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:
L >= 1
M >= 1
L + M <= A.length <= 1000
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;
}