong> The optimal pairing is (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
	<li><code>nums.length == 2 * n</code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

'''class Solution:
    def arrayPairSum(self, nums: List[int]) -> int:
        nums.sort()
        return sum(nums[::2])'''

class Solution:
    def arrayPairSum(self, nums: List[int]) -> int:
        K = 10000
        # Store the frequency of each element
        element_to_count = [0] * (2 * K + 1)
        for element in nums:
            # Add K to element to offset negative values
            element_to_count[element + K] += 1
            
        # Initialize sum to zero
        max_sum = 0
        is_even_index = True
        for element in range(2 * K + 1):
            while element_to_count[element] > 0 :
                # Add element if it is at even index
                if is_even_index:
                    max_sum += element - K
                # Flip the value (one to zero or zero to one)
                is_even_index = not is_even_index
                # Decrement the frequency count
                element_to_count[element] -= 1
        return max_sum
20000