leetcode - 2461. Maximum Sum of Distinct Subarrays With Length K

Description

You are given an integer array nums and an integer k. Find the maximum subarray sum of all the subarrays of nums that meet the following conditions:

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The length of the subarray is k, and
All the elements of the subarray are distinct.
Return the maximum subarray sum of all the subarrays that meet the conditions. If no subarray meets the conditions, return 0.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

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Input: nums = [1,5,4,2,9,9,9], k = 3
Output: 15
Explanation: The subarrays of nums with length 3 are:
- [1,5,4] which meets the requirements and has a sum of 10.
- [5,4,2] which meets the requirements and has a sum of 11.
- [4,2,9] which meets the requirements and has a sum of 15.
- [2,9,9] which does not meet the requirements because the element 9 is repeated.
- [9,9,9] which does not meet the requirements because the element 9 is repeated.
We return 15 because it is the maximum subarray sum of all the subarrays that meet the conditions

Example 2:

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Input: nums = [4,4,4], k = 3
Output: 0
Explanation: The subarrays of nums with length 3 are:
- [4,4,4] which does not meet the requirements because the element 4 is repeated.
We return 0 because no subarrays meet the conditions.

Constraints:

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1 <= k <= nums.length <= 10^5
1 <= nums[i] <= 10^5

Solution

Sliding window (TLE)

Use a sliding window to store all the elements we have visited. When the current element is in the sliding window, pop from left until it doesn't exist.

Time complexity: o ( n 2 ) o(n^2) o(n2)

Space complexity: o ( n ) o(n) o(n)

Sliding window + set

Previous solution will exceed the time limit because list's in has o ( n ) o(n) o(n) complexity. To mitigate this, we could use an additional set to store all the elements.

Time complexity: o ( n ) o(n) o(n)

Space complexity: o ( n ) o(n) o(n)

Code

Sliding window (TLE)

python3 复制代码

class Solution:
    def maximumSubarraySum(self, nums: List[int], k: int) -> int:
        sliding_window = collections.deque([])
        cur_sum = 0
        max_sum = 0
        for each_num in nums:
            while each_num in sliding_window or len(sliding_window) == k:
                pop_ele = sliding_window.popleft()
                cur_sum -= pop_ele
            cur_sum += each_num
            sliding_window.append(each_num)
            if len(sliding_window) == k:
                max_sum = max(max_sum, cur_sum)
        return max_sum

Sliding window + set

python3 复制代码

class Solution:
    def maximumSubarraySum(self, nums: List[int], k: int) -> int:
        sliding_window = collections.deque([])
        window_ele = set()
        cur_sum = 0
        max_sum = 0
        for each_num in nums:
            while each_num in window_ele or len(sliding_window) == k:
                pop_ele = sliding_window.popleft()
                window_ele.remove(pop_ele)
                cur_sum -= pop_ele
            cur_sum += each_num
            sliding_window.append(each_num)
            window_ele.add(each_num)
            if len(sliding_window) == k:
                max_sum = max(max_sum, cur_sum)
        return max_sum