How do you solve Filter Elements from Array on LeetCode?

Filter Elements from Array (LeetCode #2634) 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 truthy and falsy values is crucial for filtering.

What is the time complexity of Filter Elements from Array?

The optimal solution for Filter Elements from Array runs in O(n) time and uses O(n) space. The time complexity remains O(n) as we still traverse the array once. The space complexity is O(n) for the new filtered array.

What is the brute force solution for Filter Elements from Array?

The brute force approach for Filter Elements from Array is Brute Force, with O(n) time complexity and O(n) space complexity. The brute force approach involves iterating through each element of the array and applying the filtering function to determine if it should be included in the result. This is straightforward but inefficient as it may involve unnecessary checks. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Filter Elements from Array?

Clarify the requirements of the filtering function before coding. Think about edge cases, such as empty arrays or all elements being filtered out. Practice writing clean and efficient code without relying on built-in methods.

What are common mistakes when solving Filter Elements from Array?

Not understanding how to define the filtering function correctly. Overcomplicating the solution when a simple loop suffices.

#2634

Filter Elements from Array

Easy

ArrayFunctional Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal solution is similar to the brute force approach but focuses on clarity and efficiency. It uses a single loop to filter elements based on the provided function, ensuring we only traverse the array once.

⚙️

Algorithm

5 steps

1Step 1: Initialize an empty array called filteredArr.
2Step 2: Loop through each element of arr using its index.
3Step 3: For each element, call the function fn with the element and its index.
4Step 4: If the result of fn is truthy, add the element to filteredArr.
5Step 5: Return filteredArr after the loop.

solution.py2 lines

1def filter(arr, fn):
2    return [arr[i] for i in range(len(arr)) if fn(arr[i], i)]

ℹ

Complexity note: The time complexity remains O(n) as we still traverse the array once. The space complexity is O(n) for the new filtered array.

1Understanding truthy and falsy values is crucial for filtering.
2Using a single loop is efficient for this problem.

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