##### Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/1031.html

# 1031. Maximum Sum of Two Non-Overlapping Subarrays (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.


Note:

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

Related Topics:
Array

## Solution 1.

• class Solution {
public int maxSumTwoNoOverlap(int[] A, int L, int M) {
int length = A.length;
if (L == M) {
int[] rightMaxSums = new int[length - M + 1];
int rightSum = 0;
for (int i = 0; i < M; i++)
rightSum += A[length - i - 1];
rightMaxSums[length - M] = rightSum;
for (int i = length - M - 1; i >= L; i--) {
rightSum -= A[i + M];
rightSum += A[i];
rightMaxSums[i] = Math.max(rightMaxSums[i + 1], rightSum);
}
int maxSum = 0;
int leftSum = 0;
for (int i = 0; i < L; i++)
leftSum += A[i];
maxSum = leftSum + rightMaxSums[L];
int end = length - M;
for (int i = L; i < end; i++) {
leftSum -= A[i - L];
leftSum += A[i];
maxSum = Math.max(maxSum, leftSum + rightMaxSums[i + 1]);
}
return maxSum;
} else {
int[] rightMaxSums1 = new int[length - M + 1];
int rightSum1 = 0;
for (int i = 0; i < M; i++)
rightSum1 += A[length - i - 1];
rightMaxSums1[length - M] = rightSum1;
for (int i = length - M - 1; i >= L; i--) {
rightSum1 -= A[i + M];
rightSum1 += A[i];
rightMaxSums1[i] = Math.max(rightMaxSums1[i + 1], rightSum1);
}
int maxSum1 = 0;
int leftSum1 = 0;
for (int i = 0; i < L; i++)
leftSum1 += A[i];
maxSum1 = leftSum1 + rightMaxSums1[L];
int end1 = length - M;
for (int i = L; i < end1; i++) {
leftSum1 -= A[i - L];
leftSum1 += A[i];
maxSum1 = Math.max(maxSum1, leftSum1 + rightMaxSums1[i + 1]);
}
int[] rightMaxSums2 = new int[length - L + 1];
int rightSum2 = 0;
for (int i = 0; i < L; i++)
rightSum2 += A[length - i - 1];
rightMaxSums2[length - L] = rightSum2;
for (int i = length - L - 1; i >= M; i--) {
rightSum2 -= A[i + L];
rightSum2 += A[i];
rightMaxSums2[i] = Math.max(rightMaxSums2[i + 1], rightSum2);
}
int maxSum2 = 0;
int leftSum2 = 0;
for (int i = 0; i < M; i++)
leftSum2 += A[i];
maxSum2 = leftSum2 + rightMaxSums2[M];
int end2 = length - L;
for (int i = M; i < end2; i++) {
leftSum2 -= A[i - M];
leftSum2 += A[i];
maxSum2 = Math.max(maxSum2, leftSum2 + rightMaxSums2[i + 1]);
}
return Math.max(maxSum1, maxSum2);
}
}
}

############

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;
}
}

• // OJ: https://leetcode.com/problems/maximum-sum-of-two-non-overlapping-subarrays/
// Time: O(N)
// Space: O(N)
class Solution {
private:
int get(vector<int> &v, int i) {
return (i >= 0 && i < v.size()) ? v[i] : 0;
}
public:
int maxSumTwoNoOverlap(vector<int>& A, int L, int M) {
int N = A.size(), ans = 0;
partial_sum(A.begin(), A.end(), A.begin());
vector<int> maxLeft(N, 0), maxRight(N, 0);
for (int i = L - 1; i < N; ++i) maxLeft[i] = max(get(maxLeft, i - 1), A[i] - get(A, i - L));
for (int i = N - L; i >= 0; --i) maxRight[i] = max(get(maxRight, i + 1), A[i + L - 1] - get(A, i - 1));
for (int i = M - 1; i < N; ++i) {
int sum = A[i] - get(A, i - M)
+ max(get(maxLeft, i - M), get(maxRight, i + 1));
ans = max(ans, sum);
}
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

############

# 1031. Maximum Sum of Two Non-Overlapping Subarrays
# https://leetcode.com/problems/maximum-sum-of-two-non-overlapping-subarrays/

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


• 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
}

func max(a, b int) int {
if a > b {
return a
}
return b
}