LeetCode [1031] Maximum Sum of Two Non-Overlapping Subarrays

 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

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class Solution { public int maxSumTwoNoOverlap(int[] A, int L, int M) { int n = A.length; for(int i=1; i<n; ++i) A[i] += A[i-1]; int ret = A[L+M-1], Lmax = A[L-1], Mmax = A[M-1]; for(int i=L+M; i<n; ++i) { Lmax = Math.max(Lmax, A[i-M] - A[i-L-M]); Mmax = Math.max(Mmax, A[i-L] - A[i-L-M]); ret = Math.max(ret, Math.max(Lmax + A[i] - A[i-M], Mmax + A[i] - A[i-L])); } return ret; } }