How do you solve K Divisible Elements Subarrays on LeetCode?

K Divisible Elements Subarrays (LeetCode #2261) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Using a set to store subarrays helps in avoiding duplicates.

What is the brute force solution for K Divisible Elements Subarrays?

The brute force approach for K Divisible Elements Subarrays is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach involves generating all possible subarrays and checking each one to see if it meets the criteria of having at most k elements divisible by p. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n²).

What are the interview tips for K Divisible Elements Subarrays?

Always clarify the problem constraints and edge cases before diving into coding. Think about the efficiency of your approach and be ready to discuss improvements. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving K Divisible Elements Subarrays?

Not using a set to track distinct subarrays, leading to incorrect counts. Failing to check the count of divisible elements before adding subarrays.

#2261

K Divisible Elements Subarrays

Medium

ArrayHash TableTrieRolling HashHash FunctionEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n²)Space O(n)

The optimal solution uses a sliding window approach to efficiently count the number of divisible elements while generating subarrays. This reduces the need for nested loops and allows us to keep track of distinct subarrays using a hash set.

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Algorithm

5 steps

1Step 1: Initialize a set to store distinct subarrays.
2Step 2: Use a sliding window approach to traverse the array.
3Step 3: Maintain a count of elements divisible by p within the window.
4Step 4: For each valid window, generate subarrays and add them to the set.
5Step 5: Return the size of the set as the result.

solution.py12 lines

1def countDistinct(nums, k, p):
2    distinct_subarrays = set()
3    n = len(nums)
4    for start in range(n):
5        count = 0
6        for end in range(start, n):
7            if nums[end] % p == 0:
8                count += 1
9            if count > k:
10                break
11            distinct_subarrays.add(tuple(nums[start:end + 1]))
12    return len(distinct_subarrays)

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops, but we avoid unnecessary checks by breaking early when the count exceeds k. The space complexity is O(n) for storing distinct subarrays.

1Using a set to store subarrays helps in avoiding duplicates.
2Early termination of loops can significantly reduce unnecessary computations.

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