# 1749. Maximum Absolute Sum of Any Subarray

## Description

You are given an integer array nums. The absolute sum of a subarray [numsl, numsl+1, ..., numsr-1, numsr] is abs(numsl + numsl+1 + ... + numsr-1 + numsr).

Return the maximum absolute sum of any (possibly empty) subarray of nums.

Note that abs(x) is defined as follows:

• If x is a negative integer, then abs(x) = -x.
• If x is a non-negative integer, then abs(x) = x.

Example 1:

Input: nums = [1,-3,2,3,-4]
Output: 5
Explanation: The subarray [2,3] has absolute sum = abs(2+3) = abs(5) = 5.


Example 2:

Input: nums = [2,-5,1,-4,3,-2]
Output: 8
Explanation: The subarray [-5,1,-4] has absolute sum = abs(-5+1-4) = abs(-8) = 8.


Constraints:

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

## Solutions

Solution 1: Dynamic Programming

We define $f[i]$ to represent the maximum value of the subarray ending with $nums[i]$, and define $g[i]$ to represent the minimum value of the subarray ending with $nums[i]$. Then the state transition equation of $f[i]$ and $g[i]$ is as follows:

\begin{aligned} f[i] &= \max(f[i - 1], 0) + nums[i] \\ g[i] &= \min(g[i - 1], 0) + nums[i] \end{aligned}
 The final answer is the maximum value of $max(f[i], g[i] )$.

Since $f[i]$ and $g[i]$ are only related to $f[i - 1]$ and $g[i - 1]$, we can use two variables to replace the array, reducing the space complexity to $O(1)$.

Time complexity $O(n)$, space complexity $O(1)$, where $n$ is the length of the array $nums$.

• class Solution {
public int maxAbsoluteSum(int[] nums) {
int f = 0, g = 0;
int ans = 0;
for (int x : nums) {
f = Math.max(f, 0) + x;
g = Math.min(g, 0) + x;
ans = Math.max(ans, Math.max(f, Math.abs(g)));
}
return ans;
}
}

• class Solution {
public:
int maxAbsoluteSum(vector<int>& nums) {
int f = 0, g = 0;
int ans = 0;
for (int& x : nums) {
f = max(f, 0) + x;
g = min(g, 0) + x;
ans = max({ans, f, abs(g)});
}
return ans;
}
};

• class Solution:
def maxAbsoluteSum(self, nums: List[int]) -> int:
f = g = 0
ans = 0
for x in nums:
f = max(f, 0) + x
g = min(g, 0) + x
ans = max(ans, f, abs(g))
return ans


• func maxAbsoluteSum(nums []int) (ans int) {
var f, g int
for _, x := range nums {
f = max(f, 0) + x
g = min(g, 0) + x
ans = max(ans, max(f, abs(g)))
}
return
}

func abs(x int) int {
if x < 0 {
return -x
}
return x
}

• function maxAbsoluteSum(nums: number[]): number {
let f = 0;
let g = 0;
let ans = 0;
for (const x of nums) {
f = Math.max(f, 0) + x;
g = Math.min(g, 0) + x;
ans = Math.max(ans, f, -g);
}
return ans;
}


• impl Solution {
pub fn max_absolute_sum(nums: Vec<i32>) -> i32 {
let mut f = 0;
let mut g = 0;
let mut ans = 0;
for x in nums {
f = i32::max(f, 0) + x;
g = i32::min(g, 0) + x;
ans = i32::max(ans, f.max(-g));
}
ans
}
}