MaximumSumCircullarSubarray.java

#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

#Hint - 1
#For those of you who are familiar with the Kadane's algorithm, think in terms of that. For the newbies, Kadane's algorithm is used to finding the maximum sum subarray from a given array. This problem is a twist on that idea and it is advisable to read up on that algorithm first before starting this problem. Unless you already have a great algorithm brewing up in your mind in which case, go right ahead!

#Hint - 2 
#What is an alternate way of representing a circular array so that it appears to be a straight array? Essentially, there are two cases of this problem that we need to take care of. Let's look at the figure below to understand those two cases: 

#Hint - 3
#The first case can be handled by the good old Kadane's algorithm. However, is there a smarter way of going about handling the second case as well?

class Solution {
    public boolean chkNegative(int[] A, int n)
    {
        for(int i=0; i<n; i++)
            if(A[i]>=0)
                return false;
        return true;
    }
    public int maxSubarraySumCircular(int[] A) {
        int maxSum = Integer.MIN_VALUE, invertedSum = maxSum;
        int sum = 0,wrap=0;
        int n = A.length;
        if(chkNegative(A,n))
        {
            for(int i=0; i<n; i++)
            {
               if(A[i]>maxSum)
                   maxSum=A[i];
            }
            return maxSum;
        }
        for(int i = 0; i<n; i++)
        {  
            sum+=A[i];
            wrap+=A[i];
            A[i]=-A[i];
            if(sum>maxSum)
                maxSum=sum;
            if(sum<0)
                sum=0;
        }
        sum=0;
        for(int i = 0; i<n; i++)
        {  
            sum+=A[i];
            if(sum>invertedSum)
                invertedSum=sum;
            if(sum<0)
                sum=0;   
        }
        return maxSum>wrap+invertedSum? maxSum : wrap+invertedSum;
    }
}