You are given an array nums of n integers and two integers k and x.
The x-sum of an array is calculated by the following procedure:
x most frequent elements. If two elements have the same number of occurrences, the element with the bigger value is considered more frequent.Note that if an array has less than x distinct elements, its x-sum is the sum of the array.
Return an integer array answer of length n - k + 1 where answer[i] is the x-sum of the subarray nums[i..i + k - 1].
Example 1:
Input: nums = [1,1,2,2,3,4,2,3], k = 6, x = 2
Output: [6,10,12]
Explanation:
[1, 1, 2, 2, 3, 4], only elements 1 and 2 will be kept in the resulting array. Hence, answer[0] = 1 + 1 + 2 + 2.[1, 2, 2, 3, 4, 2], only elements 2 and 4 will be kept in the resulting array. Hence, answer[1] = 2 + 2 + 2 + 4. Note that 4 is kept in the array since it is bigger than 3 and 1 which occur the same number of times.[2, 2, 3, 4, 2, 3], only elements 2 and 3 are kept in the resulting array. Hence, answer[2] = 2 + 2 + 2 + 3 + 3.Example 2:
Input: nums = [3,8,7,8,7,5], k = 2, x = 2
Output: [11,15,15,15,12]
Explanation:
Since k == x, answer[i] is equal to the sum of the subarray nums[i..i + k - 1].
Constraints:
1 <= n == nums.length <= 501 <= nums[i] <= 501 <= x <= k <= nums.lengthThis approach involves iterating over each possible k-length subarray, counting the occurrences of elements, and then selecting the top x elements by sorting the count and element value.
The sorted elements are then summed to determine the x-sum for each subarray. This method is simple but not optimal for larger input sizes.
The solution iterates over each k-length subarray, calculates the frequency of each element in a frequency array, then sorts this array to find the top x frequent elements using their values to break ties. It then calculates the sum of these elements and stores it in the result array.
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Time Complexity: O((n-k+1)*51^2), where 51 is the possible range of elements. Space Complexity: O(51) for storing counts.
This approach employs a sliding window to maintain the frequency count between subarrays. This allows the reuse of previously calculated frequencies, significantly reducing redundant work by removing the frequency of the outgoing element while adding the frequency of the new incoming element.
This C implementation uses a sliding window on the k-length subarray by adjusting frequency counts as the window slides. This reduces computational overhead as the program recalculates less frequently than recalculating frequency from scratch for each subarray.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n*51^2), where n is the number of elements. Space Complexity: O(51).
| Approach | Complexity |
|---|---|
| Brute Force with Sorting | Time Complexity: O((n-k+1)*51^2), where 51 is the possible range of elements. Space Complexity: O(51) for storing counts. |
| Optimized Sliding Window | Time Complexity: O(n*51^2), where n is the number of elements. Space Complexity: O(51). |
Longest Subarray with sum K | Brute - Better - Optimal | Generate Subarrays • take U forward • 891,854 views views
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