2461. Maximum Sum of Distinct Subarrays With Length K

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

Complexity:

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

C++

• class Solution {
  public:
      long long maximumSubarraySum(vector<int>& nums, int k) {
          int n = nums.size();
          unordered_map<int, int> cnt;
          long long s = 0, ans = 0;
          for (int i = 0; i < k; ++i) {
              cnt[nums[i]]++;
              s += nums[i];
          }
          for (int i = k; i < n; ++i) {
              if (cnt.size() == k) ans = max(ans, s);
              cnt[nums[i]]++;
              cnt[nums[i - k]]--;
              if (cnt[nums[i - k]] == 0) cnt.erase(nums[i - k]);
              s += nums[i];
              s -= nums[i - k];
          }
          if (cnt.size() == k) ans = max(ans, s);
          return ans;
      }
  };

JAVA 

• class Solution {
      public long maximumSubarraySum(int[] nums, int k) {
          int n = nums.length;
          Map<Integer, Integer> cnt = new HashMap<>(k);
          long s = 0;
          for (int i = 0; i < k; ++i) {
              cnt.put(nums[i], cnt.getOrDefault(nums[i], 0) + 1);
              s += nums[i];
          }
          long ans = 0;
          for (int i = k; i < n; ++i) {
              if (cnt.size() == k) {
                  ans = Math.max(ans, s);
              }
              cnt.put(nums[i], cnt.getOrDefault(nums[i], 0) + 1);
              cnt.put(nums[i - k], cnt.getOrDefault(nums[i - k], 0) - 1);
              if (cnt.get(nums[i - k]) == 0) {
                  cnt.remove(nums[i - k]);
              }
              s += nums[i];
              s -= nums[i - k];
          }
          if (cnt.size() == k) {
              ans = Math.max(ans, s);
          }
          return ans;
      }
  }

PYTHON

• class Solution:
      def maximumSubarraySum(self, nums: list[int], k: int) -> int:
          from collections import defaultdict
  
          n = len(nums)
          cnt = defaultdict(int)
          s = 0
          ans = 0
  
          for i in range(k):
              cnt[nums[i]] += 1
              s += nums[i]
  
          for i in range(k, n):
              if len(cnt) == k:
                  ans = max(ans, s)
              cnt[nums[i]] += 1
              cnt[nums[i - k]] -= 1
              if cnt[nums[i - k]] == 0:
                  del cnt[nums[i - k]]
              s += nums[i]
              s -= nums[i - k]
  
          if len(cnt) == k:
              ans = max(ans, s)
  
          return ans