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:
x is a negative integer, then abs(x) = -x.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] <= 104class Solution {
public int maxAbsoluteSum(int[] nums) {
int minPrefixSum = 0, maxPrefixSum = 0;
int prefixSum = 0;
for (int i = 0; i < nums.length; i++) {
prefixSum += nums[i];
minPrefixSum = Math.min(minPrefixSum, prefixSum);
maxPrefixSum = Math.max(maxPrefixSum, prefixSum);
}
return maxPrefixSum - minPrefixSum;
}
}