Day-15_Maximum_Sum_Circular_Subarray.py

'''

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
   Hide 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!
   Hide 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:

   Hide 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:
    def maxSubarraySumCircular(self, A: List[int]) -> int:
        
        array_sum = 0
        
        local_min_sum, global_min_sum = 0, float('inf')
        local_max_sum, global_max_sum = 0, float('-inf')
        
        for number in A:
            
            local_min_sum = min( local_min_sum + number, number )
            global_min_sum = min( global_min_sum, local_min_sum )
            
            local_max_sum = max( local_max_sum + number, number )
            global_max_sum = max( global_max_sum, local_max_sum )
            
            array_sum += number
        
        
        
        # global_max_sum denotes the maximum subarray sum without crossing boundary
        # arry_sum - global_min_sum denotes the maximum subarray sum with crossing boundary
        
        if global_max_sum > 0:
            return max( array_sum - global_min_sum, global_max_sum )
        else:
            # corner case handle for all number are negative
            return global_max_sum