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

class Solution {
    public int maxSumTwoNoOverlap(int[] A, int L, int M) {
        if (A == null || A.length == 0) {
            return 0;
        }
        int n = A.length;
        int[] preSum = new int[n + 1];
        for (int i = 0; i < n; i++)
            preSum[i + 1] = A[i] + preSum[i];
        int lMax = preSum[L], mMax = preSum[M];
        int res = preSum[L + M];
        for (int i = L + M; i <= n; i++) {
            // case 1: L subarray is always before M subarray
            lMax = Math.max(lMax, preSum[i - M] - preSum[i - M - L]);
            // case 2: M subarray is always before L subarray
            mMax = Math.max(mMax, preSum[i - L] - preSum[i - M - L]);
            // compare two cases and update res
            res = Math.max(res, Math.max(lMax + preSum[i] - preSum[i - M], mMax + preSum[i] - preSum[i - L]));
        }
        return res;
    }
}