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:
nums.length == n1 <= n <= 1051 <= nums[i] <= 1091 <= x <= k <= nums.lengthWe use a sliding window technique to generate subarrays of length k. For each subarray, we calculate the frequency of each element and use a priority queue to identify the top x frequent elements. The priority queue helps us efficiently keep track of the most frequent elements as we slide the window across the array.
We iterate through each possible starting index of the subarray using a loop. For each subarray, we count the occurrences of each number with a Counter. Using nlargest from the heapq module, we extract the top x elements sorted by frequency and then by value. These elements' contributions to the x-sum are calculated and added to the result list.
JavaScript
Time Complexity: O((n-k+1)*k*logx), where logx is for maintaining the heap of the top x elements.
Space Complexity: O(k), for maintaining the counter.
This approach further optimizes by maintaining a frequency map which is updated as the window slides. Instead of recomputing the entire frequency map for each subarray, we adjust the map incrementally by adding a new element and removing the element that's no longer in the window. This reduces redundant calculations and improves efficiency.
This solution maintains a frequency map along with a map sorting the frequencies in descending order (to identify top x elements). As the window slides, the frequency map is incrementally updated to reflect the addition of the new number and the removal of the old number outside the window. This avoids full recomputation of frequencies.
Java
Time Complexity: O(n*logk) due to logarithmic operations on the map.
Space Complexity: O(k), for storing frequency counts.
| Approach | Complexity |
|---|---|
| Sliding Window with Priority Queue | Time Complexity: O((n-k+1)*k*logx), where logx is for maintaining the heap of the top x elements. |
| Optimized Sliding Window with HashMap | Time Complexity: O(n*logk) due to logarithmic operations on the map. |
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