Formatted question description: https://leetcode.ca/all/918.html

# 918. Maximum Sum Circular Subarray (Medium)

Given a circular array C of integers represented by A, find the maximum possible sum of a non-empty subarray of C.

Here, a circular array means the end of the array connects to the beginning of the array.  (Formally, C[i] = A[i] when 0 <= i < A.length, and C[i+A.length] = C[i] when i >= 0.)

Also, a subarray may only include each element of the fixed buffer A at most once.  (Formally, for a subarray C[i], C[i+1], ..., C[j], there does not exist i <= k1, k2 <= j with k1 % A.length = k2 % A.length.)

Example 1:

Input: [1,-2,3,-2]
Output: 3
Explanation: Subarray  has maximum sum 3


Example 2:

Input: [5,-3,5]
Output: 10
Explanation: Subarray [5,5] has maximum sum 5 + 5 = 10


Example 3:

Input: [3,-1,2,-1]
Output: 4
Explanation: Subarray [2,-1,3] has maximum sum 2 + (-1) + 3 = 4


Example 4:

Input: [3,-2,2,-3]
Output: 3
Explanation: Subarray  and [3,-2,2] both have maximum sum 3


Example 5:

Input: [-2,-3,-1]
Output: -1
Explanation: Subarray [-1] has maximum sum -1


Note:

1. -30000 <= A[i] <= 30000
2. 1 <= A.length <= 30000

Related Topics:
Array

## Solution 1.

// OJ: https://leetcode.com/problems/maximum-sum-circular-subarray/

// Time: O(N)
// Space: O(N)
class Solution {
public:
int maxSubarraySumCircular(vector<int>& A) {
int N = A.size(), sum = 0, ans = INT_MIN;
vector<int> p(2 * N + 1);
for (int i = 0; i < 2 * N; ++i) {
p[i + 1] = (sum += A[i % N]);
}
deque<int> q;
for (int i = 0; i < 2 * N + 1; ++i) {
if (i >= N && q.front() == i - N - 1) q.pop_front();
if (q.size()) ans = max(ans, p[i] - p[q.front()]);
while (q.size() && p[q.back()] >= p[i]) q.pop_back();
q.push_back(i);
}
return ans;
}
};


Java

class Solution {
public int maxSubarraySumCircular(int[] A) {
if (A == null || A.length == 0)
return 0;
int sum = A;
int maxNum = A;
int newMaxSum = A, maxSum = A;
int newMinSum = A, minSum = A;
int length = A.length;
for (int i = 1; i < length; i++) {
sum += A[i];
maxNum = Math.max(maxNum, A[i]);
newMaxSum = Math.max(newMaxSum + A[i], A[i]);
maxSum = Math.max(maxSum, newMaxSum);
newMinSum = Math.min(newMinSum + A[i], A[i]);
minSum = Math.min(minSum, newMinSum);
}
if (maxNum <= 0)
return maxNum;
else
return Math.max(maxSum, sum - minSum);
}
}