How do you solve Count of Interesting Subarrays on LeetCode?

Count of Interesting Subarrays (LeetCode #2845) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding how to use prefix sums can greatly reduce the complexity of counting problems.

What is the time complexity of Count of Interesting Subarrays?

The optimal solution for Count of Interesting Subarrays runs in O(n) time and uses O(n) space. The time complexity is O(n) because we traverse the array once. The space complexity is O(n) due to the hashmap storing prefix sums.

What is the brute force solution for Count of Interesting Subarrays?

The brute force approach for Count of Interesting Subarrays is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible subarray to see if it meets the interesting condition. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count of Interesting Subarrays?

Count of Interesting Subarrays 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 of Interesting Subarrays?

Always consider edge cases and test your solution with them. Explain your thought process clearly; it helps the interviewer understand your approach. Practice writing clean and efficient code to improve your speed during interviews.

What are common mistakes when solving Count of Interesting Subarrays?

Not correctly updating the hashmap or prefix sums, leading to incorrect counts. Failing to account for edge cases, such as when the entire array is interesting.

#2845

Count of Interesting Subarrays

Medium

ArrayHash TablePrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Using prefix sums and a hashmap allows us to efficiently count interesting subarrays by tracking the number of valid counts seen so far. This reduces the need to check every subarray explicitly.

⚙️

Algorithm

3 steps

1Step 1: Create a prefix sum array to track counts of indices where nums[i] % modulo == k.
2Step 2: Use a hashmap to count occurrences of each prefix sum value.
3Step 3: For each prefix sum, check how many times the adjusted count (current count - k) has been seen in the hashmap to find valid subarrays.

solution.py11 lines

1def countInterestingSubarrays(nums, modulo, k):
2    count = 0
3    prefix_count = {0: 1}
4    current_count = 0
5    for num in nums:
6        if num % modulo == k:
7            current_count += 1
8        if current_count - k in prefix_count:
9            count += prefix_count[current_count - k]
10        prefix_count[current_count] = prefix_count.get(current_count, 0) + 1
11    return count

ℹ

Complexity note: The time complexity is O(n) because we traverse the array once. The space complexity is O(n) due to the hashmap storing prefix sums.

1Understanding how to use prefix sums can greatly reduce the complexity of counting problems.
2Using a hashmap to track counts allows for quick lookups and updates, facilitating efficient solutions.

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