2552. Maximum Sum Of Distinct Subarrays With Length K¶

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2552. Maximum Sum of Distinct Subarrays With Length K

Medium

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 maximumSubarraySum(int[] nums, int k) {
        int n = nums.length;
        HashMap<Integer, Integer> map = new HashMap<>();
        long maxi_sum = 0, current_sum = 0;
        for (int i = 0; i < k; i++) {
            current_sum += nums[i];
            map.put(nums[i], map.getOrDefault(nums[i], 0) + 1);
        }
        if (map.size() == k) maxi_sum = current_sum;
        int start = 0;
        for (int i = k; i < n; i++) {
            current_sum += nums[i];
            current_sum -= nums[start];
            map.put(nums[i], map.getOrDefault(nums[i], 0) + 1);
            map.put(nums[start], map.getOrDefault(nums[start], 0) -1);
            if (map.getOrDefault(nums[start], 0) == 0) map.remove(nums[start]);
            if (map.size() == k) maxi_sum = Math.max(maxi_sum, current_sum);
            start++;
        }
        return maxi_sum;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here