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May-15 2020 Challenge

15. 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:
  1. -30000 <= A[i] <= 30000
  2. 1 <= A.length <= 30000

Solution in C++:


class Solution {
public:
    int maxSubarraySumCircular(vector<int>& A) {
        int nocircle= help(A), sum= 0;
        if(nocircle < 0)
            return nocircle;
        
        vector<int> tmp= A;
        for(auto &n: tmp){
            sum+= n;
            n*= -1;
        }   
        
        int circle= sum+ help(tmp);
        return max(circle, nocircle);
    }
    int help(vector<int>& A){
        int max_endhere= 0, max_sofar= INT_MIN;
        for(int i= 0; i< A.size(); i++){
            if(max_endhere < 0)
                max_endhere= A[i];
            else
                max_endhere+= A[i];
            
            max_sofar= max(max_sofar, max_endhere);
        }
        return max_sofar;
    }
};

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