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
Solution:
class Solution {
public int maxAbsoluteSum(int[] nums) {
int n = nums.length;
int minSoFar = 0;
int maxSoFar = 0;
int prefix = 0, min = 0, max = 0;
for (int i = 0; i < n; i ++) {
prefix += nums[i];
// System.out.println(prefix + ", " + minSoFar + ", " + maxSoFar);
max = Math.max(max, prefix - minSoFar);
min = Math.min(min, prefix - maxSoFar);
minSoFar = Math.min(minSoFar, prefix);
maxSoFar = Math.max(maxSoFar, prefix);
}
return Math.max(max, Math.abs(min));
}
}