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2461. Maximum Sum of Distinct Subarrays With Length K

Difficulty Topics

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:

  • 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:

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:

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:

  • 1 <= k <= nums.length <= 105
  • 1 <= nums[i] <= 105

Solution

maximum-sum-of-distinct-subarrays-with-length-k.py
class Solution:
    def maximumSubarraySum(self, nums: List[int], k: int) -> int:
        N = len(nums)
        i = 0
        curr = set()
        currSum = 0
        res = 0

        for j, x in enumerate(nums):
            while j - i + 1 > k or x in curr:
                curr.remove(nums[i])
                currSum -= nums[i]
                i += 1

            curr.add(x)
            currSum += x
            if j - i + 1 == k:
                res = max(res, currSum)

        return res