LeetCode-in-Java
1031. Maximum Sum of Two Non-Overlapping Subarrays
Medium
Given an integer array nums and two integers firstLen and secondLen, return the maximum sum of elements in two non-overlapping subarrays with lengths firstLen and secondLen.
The array with length firstLen could occur before or after the array with length secondLen, but they have to be non-overlapping.
A subarray is a contiguous part of an array.
Example 1:
Input: nums = [0,6,5,2,2,5,1,9,4], firstLen = 1, secondLen = 2
Output: 20
Explanation: One choice of subarrays is [9] with length 1, and [6,5] with length 2.
Example 2:
Input: nums = [3,8,1,3,2,1,8,9,0], firstLen = 3, secondLen = 2
Output: 29
Explanation: One choice of subarrays is [3,8,1] with length 3, and [8,9] with length 2.
Example 3:
Input: nums = [2,1,5,6,0,9,5,0,3,8], firstLen = 4, secondLen = 3
Output: 31
Explanation: One choice of subarrays is [5,6,0,9] with length 4, and [3,8] with length 3.
Constraints:
1 <= firstLen, secondLen <= 10002 <= firstLen + secondLen <= 1000firstLen + secondLen <= nums.length <= 10000 <= nums[i] <= 1000
Solution
public class Solution {
public int maxSumTwoNoOverlap(int[] nums, int f, int s) {
int sum = 0;
int n = nums.length;
int[] pref = new int[n];
int[] suff = new int[n];
for (int i = 0; i < n; i++) {
sum += nums[i];
if (i < f - 1) {
continue;
}
pref[i] = Math.max(i > 0 ? pref[i - 1] : 0, sum);
sum -= nums[i + 1 - f];
}
sum = 0;
for (int i = n - 1; i >= 0; i--) {
sum += nums[i];
if (i > n - f) {
continue;
}
suff[i] = Math.max(i < n - 1 ? suff[i + 1] : 0, sum);
sum -= nums[i + f - 1];
}
sum = 0;
for (int i = 0; i < s - 1; i++) {
sum += nums[i];
}
int ans = Integer.MIN_VALUE;
for (int i = s - 1; i < n; i++) {
sum += nums[i];
if (i >= s) {
ans = Math.max(ans, pref[i - s] + sum);
}
if (i < n - 1) {
ans = Math.max(ans, suff[i + 1] + sum);
}
sum -= nums[i + 1 - s];
}
return ans;
}
}