918. Maximum Sum Circular Subarray.cpp

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

/*
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] <= 30000
1 <= A.length <= 30000
*/

/*
Result:
Runtime: 156 ms, faster than 46.77% of C++ online submissions for Maximum Sum Circular Subarray.
Memory Usage: 39.9 MB, less than 8.33% of C++ online submissions for Maximum Sum Circular Subarray.
*/

class Solution {
public:
    int maxSubarraySumCircular(vector<int>& A) {
        int sum = A[0];
        int ans = A[0];

        int i = 1;
        
        int min_sum = A[0];
        int min_ = 30000;
        int all_sum = A[0];
        
        while(i < A.size()){
            if(sum < 0){
                sum = A[i];
            }
            else{
                sum += A[i];
            }
            if(min_sum > 0){
                min_sum = A[i];
            }
            else{
                min_sum += A[i];
            }
            min_ = min(min_sum, min_);
            ans = max(sum, ans);
            all_sum += A[i];
            i ++;
        }
        
        if(min_ != all_sum){
            return max(ans, all_sum - min_);
        }
        return ans;
        
    }
};