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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 $O(1)$.

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class Solution:
    def maxSumOfThreeSubarrays(self, nums: List[int], k: int) -> List[int]:
        s = s1 = s2 = s3 = 0
        mx1 = mx12 = 0
        idx1, idx12 = 0, ()
        ans = []
        for i in range(k * 2, len(nums)):
            s1 += nums[i - k * 2]
            s2 += nums[i - k]
            s3 += nums[i]
            if i >= k * 3 - 1:
                if s1 > mx1:
                    mx1 = s1
                    idx1 = i - k * 3 + 1
                if mx1 + s2 > mx12:
                    mx12 = mx1 + s2
                    idx12 = (idx1, i - k * 2 + 1)
                if mx12 + s3 > s:
                    s = mx12 + s3
                    ans = [*idx12, i - k + 1]
                s1 -= nums[i - k * 3 + 1]
                s2 -= nums[i - k * 2 + 1]
                s3 -= nums[i - k + 1]
        return ans

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class Solution {
    public int[] maxSumOfThreeSubarrays(int[] nums, int k) {
        int[] ans = new int[3];
        int s = 0, s1 = 0, s2 = 0, s3 = 0;
        int mx1 = 0, mx12 = 0;
        int idx1 = 0, idx121 = 0, idx122 = 0;
        for (int i = k * 2; i < nums.length; ++i) {
            s1 += nums[i - k * 2];
            s2 += nums[i - k];
            s3 += nums[i];
            if (i >= k * 3 - 1) {
                if (s1 > mx1) {
                    mx1 = s1;
                    idx1 = i - k * 3 + 1;
                }
                if (mx1 + s2 > mx12) {
                    mx12 = mx1 + s2;
                    idx121 = idx1;
                    idx122 = i - k * 2 + 1;
                }
                if (mx12 + s3 > s) {
                    s = mx12 + s3;
                    ans = new int[] {idx121, idx122, i - k + 1};
                }
                s1 -= nums[i - k * 3 + 1];
                s2 -= nums[i - k * 2 + 1];
                s3 -= nums[i - k + 1];
            }
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
        vector<int> ans(3);
        int s