Today I have done leetcode's November LeetCoding Challenge with cpp.

November LeetCoding Challenge 19

Description

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

Solution

class Solution {
public:
  long long maximumSubarraySum(vector<int>& nums, int k) {
    long long answer = 0;
    long long current = 0;
    map<int, int> count;
    for(int i = 0; i < k; ++i) {
      current += nums[i];
      count[nums[i]] += 1;
    }
    if(count.size() == k) answer = current;
    for(int i = k; i < nums.size(); ++i) {
      count[nums[i]] += 1;
      count[nums[i - k]] -= 1;
      current += nums[i];
      current -= nums[i - k];
      if(!count[nums[i - k]]) count.erase(nums[i - k]);
      if(count.size() == k) answer = max(answer, current);
    }
    return answer;
  }
};