Question

689. Maximum Sum of 3 Non-Overlapping Subarrays

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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 $O(1)$.

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

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;
  }
}

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;
  }
};

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
}

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;
}

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:
  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

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;
  }
};

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
}