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:
k, andReturn 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 <= 1051 <= nums[i] <= 105This approach uses a hash map (or dictionary) to count occurrences of elements in the data. By iterating through the data once to populate the hash map, we can achieve efficient lookups. The main idea is to traverse the input list, counting occurrences of each element, and storing these counts in a hash map for quick access in the future.
This C program sets up an array to count the frequency of each element in a list. It assumes input values are non-negative and less than MAX. We use a simple array count to store the tally of each integer that appears in the input list.
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Time Complexity: O(n) for traversing the list once.
Space Complexity: O(k), where k is the range of input values (determined by MAX).
By sorting the input list, elements of the same value will be grouped together. We can then iterate through the sorted list to count the occurrences of each element. This takes advantage of the property of sorting to simplify the counting process.
This C program sorts the list of elements using qsort(), then iterates over the sorted list to count consecutive occurrences of each number. It outputs the frequency of each unique element.
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Time Complexity: O(n log n) due to sorting.
Space Complexity: O(1) additional space for counting.
| Approach | Complexity |
|---|---|
| Approach 1: Hash Map for Counting | Time Complexity: O(n) for traversing the list once. |
| Approach 2: Sorting | Time Complexity: O(n log n) due to sorting. |
Maximum Subarray - Amazon Coding Interview Question - Leetcode 53 - Python • NeetCode • 605,143 views views
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