2461. Maximum Sum of Distinct Subarrays With Length K

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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:

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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:

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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 <= 10^5
  • 1 <= nums[i] <= 10^5

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    long long maximumSubarraySum(vector<int>& nums, int k) {
        long long res = 0, cur = 0;
        int right = 0, n = nums.size();
        unordered_map<int, int> m;
        while (right < n) {
            int num = nums[right];
            m[num]++;
            cur += num;
            if (right >= k - 1) {
                if (m.size() == k) {
                    res = max(res, cur);
                }
                int l = nums[right - k + 1];
                cur -= l;
                m[l]--;
                if (m[l] == 0) {
                    m.erase(l);
                }
            }
            right++;
        }
        return res;
    }
};