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 [3] 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 [3] 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:
-30000 <= A[i] <= 300001 <= A.length <= 30000
Runtime: 76 ms, faster than 100.00% of C++ online submissions for Maximum Sum Circular Subarray.
Memory Usage: 13.2 MB, less than 0.77% of C++ online submissions for Maximum Sum Circular Subarray.
class Solution { public: int maxSubarraySumCircular(vector<int>& A) { int total = 0; int curmax = 0, maxsum = INT32_MIN; int curmin = 0, minsum = INT32_MAX; for(int a : A) { curmax = max(curmax + a, a); maxsum = max(maxsum, curmax); curmin = min(curmin + a, a); minsum = min(minsum, curmin); total += a; } return maxsum > 0 ? max(maxsum, total - minsum) : maxsum; } };