README_EN.md

689. Maximum Sum of 3 Non-Overlapping Subarrays

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Description

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

Python3

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

Java

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;
    }
}

C++

class Solution {
public:
    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
        vector<int> ans(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.size(); ++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 = {idx121, idx122, i - k + 1};
                }
                s1 -= nums[i - k * 3 + 1];
                s2 -= nums[i - k * 2 + 1];
                s3 -= nums[i - k + 1];
            }
        }
        return ans;
    }
};

Go

func maxSumOfThreeSubarrays(nums []int, k int) []int {
	ans := make([]int, 3)
	s, s1, s2, s3 := 0, 0, 0, 0
	mx1, mx12 := 0, 0
	idx1, idx121, idx122 := 0, 0, 0
	for i := k * 2; i < len(nums); 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 = []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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