## 689. Maximum Sum of 3 Non-Overlapping Subarrays

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).

## Rust Solution

``````struct Solution;

impl Solution {
fn max_sum_of_three_subarrays(nums: Vec<i32>, k: i32) -> Vec<i32> {
let k = k as usize;
let sums: Vec<i32> = nums.windows(k).map(|w| w.iter().sum()).collect();
let n = sums.len();
let mut left = vec![];
let mut left_max = 0;
let mut left_index = 0;
for i in 0..n {
if sums[i] > left_max {
left_max = sums[i];
left_index = i;
}
left.push((left_max, left_index));
}

let mut right = vec![];
let mut right_max = 0;
let mut right_index = n;
for i in (0..n).rev() {
if sums[i] >= right_max {
right_max = sums[i];
right_index = i;
}
right.push((right_max, right_index));
}
right.reverse();
let mut mid_max = 0;
let mut res = vec![0, 0, 0];
for i in k..n - k {
let sum_3 = sums[i] + left[i - k].0 + right[i + k].0;
if sum_3 > mid_max {
mid_max = sum_3;
res = vec![left[i - k].1, i, right[i + k].1];
}
}
res.into_iter().map(|x| x as i32).collect()
}
}

#[test]
fn test() {
let nums = vec![1, 2, 1, 2, 6, 7, 5, 1];
let k = 2;
let res = vec![0, 3, 5];
assert_eq!(Solution::max_sum_of_three_subarrays(nums, k), res);
}
``````

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