1749. Maximum Absolute Sum of Any Subarray

Problem

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 <= 10^5

• -10^4 <= nums[i] <= 10^4

Solution (Java)

class Solution {
    public int maxAbsoluteSum(int[] nums) {
        int min = 0;
        int max = 0;
        int s = 0;
        for (int num : nums) {
            s += num;
            min = Math.min(min, s);
            max = Math.max(max, s);
        }
        return max - min;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).