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.length
- 1 <= n <= 3 * 104
- -3 * 104 <= nums[i] <= 3 * 104
Solution
class Solution {
public int maxSubarraySumCircular(int[] nums) {
return maxsum(nums);
}
public static int maxsum(int[] arr)
{
int linear_sum=kadens(arr);
int totalsum=0;
for(int i=0;i<arr.length;i++)
{
totalsum+=arr[i];
arr[i]=arr[i]*-1;
}
int mid_sum=kadens(arr);
int curr_sum=totalsum+mid_sum;
if(curr_sum==0)
{
return linear_sum;
}
return Math.max(linear_sum,curr_sum);
}
public static int kadens(int[] arr)
{
int ans=Integer.MIN_VALUE;
int sum=0;
for(int i=0;i<arr.length;i++)
{
sum=sum+arr[i];
ans=Math.max(ans,sum);
if(sum<0)
{
sum=0;
}
}
return ans;
}
}