How do you solve Find X-Sum of All K-Long Subarrays I on LeetCode?

Find X-Sum of All K-Long Subarrays I (LeetCode #3318) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log x) time complexity and O(n) space complexity. Key insight: Using a frequency map helps in counting occurrences efficiently.

What is the brute force solution for Find X-Sum of All K-Long Subarrays I?

The brute force approach for Find X-Sum of All K-Long Subarrays I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves calculating the x-sum for each subarray of size k by counting the occurrences of each element in the subarray. This method is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n log x).

What algorithm or data structure is used to solve Find X-Sum of All K-Long Subarrays I?

Find X-Sum of All K-Long Subarrays I uses the following concepts: Array, Hash Table, Sliding Window, Heap (Priority Queue). The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find X-Sum of All K-Long Subarrays I?

Explain your thought process clearly while coding. Consider edge cases, such as when k equals x or when there are fewer distinct elements than x. Practice optimizing your solution after writing the brute force version.

What are common mistakes when solving Find X-Sum of All K-Long Subarrays I?

Not updating the frequency map correctly when sliding the window. Failing to handle cases where there are fewer than x distinct elements.

#3318

Find X-Sum of All K-Long Subarrays I

Easy

ArrayHash TableSliding WindowHeap (Priority Queue)Hash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log x)
Space	O(1)	O(n)

💡

Intuition

Time O(n log x)Space O(n)

The optimal solution uses a sliding window approach to efficiently maintain the count of elements as we move through the array. This avoids recalculating counts from scratch for each subarray.

⚙️

Algorithm

4 steps

1Step 1: Initialize a frequency map to count occurrences of elements in the first subarray.
2Step 2: For each subsequent subarray, update the frequency map by removing the element that is sliding out and adding the new element that is sliding in.
3Step 3: After updating the frequency map, determine the top x most frequent elements using a max-heap to efficiently get the largest elements.
4Step 4: Calculate the x-sum for the current subarray and store it in the answer array.

solution.py19 lines

1from collections import defaultdict
2import heapq
3
4def x_sum(nums, k, x):
5    n = len(nums)
6    answer = []
7    count = defaultdict(int)
8    for i in range(k):
9        count[nums[i]] += 1
10    for i in range(n - k + 1):
11        if i > 0:
12            count[nums[i - 1]] -= 1
13            count[nums[i + k - 1]] += 1
14        freq_heap = [(-freq, num) for num, freq in count.items() if freq > 0]
15        heapq.heapify(freq_heap)
16        top_x = heapq.nlargest(x, freq_heap)
17        x_sum_value = sum(-num * freq for freq, num in top_x)
18        answer.append(x_sum_value)
19    return answer

ℹ

Complexity note: The time complexity is O(n log x) because we maintain a frequency map and use a heap to get the top x elements, which is efficient compared to recalculating counts from scratch. The space complexity is O(n) due to the frequency map.

1Using a frequency map helps in counting occurrences efficiently.
2Maintaining a sliding window allows for quick updates as we move through the array.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.