Maximum Subarray Sum After One Operation
<p>You are given an integer array <code>nums</code>. You must perform <strong>exactly one</strong> operation&nbsp;where you can <strong>replace</strong> one&nbsp;element <code>nums[i]</code> with <code>nums[i] * nums[i]</code>.&nbsp;</p>

<p>Return <em>the <strong>maximum</strong> possible subarray sum after <strong>exactly&nbsp;one</strong> operation</em>. The subarray must be non-empty.</p>

<p>&nbsp;</p>

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

<pre>

<strong>Input:</strong> nums = [2,-1,-4,-3]

<strong>Output:</strong> 17

<strong>Explanation:</strong> You can perform the operation on index 2 (0-indexed) to make nums = [2,-1,<strong>16</strong>,-3]. Now, the maximum subarray sum is 2 + -1 + 16 = 17.</pre>

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

<pre>

<strong>Input:</strong> nums = [1,-1,1,1,-1,-1,1]

<strong>Output:</strong> 4

<strong>Explanation:</strong> You can perform the operation on index 1 (0-indexed) to make nums = [1,<strong>1</strong>,1,1,-1,-1,1]. Now, the maximum subarray sum is 1 + 1 + 1 + 1 = 4.</pre>

<p>&nbsp;</p>

<p><strong>Constraints:</strong></p>

<ul>

    <li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>

    <li><code>-10<sup>4</sup>&nbsp;&lt;= nums[i] &lt;= 10<sup>4</sup></code></li>

</ul>
Array; Dynamic Programming
class Solution:
    def maxSumAfterOperation(self, nums: List[int]) -> int:
        f = g = 0
        ans = -inf
        for x in nums:
            ff = max(f, 0) + x
            gg = max(max(f, 0) + x * x, g + x)
            f, g = ff, gg
            ans = max(ans, f, g)
        return ans
