content: package codingquestions.leetcode;
  
  /**
   * 2461. Maximum Sum of Distinct Subarrays With Length K
   * Solved
   * Medium
   * Topics
   * Companies
   * Hint
   * 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:
   * <p>
   * 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.
   * <p>
   * A subarray is a contiguous non-empty sequence of elements within an array.
   * <p>
   * <p>
   * <p>
   * Example 1:
   * <p>
   * 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:
   * <p>
   * 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.
   * <p>
   * <p>
   * Constraints:
   * <p>
   * 1 <= k <= nums.length <= 105
   * 1 <= nums[i] <= 105
   */
  
  public class MaximumSumofDistinctSubarraysWithLengthK {
  
      public static void main(String[] args) {
          MaximumSumofDistinctSubarraysWithLengthK msd = new MaximumSumofDistinctSubarraysWithLengthK();
          System.out.println(msd.maximumSubarraySum(new int[]{1, 5, 4, 2, 9, 9, 9}, 3));
      }
  
      public long maximumSubarraySum(int[] nums, int k) {
          int maxNum = 0;
          for (int num : nums) {
              maxNum = Math.max(maxNum, num);
          }
          int[] counts = new int[maxNum + 1];
          int dupCount = 0;
          long totalSum = 0;
          long curSum = 0;
          for (int i = 0; i < k; i++) {
              if (counts[nums[i]] >= 1) {
                  dupCount++;
              }
              counts[nums[i]]++;
              curSum += nums[i];
          }
          if (dupCount == 0) {
              totalSum = curSum;
          }
          for (int i = k; i < nums.length; i++) {
              if (counts[nums[i]] >= 1) {
                  dupCount++;
              }
              counts[nums[i]]++;
              curSum += nums[i];
              if (counts[nums[i - k]] > 1) {
                  dupCount--;
              }
              counts[nums[i - k]]--;
              curSum -= nums[i - k];
              if (dupCount == 0) {
                  totalSum = Math.max(totalSum, curSum);
              }
          }
          return totalSum;
      }
  }
  