Sliding Window Technique - Part 7

There is usually one caveat on sliding window approaches: usually after the main loop, you still need to make one extra check for the very last case. Treat it as a corner case, albeit critical for the correctness of the code. Talking about this line here. Code is down below, cheers, ACC.

Maximum Sum of Distinct Subarrays With Length K - LeetCode

2461. Maximum Sum of Distinct Subarrays With Length K

Medium

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

Accepted

14,455

Submissions

44,090

public long MaximumSubarraySum(int[] nums, int k)
{
    //Sliding window approach
    long maxSum = 0;
    long currentSum = 0;
    Hashtable uniqueElements = new Hashtable();
    for (int i = 0; i < nums.Length; i++)
    {
        if (i < k)
        {
            if (!uniqueElements.ContainsKey(nums[i])) 
                uniqueElements.Add(nums[i], 0);
            uniqueElements[nums[i]] = (int)uniqueElements[nums[i]] + 1;
            currentSum += nums[i];
        }
        else
        {
            maxSum = (uniqueElements.Count == k && currentSum >= maxSum) ? currentSum : maxSum;

            int firstVal = nums[i - k];
            if (uniqueElements.ContainsKey(firstVal))
            {
                uniqueElements[firstVal] = (int)uniqueElements[firstVal] - 1;
                if ((int)uniqueElements[firstVal] <= 0) 
                    uniqueElements.Remove(firstVal);
                currentSum -= firstVal;
            }
            if (!uniqueElements.ContainsKey(nums[i])) 
                uniqueElements.Add(nums[i], 0);
            uniqueElements[nums[i]] = (int)uniqueElements[nums[i]] + 1;
            currentSum += nums[i];
        }
    }

    //Last, corner case
    maxSum = (uniqueElements.Count == k && currentSum >= maxSum) ? currentSum : maxSum;

    return maxSum;
}