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:

• 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

 

Solutions

Solution: Sliding Window

• Time complexity: O(n)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number}
 */
const maximumSubarraySum = function (nums, k) {
  const n = nums.length;
  const memo = new Set();
  let left = 0;
  let sum = 0;
  let length = 0;
  let result = 0;

  for (let index = 0; index < n; index++) {
    const num = nums[index];

    sum += num;
    length += 1;

    while (memo.has(num) || length > k) {
      sum -= nums[left];
      length -= 1;
      memo.delete(nums[left]);
      left += 1;
    }
    memo.add(num);
    if (length < k) continue;

    result = Math.max(sum, result);
  }
  return result;
};