Maximum Sum of Two Non-Overlapping Subarrays

LeetCode Q 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

Solution :

We divide this problem into two cases. One is L is in the left of M, another is L is in the right of M. For each case, we find its maximum sum.

Code:

public int maxSumTwoNoOverlap(int[] A, int L, int M) {
	return Math.max(maxSumHelper(A, L, M), maxSumHelper(A, M, L));
}

private int maxSumHelper(int[] A, int L, int M) {
	int Lsum = 0, Msum =0, Lmax = 0, Mmax = 0, res = 0;

	for (int i = 0; i < A.length; i++) {
		Msum += A[i];
		if (i >= M) Msum -= A[i - M];
		if (i >= M) Lsum += A[i - M];
		if (i >= M + L) Lsum -= A[i - L - M];

		Lmax = Math.max(Lmax, Lsum);
		res = Math.max(res, Lmax + Msum);
	}

	return res;
}