Maximum sum of two non overlapping subarrays

Time O(N); Space: O(1); medium

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.

Notes:

  • L >= 1

  • M >= 1

  • L + M <= len(A) <= 1000

  • 0 <= A[i] <= 1000

[1]:

class Solution1(object):
    def maxSumTwoNoOverlap(self, A, L, M):
        """
        :type A: List[int]
        :type L: int
        :type M: int
        :rtype: int
        """
        for i in range(1, len(A)):
            A[i] += A[i-1]

        result, L_max, M_max = A[L+M-1], A[L-1], A[M-1]
        for i in range(L+M, len(A)):
            L_max = max(L_max, A[i-M] - A[i-L-M])
            M_max = max(M_max, A[i-L] - A[i-L-M])
            result = max(result,
                         L_max + A[i] - A[i-M],
                         M_max + A[i] - A[i-L])
        return result

[3]:

s = Solution1()
A = [0, 6, 5, 2, 2, 5, 1, 9, 4]
L = 1
M = 2
assert s.maxSumTwoNoOverlap(A, L, M) == 20
A = [3, 8, 1, 3, 2, 1, 8, 9, 0]
L = 3
M = 2
assert s.maxSumTwoNoOverlap(A, L, M) == 29
A = [2, 1, 5, 6, 0, 9, 5, 0, 3, 8]
L = 4
M = 3
assert s.maxSumTwoNoOverlap(A, L, M) == 31