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;
}

689 - Maximum Sum of 3 Non-Overlapping Subarrays

689. Maximum Sum of 3 Non-Overlapping Subarrays

Description

Solutions

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689. Maximum Sum of 3 Non-Overlapping Subarrays

Description

Solutions

All Problems

All Solutions