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:
Explanation
Lsum, sum of the last L elements Msum, sum of the last M elements
Lmax, max sum of contiguous L elements before the last M elements. Mmax, max sum of contiguous M elements before the last L elements/
class Solution {
public int maxSumTwoNoOverlap(int[] A, int L, int M) {
int res = 0, Lsum = 0, Lmax = 0, Msum = 0, Mmax = 0;
for (int i = 0; i < A.length; ++i) {
Msum += A[i];
if (i - M >= 0) Msum -= A[i - M];
if (i - M >= 0) Lsum += A[i - M];
if (i - M - L >= 0) Lsum -= A[i - L - M];
Lmax = Math.max(Lmax, Lsum);
res = Math.max(res, Lmax + Msum);
}
Lsum = Lmax = Msum = Mmax = 0;
for (int i = 0; i < A.length; ++i) {
Lsum += A[i];
if (i - L >= 0) Lsum -= A[i - L];
if (i - L >= 0) Msum += A[i - L];
if (i - M - L >= 0) Msum -= A[i - L - M];
Mmax = Math.max(Mmax, Msum);
res = Math.max(res, Mmax + Lsum);
}
return res;
}
}