# 1186. Maximum Subarray Sum with One Deletion

## Description

Given an array of integers, return the maximum sum for a non-empty subarray (contiguous elements) with at most one element deletion. In other words, you want to choose a subarray and optionally delete one element from it so that there is still at least one element left and the sum of the remaining elements is maximum possible.

Note that the subarray needs to be non-empty after deleting one element.

Example 1:

Input: arr = [1,-2,0,3]
Output: 4
Explanation: Because we can choose [1, -2, 0, 3] and drop -2, thus the subarray [1, 0, 3] becomes the maximum value.

Example 2:

Input: arr = [1,-2,-2,3]
Output: 3
Explanation: We just choose [3] and it's the maximum sum.


Example 3:

Input: arr = [-1,-1,-1,-1]
Output: -1
Explanation: The final subarray needs to be non-empty. You can't choose [-1] and delete -1 from it, then get an empty subarray to make the sum equals to 0.


Constraints:

• 1 <= arr.length <= 105
• -104 <= arr[i] <= 104

## Solutions

Solution 1: Preprocessing + Enumeration

We can first preprocess the array $arr$ to find the maximum subarray sum ending at and starting from each element, and store them in the arrays $left$ and $right$ respectively.

If we do not delete any element, then the maximum subarray sum is the maximum value in $left[i]$ or $right[i]$. If we delete one element, we can enumerate each position $i$ in $[1..n-2]$, calculate the value of $left[i-1] + right[i+1]$, and take the maximum value.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $arr$.

• class Solution {
public int maximumSum(int[] arr) {
int n = arr.length;
int[] left = new int[n];
int[] right = new int[n];
int ans = -(1 << 30);
for (int i = 0, s = 0; i < n; ++i) {
s = Math.max(s, 0) + arr[i];
left[i] = s;
ans = Math.max(ans, left[i]);
}
for (int i = n - 1, s = 0; i >= 0; --i) {
s = Math.max(s, 0) + arr[i];
right[i] = s;
}
for (int i = 1; i < n - 1; ++i) {
ans = Math.max(ans, left[i - 1] + right[i + 1]);
}
return ans;
}
}

• class Solution {
public:
int maximumSum(vector<int>& arr) {
int n = arr.size();
int left[n];
int right[n];
for (int i = 0, s = 0; i < n; ++i) {
s = max(s, 0) + arr[i];
left[i] = s;
}
for (int i = n - 1, s = 0; ~i; --i) {
s = max(s, 0) + arr[i];
right[i] = s;
}
int ans = *max_element(left, left + n);
for (int i = 1; i < n - 1; ++i) {
ans = max(ans, left[i - 1] + right[i + 1]);
}
return ans;
}
};

• class Solution:
def maximumSum(self, arr: List[int]) -> int:
n = len(arr)
left = [0] * n
right = [0] * n
s = 0
for i, x in enumerate(arr):
s = max(s, 0) + x
left[i] = s
s = 0
for i in range(n - 1, -1, -1):
s = max(s, 0) + arr[i]
right[i] = s
ans = max(left)
for i in range(1, n - 1):
ans = max(ans, left[i - 1] + right[i + 1])
return ans


• func maximumSum(arr []int) int {
n := len(arr)
left := make([]int, n)
right := make([]int, n)
for i, s := 0, 0; i < n; i++ {
s = max(s, 0) + arr[i]
left[i] = s
}
for i, s := n-1, 0; i >= 0; i-- {
s = max(s, 0) + arr[i]
right[i] = s
}
ans := slices.Max(left)
for i := 1; i < n-1; i++ {
ans = max(ans, left[i-1]+right[i+1])
}
return ans
}

• function maximumSum(arr: number[]): number {
const n = arr.length;
const left: number[] = Array(n).fill(0);
const right: number[] = Array(n).fill(0);
for (let i = 0, s = 0; i < n; ++i) {
s = Math.max(s, 0) + arr[i];
left[i] = s;
}
for (let i = n - 1, s = 0; i >= 0; --i) {
s = Math.max(s, 0) + arr[i];
right[i] = s;
}
let ans = Math.max(...left);
for (let i = 1; i < n - 1; ++i) {
ans = Math.max(ans, left[i - 1] + right[i + 1]);
}
return ans;
}


• impl Solution {
pub fn maximum_sum(arr: Vec<i32>) -> i32 {
let n = arr.len();
let mut left = vec![0; n];
let mut right = vec![0; n];
let mut s = 0;
for i in 0..n {
s = (s.max(0)) + arr[i];
left[i] = s;
}
s = 0;
for i in (0..n).rev() {
s = (s.max(0)) + arr[i];
right[i] = s;
}
let mut ans = *left.iter().max().unwrap();
for i in 1..n - 1 {
ans = ans.max(left[i - 1] + right[i + 1]);
}
ans
}
}