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1031. Maximum Sum of Two Non-Overlapping Subarrays

Difficulty Topics

Description

Given an integer array nums and two integers firstLen and secondLen, return the maximum sum of elements in two non-overlapping subarrays with lengths firstLen and secondLen.

The array with length firstLen could occur before or after the array with length secondLen, but they have to be non-overlapping.

A subarray is a contiguous part of an array.

 

Example 1:

Input: nums = [0,6,5,2,2,5,1,9,4], firstLen = 1, secondLen = 2
Output: 20
Explanation: One choice of subarrays is [9] with length 1, and [6,5] with length 2.

Example 2:

Input: nums = [3,8,1,3,2,1,8,9,0], firstLen = 3, secondLen = 2
Output: 29
Explanation: One choice of subarrays is [3,8,1] with length 3, and [8,9] with length 2.

Example 3:

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

 

Constraints:

  • 1 <= firstLen, secondLen <= 1000
  • 2 <= firstLen + secondLen <= 1000
  • firstLen + secondLen <= nums.length <= 1000
  • 0 <= nums[i] <= 1000

Solution

maximum-sum-of-two-non-overlapping-subarrays.py
class Solution:
    def maxSumTwoNoOverlap(self, nums: List[int], firstLen: int, secondLen: int) -> int:
        n = len(nums)
        prefix = [0] + list(accumulate(nums))
        res = 0

        for i in range(firstLen, n + 1):
            first = prefix[i] - prefix[i - firstLen]
            second = 0

            for j in range(secondLen, i - firstLen + 1):
                second = max(second, prefix[j] - prefix[j - secondLen])

            for j in range(secondLen + i + 1, n + 1):
                second = max(second, prefix[j] - prefix[j - secondLen])

            res = max(res, first + second)

        return res