1031. Maximum Sum of Two Non-Overlapping Subarrays

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

int maxSumTwoNoOverlap(vector<int>& A, int L, int M) { // time: O(n); space: O(n)
    int n = A.size();
    vector<int> sums(n + 1, 0); // accumulated sum array
    for (int i = 1; i <= n; ++i) sums[i] = sums[i - 1] + A[i - 1];
    // L_M: max sum of L continuous elements before the last M elements
    // M_L: max sum of M continuous elements before the last L elements
    int res = sums[L + M], L_M = sums[L], M_L = sums[M];
    for (int i = L + M + 1; i <= n; ++i) {
        L_M = max(L_M, sums[i - M] - sums[i - L - M]);
        M_L = max(M_L, sums[i - L] - sums[i - L - M]);
        res = max({res, L_M + sums[i] - sums[i - M], M_L + sums[i] - sums[i - L]});
    }
    return res;
}