How do you solve Count Complete Subarrays in an Array on LeetCode?

Count Complete Subarrays in an Array (LeetCode #2799) 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 distinct elements is crucial to solving the problem.

What is the brute force solution for Count Complete Subarrays in an Array?

The brute force approach for Count Complete Subarrays in an Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible subarray to see if it contains all distinct elements of the original array. 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 Complete Subarrays in an Array?

Count Complete Subarrays in an Array uses the following concepts: Array, Hash Table, Sliding Window. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Complete Subarrays in an Array?

Always clarify the definition of a complete subarray before starting. Think about edge cases, such as arrays with all identical elements. Practice explaining your thought process while coding to simulate the interview environment.

What are common mistakes when solving Count Complete Subarrays in an Array?

Not correctly managing the distinct count when modifying the window. Confusing the conditions for counting complete subarrays.

#2799

Count Complete Subarrays in an Array

Medium

ArrayHash TableSliding WindowHash 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)

The optimal approach uses a sliding window technique to efficiently count complete subarrays. Instead of checking every subarray, we expand and contract our window to maintain the required number of distinct elements.

⚙️

Algorithm

4 steps

1Step 1: Use a hashmap to count occurrences of elements in the current window.
2Step 2: Expand the window by moving the end pointer and adding elements to the hashmap.
3Step 3: When the number of distinct elements matches the total distinct count, count all valid subarrays ending at the current end pointer.
4Step 4: Contract the window from the start until the number of distinct elements is less than required.

solution.py28 lines

1# Full working Python code
2
3def count_complete_subarrays(nums):
4    total_distinct = len(set(nums))
5    count = 0
6    left = 0
7    freq_map = {}
8    distinct_count = 0
9
10    for right in range(len(nums)):
11        if nums[right] in freq_map:
12            freq_map[nums[right]] += 1
13        else:
14            freq_map[nums[right]] = 1
15            distinct_count += 1
16
17        while distinct_count == total_distinct:
18            count += len(nums) - right
19            freq_map[nums[left]] -= 1
20            if freq_map[nums[left]] == 0:
21                del freq_map[nums[left]]
22                distinct_count -= 1
23            left += 1
24
25    return count
26
27# Example usage
28print(count_complete_subarrays([1, 3, 1, 2, 2]))  # Output: 4

ℹ

Complexity note: The time complexity is O(n) because each element is processed at most twice (once when expanding and once when contracting the window). The space complexity is O(n) due to the hashmap storing the frequency of elements.

1Understanding distinct elements is crucial to solving the problem.
2Using a sliding window approach is more efficient than brute force for counting subarrays.

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