1749

1749. Maximum Absolute Sum of Any Subarray (Medium)

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

Related Topics:
Greedy

Similar Questions:

• Maximum Subarray (Easy)

Solution 1.

// OJ: https://leetcode.com/problems/maximum-absolute-sum-of-any-subarray/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int maxAbsoluteSum(vector<int>& A) {
        int ans = 0, mn = 0, mx = 0, sum = 0;
        for (int n : A) {
            sum += n;
            ans = max({ ans, mx - sum, sum - mn });
            mx = max(mx, sum);
            mn = min(mn, sum);
        }
        return ans;
    }
};