Formatted question description: https://leetcode.ca/all/1746.html

# 1746. Maximum Subarray Sum After One Operation

Medium

## Description

You are given an integer array nums. You must perform exactly one operation where you can replace one element nums[i] with nums[i] * nums[i].

Return the maximum possible subarray sum after exactly one operation. The subarray must be non-empty.

Example 1:

Input: nums = [2,-1,-4,-3]

Output: 17

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

Example 2:

Input: nums = [1,-1,1,1,-1,-1,1]

Output: 4

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

Constraints:

• 1 <= nums.length <= 10^5
• -10^4 <= nums[i] <= 10^4

## Solution

Use dynamic programming. Let n be the length of nums. Create a 2D array dp of n rows and 3 columns, where

• dp[i][0] represents the maximum subarray sum that ends at index i when no operation is performed,
• dp[i][1] represents the maximum subarray sum that ends at index i when the one operation is performed at index i, and
• dp[i][2] represents the maximum subarray sum that ends at index i when the one operation is performed before index i.

Initially, dp[0][0] = nums[0], dp[0][1] = nums[0] * nums[0] and dp[0][2] = Integer.MIN_VALUE, which means dp[0][2] is an impossible value.

Loop over i from 1 to n - 1. For each i, there is dp[i][0] = Math.max(dp[i - 1][0], 0) + nums[i], dp[i][1] = Math.max(dp[i - 1][0], 0) + nums[i] * nums[i] and dp[i][2] = Math.max(Math.max(dp[i - 1][1], dp[i - 1][2]), 0) + nums[i].

Finally, return the maximum value in the last two columns of dp.

• class Solution {
public int maxSumAfterOperation(int[] nums) {
int length = nums.length;
int[][] dp = new int[length][3];
dp[0][0] = nums[0];
dp[0][1] = nums[0] * nums[0];
dp[0][2] = Integer.MIN_VALUE;
int max = dp[0][1];
for (int i = 1; i < length; i++) {
dp[i][0] = Math.max(dp[i - 1][0], 0) + nums[i];
dp[i][1] = Math.max(dp[i - 1][0], 0) + nums[i] * nums[i];
dp[i][2] = Math.max(Math.max(dp[i - 1][1], dp[i - 1][2]), 0) + nums[i];
int curMax = Math.max(dp[i][1], dp[i][2]);
max = Math.max(max, curMax);
}
return max;
}
}

from typing import List

class Solution:
def maximumSum(self, arr: List[int]) -> int:
dp = []
dp = [[0]*3 for i in range(len(arr))]
dp[0][0] = arr[0]
dp[0][1] = arr[0] * arr[0]
dp[0][2] = float('-inf')

result = float('-inf')
for i in range(1, len(arr)):
dp[i][0] = arr[i] + max(0, dp[i - 1][0])
dp[i][1] = arr[i] * arr[i] + max(0, dp[i - 1][0])
dp[i][2] = arr[i] + max(0, dp[i - 1][1], dp[i - 1][2])

result = max(result, dp[i][0], dp[i][1], dp[i][2])

return result

if __name__ == "__main__":
print(Solution().maximumSum([2,-1,-4,-3]))


• Todo

• print("Todo!")