Description

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:

1. L >= 1
2. M >= 1
3. L + M <= A.length <= 1000
4. 0 <= A[i] <= 1000

Solutions

1. Array

第一次做还是比较棘手的，比较多数组的操作！

# Time: O(n)
# Space: O(1)
class Solution:
def maxSumTwoNoOverlap(self, A: List[int], L: int, M: int) -> int:
res = Lsum = Lmax = Msum = Mmax = 0
n = len(A)
# L left, M right
for i in range(n):
# add number value to M part
Msum += A[i]
# Keep M numbers
if i - M >= 0:
Msum -= A[i - M]
# add number value to L part
if i - M >= 0:
Lsum += A[i - M]
# Keep L numbers
if i - M - L >= 0:
Lsum -= A[i - M - L]
# L find max, M move forward
Lmax = max(Lmax, Lsum)
res = max(res, Lmax + Msum)
Lsum = Lmax = Msum = Mmax = 0
# M left, L right
for i in range(n):
# add number to L part
Lsum += A[i]
# keep L numbers
if i - L >= 0:
Lsum -= A[i - L]
if i - L >= 0:
Msum += A[i - L]
if i - L - M >= 0:
Msum -= A[i - L - M]
Mmax = max(Mmax, Msum)
res = max(res, Mmax + Lsum)
return res
# 51/51 cases passed (72 ms)
# Your runtime beats 37.21 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.7 MB)