<p>Given an integer array <code>nums</code> and an integer <code>k</code>, modify the array in the following way:</p>

<ul>
	<li>choose an index <code>i</code> and replace <code>nums[i]</code> with <code>-nums[i]</code>.</li>
</ul>

<p>You should apply this process exactly <code>k</code> times. You may choose the same index <code>i</code> multiple times.</p>

<p>Return <em>the largest possible sum of the array after modifying it in this way</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [4,2,3], k = 1
<strong>Output:</strong> 5
<strong>Explanation:</strong> Choose index 1 and nums becomes [4,-2,3].
</pre>

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

<pre>
<strong>Input:</strong> nums = [3,-1,0,2], k = 3
<strong>Output:</strong> 6
<strong>Explanation:</strong> Choose indices (1, 2, 2) and nums becomes [3,1,0,2].
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,-3,-1,5,-4], k = 2
<strong>Output:</strong> 13
<strong>Explanation:</strong> Choose indices (1, 4) and nums becomes [2,3,-1,5,4].
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>-100 &lt;= nums[i] &lt;= 100</code></li>
	<li><code>1 &lt;= k &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution:
    def largestSumAfterKNegations(self, nums: List[int], k: int) -> int:
        # Sort the array to bring negative values to the front
        nums.sort()
        
        # Flip the negative values to positive values k times
        for i in range(len(nums)):
            if nums[i] < 0 and k > 0:
                nums[i] = -nums[i]
                k -= 1
        
        # If k is still greater than 0 and is odd, flip the smallest value
        if k % 2 == 1:
            nums.sort()
            nums[0] = -nums[0]
        
        # Return the sum of the modified array
        return sum(nums)
16700