# 689. Maximum Sum of 3 Non-Overlapping Subarrays

## Description

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 interval (0-indexed). If there are multiple answers, return the lexicographically smallest one.

Example 1:

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:

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


Constraints:

• 1 <= nums.length <= 2 * 104
• 1 <= nums[i] < 216
• 1 <= k <= floor(nums.length / 3)

## Solutions

Solution 1: Sliding Window

We use a sliding window to enumerate the position of the third subarray, while maintaining the maximum sum and its position of the first two non-overlapping subarrays.

The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.

Solution 2: Preprocessing Prefix and Suffix + Enumerating Middle Subarray

We can preprocess to get the prefix sum array $s$ of the array $nums$, where $s[i] = \sum_{j=0}^{i-1} nums[j]$. Then for any $i$, $j$, $s[j] - s[i]$ is the sum of the subarray $[i, j)$.

Next, we use dynamic programming to maintain two arrays $pre$ and $suf$ of length $n$, where $pre[i]$ represents the maximum sum and its starting position of the subarray of length $k$ within the range $[0, i]$, and $suf[i]$ represents the maximum sum and its starting position of the subarray of length $k$ within the range $[i, n)$.

Then, we enumerate the starting position $i$ of the middle subarray. The sum of the three subarrays is $pre[i-1][0] + suf[i+k][0] + (s[i+k] - s[i])$, where $pre[i-1][0]$ represents the maximum sum of the subarray of length $k$ within the range $[0, i-1]$, $suf[i+k][0]$ represents the maximum sum of the subarray of length $k$ within the range $[i+k, n)$, and $(s[i+k] - s[i])$ represents the sum of the subarray of length $k$ within the range $[i, i+k)$. We find the starting positions of the three subarrays corresponding to the maximum sum.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

• class Solution {
public int[] maxSumOfThreeSubarrays(int[] nums, int k) {
int n = nums.length;
int[] s = new int[n + 1];
for (int i = 0; i < n; ++i) {
s[i + 1] = s[i] + nums[i];
}
int[][] pre = new int[n][0];
int[][] suf = new int[n][0];
for (int i = 0, t = 0, idx = 0; i < n - k + 1; ++i) {
int cur = s[i + k] - s[i];
if (cur > t) {
pre[i + k - 1] = new int[] {cur, i};
t = cur;
idx = i;
} else {
pre[i + k - 1] = new int[] {t, idx};
}
}
for (int i = n - k, t = 0, idx = 0; i >= 0; --i) {
int cur = s[i + k] - s[i];
if (cur >= t) {
suf[i] = new int[] {cur, i};
t = cur;
idx = i;
} else {
suf[i] = new int[] {t, idx};
}
}
int[] ans = new int[0];
for (int i = k, t = 0; i < n - 2 * k + 1; ++i) {
int cur = s[i + k] - s[i] + pre[i - 1][0] + suf[i + k][0];
if (cur > t) {
ans = new int[] {pre[i - 1][1], i, suf[i + k][1]};
t = cur;
}
}
return ans;
}
}

• class Solution {
public:
vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
int n = nums.size();
vector<int> s(n + 1, 0);
for (int i = 0; i < n; ++i) {
s[i + 1] = s[i] + nums[i];
}

vector<vector<int>> pre(n, vector<int>(2, 0));
vector<vector<int>> suf(n, vector<int>(2, 0));

for (int i = 0, t = 0, idx = 0; i < n - k + 1; ++i) {
int cur = s[i + k] - s[i];
if (cur > t) {
pre[i + k - 1] = {cur, i};
t = cur;
idx = i;
} else {
pre[i + k - 1] = {t, idx};
}
}

for (int i = n - k, t = 0, idx = 0; i >= 0; --i) {
int cur = s[i + k] - s[i];
if (cur >= t) {
suf[i] = {cur, i};
t = cur;
idx = i;
} else {
suf[i] = {t, idx};
}
}

vector<int> ans;
for (int i = k, t = 0; i < n - 2 * k + 1; ++i) {
int cur = s[i + k] - s[i] + pre[i - 1][0] + suf[i + k][0];
if (cur > t) {
ans = {pre[i - 1][1], i, suf[i + k][1]};
t = cur;
}
}

return ans;
}
};

• class Solution:
def maxSumOfThreeSubarrays(self, nums: List[int], k: int) -> List[int]:
n = len(nums)
s = list(accumulate(nums, initial=0))
pre = [[] for _ in range(n)]
suf = [[] for _ in range(n)]
t = idx = 0
for i in range(n - k + 1):
if (cur := s[i + k] - s[i]) > t:
pre[i + k - 1] = [cur, i]
t, idx = pre[i + k - 1]
else:
pre[i + k - 1] = [t, idx]
t = idx = 0
for i in range(n - k, -1, -1):
if (cur := s[i + k] - s[i]) >= t:
suf[i] = [cur, i]
t, idx = suf[i]
else:
suf[i] = [t, idx]
t = 0
ans = []
for i in range(k, n - 2 * k + 1):
if (cur := s[i + k] - s[i] + pre[i - 1][0] + suf[i + k][0]) > t:
ans = [pre[i - 1][1], i, suf[i + k][1]]
t = cur
return ans


• func maxSumOfThreeSubarrays(nums []int, k int) (ans []int) {
n := len(nums)
s := make([]int, n+1)
for i := 0; i < n; i++ {
s[i+1] = s[i] + nums[i]
}

pre := make([][]int, n)
suf := make([][]int, n)

for i, t, idx := 0, 0, 0; i < n-k+1; i++ {
cur := s[i+k] - s[i]
if cur > t {
pre[i+k-1] = []int{cur, i}
t, idx = cur, i
} else {
pre[i+k-1] = []int{t, idx}
}
}

for i, t, idx := n-k, 0, 0; i >= 0; i-- {
cur := s[i+k] - s[i]
if cur >= t {
suf[i] = []int{cur, i}
t, idx = cur, i
} else {
suf[i] = []int{t, idx}
}
}

for i, t := k, 0; i < n-2*k+1; i++ {
cur := s[i+k] - s[i] + pre[i-1][0] + suf[i+k][0]
if cur > t {
ans = []int{pre[i-1][1], i, suf[i+k][1]}
t = cur
}
}

return
}

• function maxSumOfThreeSubarrays(nums: number[], k: number): number[] {
const n: number = nums.length;
const s: number[] = Array(n + 1).fill(0);

for (let i = 0; i < n; ++i) {
s[i + 1] = s[i] + nums[i];
}

const pre: number[][] = Array(n)
.fill([])
.map(() => new Array(2).fill(0));
const suf: number[][] = Array(n)
.fill([])
.map(() => new Array(2).fill(0));

for (let i = 0, t = 0, idx = 0; i < n - k + 1; ++i) {
const cur: number = s[i + k] - s[i];
if (cur > t) {
pre[i + k - 1] = [cur, i];
t = cur;
idx = i;
} else {
pre[i + k - 1] = [t, idx];
}
}

for (let i = n - k, t = 0, idx = 0; i >= 0; --i) {
const cur: number = s[i + k] - s[i];
if (cur >= t) {
suf[i] = [cur, i];
t = cur;
idx = i;
} else {
suf[i] = [t, idx];
}
}

let ans: number[] = [];
for (let i = k, t = 0; i < n - 2 * k + 1; ++i) {
const cur: number = s[i + k] - s[i] + pre[i - 1][0] + suf[i + k][0];
if (cur > t) {
ans = [pre[i - 1][1], i, suf[i + k][1]];
t = cur;
}
}

return ans;
}