What is the time complexity of Count Distinct Subarrays Divisible by K in Sorted Array?

The optimal solution for Count Distinct Subarrays Divisible by K in Sorted Array runs in O(n) time and uses O(n) space. The time complexity is O(n) due to a single pass through the array. Space complexity is O(n) for the hash map storing remainders.

What is the brute force solution for Count Distinct Subarrays Divisible by K in Sorted Array?

The brute force approach for Count Distinct Subarrays Divisible by K in Sorted Array is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible subarrays and check if their sums are divisible by k. This is straightforward but inefficient for large arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Distinct Subarrays Divisible by K in Sorted Array?

Count Distinct Subarrays Divisible by K in Sorted Array uses the following concepts: Array, Hash Table, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Distinct Subarrays Divisible by K in Sorted Array?

Explain your thought process clearly. Discuss edge cases and constraints. Practice coding under time constraints.

What are common mistakes when solving Count Distinct Subarrays Divisible by K in Sorted Array?

Not handling negative remainders correctly. Forgetting to account for distinct sequences.

#3729

Count Distinct Subarrays Divisible by K in Sorted Array

Hard

ArrayHash TablePrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

Using prefix sums and a hash map allows us to efficiently track sums and their remainders, reducing the need to check each subarray individually.

⚙️

Algorithm

3 steps

1Step 1: Calculate prefix sums and their remainders when divided by k.
2Step 2: Use a hash map to track the frequency of each remainder.
3Step 3: For each unique value, count distinct subarrays based on the prefix sums and their remainders.

solution.py10 lines

1def countDistinct(nums, k):
2    prefix_sum = 0
3    count = 0
4    prefix_count = {0: 1}
5    for num in nums:
6        prefix_sum += num
7        remainder = prefix_sum % k
8        count += prefix_count.get(remainder, 0)
9        prefix_count[remainder] = prefix_count.get(remainder, 0) + 1
10    return count

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Complexity note: The time complexity is O(n) due to a single pass through the array. Space complexity is O(n) for the hash map storing remainders.

1Distinct subarrays can be counted using prefix sums and remainders.
2Using a hash map optimizes the counting process.

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