Algorithm
Problem Name: 1031. Maximum Sum of Two Non-Overlapping Subarrays
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 <= 10002 <= firstLen + secondLen <= 1000firstLen + secondLen <= nums.length <= 10000 <= nums[i] <= 1000
Code Examples
#1 Code Example with Javascript Programming
Code -
Javascript Programming
const maxSumTwoNoOverlap = function(A, L, M) {
for(let i = 1, len = A.length; i < len; i++) {
A[i] += A[i - 1]
}
let LMax = A[L - 1], MMax = A[M - 1], res = A[L + M - 1]
for(let i = L + M, len = A.length; i < len; i++) {
LMax = Math.max(LMax, A[i - M] - A[i - M - L])
MMax = Math.max(MMax, A[i - L] - A[i - M - L])
res = Math.max(res, Math.max(LMax + A[i] - A[i - M], MMax + A[i] - A[i - L]))
}
return res
}
Input
nums = [0,6,5,2,2,5,1,9,4], firstLen = 1, secondLen = 2
Output
20
#2 Code Example with Python Programming
Code -
Python Programming
class Solution:
def maxSumTwoNoOverlap(self, A: List[int], L: int, M: int) -> int:
n = len(A)
l = [0] * n
r = [0] * n
sm = 0
for i in range(M - 1):
sm += A[i]
for j in range(n - M + 1):
sm += A[j + M - 1]
for i in range(j + 1):
r[i] = max(r[i], sm)
sm -= A[j]
sm = 0
for i in range(n - 1, n - M, -1):
sm += A[i]
for i in range(n - 1, M - 2, -1):
sm += A[i - M + 1]
for j in range(i + 1, n):
l[j] = max(l[j], sm)
sm -= A[i]
sm = 0
for i in range(L - 1):
sm += A[i]
res = 0
for j in range(n - L + 1):
sm += A[j + L - 1]
if j >= M:
res = max(res, sm + l[j - 1])
if j + L < n:
res = max(res, sm + r[j + L])
sm -= A[j]
return res
Input
nums = [3,8,1,3,2,1,8,9,0], firstLen = 3, secondLen = 2
Output
29