How do you solve Number of Distinct Averages on LeetCode?

Number of Distinct Averages (LeetCode #2465) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting the array allows for efficient pairing of elements.

What is the time complexity of Number of Distinct Averages?

The optimal solution for Number of Distinct Averages runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the sorting step. The space complexity remains O(n) for storing the distinct averages.

What is the brute force solution for Number of Distinct Averages?

The brute force approach for Number of Distinct Averages is Brute Force, with O(n²) time complexity and O(n) space complexity. In this approach, we repeatedly find and remove the minimum and maximum elements from the array, calculate their average, and store these averages. This method is straightforward but inefficient due to repeated searching and removing elements. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Number of Distinct Averages?

Number of Distinct Averages uses the following concepts: Array, Hash Table, Two Pointers, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Number of Distinct Averages?

Always consider edge cases, such as arrays with duplicate values. Explain your thought process clearly while coding. Practice writing clean and efficient code, focusing on readability.

What are common mistakes when solving Number of Distinct Averages?

Not using a set to store averages, leading to incorrect counts. Forgetting to sort the array before pairing elements.

#2465

Number of Distinct Averages

Easy

ArrayHash TableTwo PointersSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

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

By sorting the array first, we can efficiently pair the smallest and largest elements to compute averages. This reduces the need for repeated searches and allows us to directly access the elements we need.

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Algorithm

5 steps

1Step 1: Sort the array.
2Step 2: Initialize a set to store distinct averages.
3Step 3: Use a loop to iterate from the start to the midpoint of the array, pairing elements from the start and end.
4Step 4: Calculate the average for each pair and add it to the set.
5Step 5: Return the size of the set.

solution.py10 lines

1# Full working Python code
2
3def distinct_averages(nums):
4    nums.sort()
5    averages = set()
6    n = len(nums)
7    for i in range(n // 2):
8        avg = (nums[i] + nums[n - 1 - i]) / 2
9        averages.add(avg)
10    return len(averages)

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Complexity note: The time complexity is O(n log n) due to the sorting step. The space complexity remains O(n) for storing the distinct averages.

1Sorting the array allows for efficient pairing of elements.
2Using a set ensures that only distinct averages are stored.

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