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maximum-sum-of-distinct-subarrays-with-length-k

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
#define ll long long
class Solution {
public:
    long long maximumSubarraySum(vector<int>& nums, int k) {
        int n = nums.size();
        ll res = 0, sum = 0;
        unordered_map<int, int> freq;
        for(int i = 0; i < k; i++) {
            ++freq[nums[i]];
            sum += nums[i];
        }

        // base case
        if(freq.size() == k)
            res = sum;

        int i = k;
        for(int i = k; i < n; i++) {
            ++freq[nums[i]];
            --freq[nums[i - k]];

            if(freq[nums[i - k]] == 0)
                freq.erase(nums[i - k]);

            sum += nums[i];
            sum -= nums[i - k];
            // 維持 freq 的長度要等於 3
            // 也就是 distinct element = 3
            if(freq.size() == k)
                res = max(res, sum);
        }
        return res;
    }
};
  • T: O(N)O(N)
  • S: O(1)O(1)