# Maximum Sum of Distinct Subarrays With Length K

Published: May 24, 2024

Medium Array Hash Table Sliding Window

## Problem 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.

Constraints:

• `1 <= k <= nums.length <= 10**5`
• `1 <= nums[i] <= 10**5`

https://leetcode.com/problems/maximum-sum-of-distinct-subarrays-with-length-k/

## Examples

``````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.
``````

## How to Solve

Use hash table and sliding window.

## Solution

``````class MaxSumOfSubArrayLengthK {
public:
long long maximumSubarraySum(vector<int>& nums, int k) {
unordered_map<int, int> seen;
long long curSum = 0, maxSum = 0;
for (int i = 0; i < k; ++i) {
seen[nums[i]]++;
curSum += nums[i];
}
if (seen.size() == k) maxSum = max(maxSum, curSum);
for (int i = k; i < nums.size(); ++i) {
curSum -= nums[i - k];
if (--seen[nums[i - k]] == 0) seen.erase(nums[i - k]);
seen[nums[i]]++;
curSum += nums[i];
if (seen.size() == k) maxSum = max(maxSum, curSum);
}
return maxSum;
}
};
``````
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## Complexities

• Time: `O(n)`
• Space: `O(k)`