1031.py

'''
Given an array A of non-negative integers, return the maximum sum of elements in two non-overlapping (contiguous) subarrays, which have lengths L and M.  (For clarification, the L-length subarray could occur before or after the M-length subarray.)

Formally, return the largest V for which V = (A[i] + A[i+1] + ... + A[i+L-1]) + (A[j] + A[j+1] + ... + A[j+M-1]) and either:

0 <= i < i + L - 1 < j < j + M - 1 < A.length, or
0 <= j < j + M - 1 < i < i + L - 1 < A.length.
 

Example 1:

Input: A = [0,6,5,2,2,5,1,9,4], L = 1, M = 2
Output: 20
Explanation: One choice of subarrays is [9] with length 1, and [6,5] with length 2.
Example 2:

Input: A = [3,8,1,3,2,1,8,9,0], L = 3, M = 2
Output: 29
Explanation: One choice of subarrays is [3,8,1] with length 3, and [8,9] with length 2.
Example 3:

Input: A = [2,1,5,6,0,9,5,0,3,8], L = 4, M = 3
Output: 31
Explanation: One choice of subarrays is [5,6,0,9] with length 4, and [3,8] with length 3.
 

Note:

L >= 1
M >= 1
L + M <= A.length <= 1000
0 <= A[i] <= 1000
'''

class Solution(object):
    def maxSumTwoNoOverlap(self, A, L, M):
        """
        :type A: List[int]
        :type L: int
        :type M: int
        :rtype: int
        """
        cumm_sum = [0]
        for index in range(len(A)):
            cumm_sum.append(cumm_sum[index]+A[index])
        result = 0
        
        def valid(index_i, index_j):
            return index_i+L <=len(A) and index_j+M <= len(A) and(index_j>=index_i+L or index_i>=index_j+M)
    
        for index_i in range(len(A)):
            for index_j in range(len(A)):
                if valid(index_i, index_j):
                    result = max(result, cumm_sum[index_i+L]-cumm_sum[index_i] + cumm_sum[index_j+M]-cumm_sum[index_j])
        return result