# 689 Maximum Sum of 3 Non-Overlapping Subarrays

### Problem

In a given array nums of positive integers, find three non-overlapping subarrays with maximum sum.

Each subarray will be of size k, and we want to maximize the sum of all 3*k entries.

Return the result as a list of indices representing the starting position of each interval (0-indexed). If there are multiple answers, return the lexicographically smallest one.

Example:

``````Input: [1,2,1,2,6,7,5,1], 2
Output: [0, 3, 5]
Explanation: Subarrays [1, 2], [2, 6], [7, 5] correspond to the starting indices [0, 3, 5].
We could have also taken [2, 1], but an answer of [1, 3, 5] would be lexicographically larger.
``````

Note: nums.length will be between 1 and 20000. nums[i] will be between 1 and 65535. k will be between 1 and floor(nums.length / 3).

### Solutions

``````class Solution {
public int[] maxSumOfThreeSubarrays(int[] nums, int k) {
int[] res = new int[3];
int[] sums = new int[nums.length + 1];
int sum = 0;
for (int i = 0; i < nums.length; i ++) {
sum += nums[i];
sums[i + 1] = sum;
}
int[][] rsums = new int[nums.length][3];
int[][] rindexes = new int[nums.length][3];
for (int i = 0; i < 3; i ++) {
for (int j = i * k; j + k - 1 < nums.length; j ++) {
int lastRound = 0;
if (i > 0 && j - 1 >= 0) {
lastRound = rsums[j - 1][i - 1];
}
int cand = getSum(sums, j, j + k - 1) + lastRound;
int pre = 0;
if (j + k - 2 >= 0) {
pre = rsums[j + k - 2][i];
}
if (cand > pre) {
rsums[j + k - 1][i] = cand;
rindexes[j + k - 1][i] = j;
}
else {
rsums[j + k - 1][i] = pre;
rindexes[j + k - 1][i] = rindexes[j + k - 2][i];
}
}
}
int tmp = rindexes[nums.length - 1][2];
res[2] = tmp;
tmp = rindexes[tmp - 1][1];
res[1] = tmp;
tmp = rindexes[tmp - 1][0];
res[0] = tmp;
return res;

}
private int getSum(int[] sums, int start, int end) {
return sums[end + 1] - sums[start];
}
}
``````