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

Approach 01:

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    long long maximumSubarraySum(vector<int>& nums, int subarraySize) {
        long maxSum = 0;
        long currentSum = 0;
        int uniqueCount = 0;
        unordered_map<int, int> frequencyMap;

        for (int i = 0; i < nums.size(); ++i) {
            currentSum += nums[i];

            if (++frequencyMap[nums[i]] == 1) // New unique element
                ++uniqueCount;

            // Remove the element that is sliding out of the window
            if (i >= subarraySize) {
                if (--frequencyMap[nums[i - subarraySize]] == 0)
                    --uniqueCount;
                currentSum -= nums[i - subarraySize];
            }

            // Check if the current subarray has exactly `subarraySize` unique
            // elements
            if (i >= subarraySize - 1 && uniqueCount == subarraySize)
                maxSum = max(maxSum, currentSum);
        }

        return maxSum;
    }
};

from typing import List

class Solution:
    def maximumSubarraySum(self, nums: List[int], subarraySize: int) -> int:
        maxSum = 0
        currentSum = 0
        uniqueCount = 0
        frequencyMap = {}

        for i in range(len(nums)):
            currentSum += nums[i]

            # Add the current number to the frequency map
            frequencyMap[nums[i]] = frequencyMap.get(nums[i], 0) + 1
            if frequencyMap[nums[i]] == 1:  # New unique element
                uniqueCount += 1

            # Remove the element sliding out of the window
            if i >= subarraySize:
                frequencyMap[nums[i - subarraySize]] -= 1
                if frequencyMap[nums[i - subarraySize]] == 0:
                    uniqueCount -= 1
                currentSum -= nums[i - subarraySize]

            # Check if the current subarray has exactly `subarraySize` unique elements
            if i >= subarraySize - 1 and uniqueCount == subarraySize:
                maxSum = max(maxSum, currentSum)

        return maxSum

Time Complexity

• Sliding Window Traversal:

  The function iterates through the array using a single loop, visiting each element once. This takes \( O(n) \), where \( n \) is the size of the array.

• Map Operations:

  Updating and accessing the frequency map (unordered_map) takes \( O(1) \) on average due to hash map operations. These operations are performed for each element, so they contribute \( O(n) \) overall.

• Overall Time Complexity:

  The overall time complexity is \( O(n) \).

Space Complexity

• Frequency Map:

  The space used by the unordered_map depends on the number of unique elements in the array. In the worst case (all elements are unique), the space complexity is \( O(k) \), where \( k \) is the subarray size.

• Auxiliary Variables:

  Variables like maxSum, currentSum, and uniqueCount take \( O(1) \) space.

• Overall Space Complexity:

  The overall space complexity is \( O(k) \), where \( k \) is the subarray size.