Leetcode #2461: Maximum Sum of Distinct Subarrays With Length K
In this guide, we solve Leetcode #2461 Maximum Sum of Distinct Subarrays With Length K in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Hash Table, Sliding Window
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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
Python Solution
class Solution:
def maximumSubarraySum(self, nums: List[int], k: int) -> int:
cnt = Counter(nums[:k])
s = sum(nums[:k])
ans = s if len(cnt) == k else 0
for i in range(k, len(nums)):
cnt[nums[i]] += 1
cnt[nums[i - k]] -= 1
if cnt[nums[i - k]] == 0:
cnt.pop(nums[i - k])
s += nums[i] - nums[i - k]
if len(cnt) == k:
ans = max(ans, s)
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.