# Maximum Sum Of 3 Non-Overlapping Subarrays Problem

## Description

LeetCode Problem 689.

Given an integer array nums and an integer k, find three non-overlapping subarrays of length k with maximum sum and return them.

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

Example 1:

``````1
2
3
4
Input: nums = [1,2,1,2,6,7,5,1], k = 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.
``````

Example 2:

``````1
2
Input: nums = [1,2,1,2,1,2,1,2,1], k = 2
Output: [0,2,4]
``````

Constraints:

• 1 <= nums.length <= 2 * 10^4
• 1 <= nums[i] <2^16
• 1 <= k <= floor(nums.length / 3)

## Sample C++ Code

``````1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
class Solution {
public:
vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
int n = nums.size();
vector<int> prefixSum(n+1, 0);

for (int i = 1; i <= n; i++) {
prefixSum[i] = prefixSum[i-1] + nums[i-1];
}

vector<int> left(n+1, 0);
vector<int> leftInd(n+1, 0);
vector<int> right(n+1, 0);
vector<int> rightInd(n+1, 0);

for (int i = k; i <= n; i++) {
left[i] = max(left[i-1], prefixSum[i] - prefixSum[i-k]);
if (left[i] == left[i-1])
leftInd[i] = leftInd[i-1];
else
leftInd[i] = i;
}

for (int i = n-k; i >= 0; i--) {
right[i] = max(right[i+1], prefixSum[i+k] - prefixSum[i]);
if (right[i] == prefixSum[i+k] - prefixSum[i])
rightInd[i] = i;
else
rightInd[i] = rightInd[i+1];
}

int maxSum = 0, a, b, c;

for (int i = k; i <= n - 2 * k; i++) {
if (maxSum < left[i] + (prefixSum[i+k] - prefixSum[i]) + right[i+k]) {
maxSum = left[i] + (prefixSum[i+k] - prefixSum[i]) + right[i+k];
a = leftInd[i] - k;
b = i;
c = rightInd[i+k];
}
}

return {a, b, c};
}
};
``````