human
Solve in Python:
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

Give me a the Python program you requested?
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#Here is the Python program you requested.

from functools import reduce

class Solution:
        
    
    def maxSumTwoNoOverlap(self, A: List[int], L: int, M: int) -> int:
        
        lsum = []
               
        csum = reduce(lambda a,b:a+b, A[:L]) 
        lsum.append(csum)
        
        for i in range(L, len(A)):
            csum += A[i] - A[i-L]
            lsum.append(csum)
            
        msum = []
        csum = reduce(lambda a,b:a+b, A[:M])
            
        msum.append(csum)
        
        for i in range(M, len(A)):
            csum += A[i] - A[i-M]
            msum.append(csum)
            
        maxSum = 0
        for i in range(len(A)-L+1):
            ls = i
            le = i + L - 1
            
            for j in range(len(A) - M+1):
                ms = j
                me = j + M - 1
                
                if (le < ms) or (me < ls):
                    maxSum = max(maxSum, lsum[ls] + msum[ms])
                    
        return maxSum