LeetCode-in-Java
918. Maximum Sum Circular Subarray
Medium
Given a circular integer array nums of length n, return the maximum possible sum of a non-empty subarray of nums.
A circular array means the end of the array connects to the beginning of the array. Formally, the next element of nums[i] is nums[(i + 1) % n] and the previous element of nums[i] is nums[(i - 1 + n) % n].
A subarray may only include each element of the fixed buffer nums at most once. Formally, for a subarray nums[i], nums[i + 1], ..., nums[j], there does not exist i <= k1, k2 <= j with k1 % n == k2 % n.
Example 1:
Input: nums = [1,-2,3,-2]
Output: 3
Explanation: Subarray [3] has maximum sum 3.
Example 2:
Input: nums = [5,-3,5]
Output: 10
Explanation: Subarray [5,5] has maximum sum 5 + 5 = 10.
Example 3:
Input: nums = [-3,-2,-3]
Output: -2
Explanation: Subarray [-2] has maximum sum -2.
Constraints:
n == nums.length1 <= n <= 3 * 104-3 * 104 <= nums[i] <= 3 * 104
Solution
public class Solution {
private int kadane(int[] nums, int sign) {
int currSum = Integer.MIN_VALUE;
int maxSum = Integer.MIN_VALUE;
for (int i : nums) {
currSum = sign * i + Math.max(currSum, 0);
maxSum = Math.max(maxSum, currSum);
}
return maxSum;
}
public int maxSubarraySumCircular(int[] nums) {
if (nums.length == 1) {
return nums[0];
}
int sumOfArray = 0;
for (int i : nums) {
sumOfArray += i;
}
int maxSumSubarray = kadane(nums, 1);
int minSumSubarray = kadane(nums, -1) * -1;
if (sumOfArray == minSumSubarray) {
return maxSumSubarray;
} else {
return Math.max(maxSumSubarray, sumOfArray - minSumSubarray);
}
}
}