# 1031. Maximum Sum of Two Non-Overlapping Subarrays

## 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

## Solutions

• class Solution {
public int maxSumTwoNoOverlap(int[] nums, int firstLen, int secondLen) {
int n = nums.length;
int[] s = new int[n + 1];
for (int i = 0; i < n; ++i) {
s[i + 1] = s[i] + nums[i];
}
int ans = 0;
for (int i = firstLen, t = 0; i + secondLen - 1 < n; ++i) {
t = Math.max(t, s[i] - s[i - firstLen]);
ans = Math.max(ans, t + s[i + secondLen] - s[i]);
}
for (int i = secondLen, t = 0; i + firstLen - 1 < n; ++i) {
t = Math.max(t, s[i] - s[i - secondLen]);
ans = Math.max(ans, t + s[i + firstLen] - s[i]);
}
return ans;
}
}

• class Solution {
public:
int maxSumTwoNoOverlap(vector<int>& nums, int firstLen, int secondLen) {
int n = nums.size();
vector<int> s(n + 1);
for (int i = 0; i < n; ++i) {
s[i + 1] = s[i] + nums[i];
}
int ans = 0;
for (int i = firstLen, t = 0; i + secondLen - 1 < n; ++i) {
t = max(t, s[i] - s[i - firstLen]);
ans = max(ans, t + s[i + secondLen] - s[i]);
}
for (int i = secondLen, t = 0; i + firstLen - 1 < n; ++i) {
t = max(t, s[i] - s[i - secondLen]);
ans = max(ans, t + s[i + firstLen] - s[i]);
}
return ans;
}
};

• class Solution:
def maxSumTwoNoOverlap(self, nums: List[int], firstLen: int, secondLen: int) -> int:
n = len(nums)
s = list(accumulate(nums, initial=0))
ans = t = 0
i = firstLen
while i + secondLen - 1 < n:
t = max(t, s[i] - s[i - firstLen])
ans = max(ans, t + s[i + secondLen] - s[i])
i += 1
t = 0
i = secondLen
while i + firstLen - 1 < n:
t = max(t, s[i] - s[i - secondLen])
ans = max(ans, t + s[i + firstLen] - s[i])
i += 1
return ans


• func maxSumTwoNoOverlap(nums []int, firstLen int, secondLen int) (ans int) {
n := len(nums)
s := make([]int, n+1)
for i, x := range nums {
s[i+1] = s[i] + x
}
for i, t := firstLen, 0; i+secondLen-1 < n; i++ {
t = max(t, s[i]-s[i-firstLen])
ans = max(ans, t+s[i+secondLen]-s[i])
}
for i, t := secondLen, 0; i+firstLen-1 < n; i++ {
t = max(t, s[i]-s[i-secondLen])
ans = max(ans, t+s[i+firstLen]-s[i])
}
return
}