LeetCode #2634 — EASY

Filter Elements from Array

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

Solve on LeetCode
The Problem

Problem Statement

Given an integer array arr and a filtering function fn, return a filtered array filteredArr.

The fn function takes one or two arguments:

  • arr[i] - number from the arr
  • i - index of arr[i]

filteredArr should only contain the elements from the arr for which the expression fn(arr[i], i) evaluates to a truthy value. A truthy value is a value where Boolean(value) returns true.

Please solve it without the built-in Array.filter method.

Example 1:

Input: arr = [0,10,20,30], fn = function greaterThan10(n) { return n > 10; }
Output: [20,30]
Explanation:
const newArray = filter(arr, fn); // [20, 30]
The function filters out values that are not greater than 10

Example 2:

Input: arr = [1,2,3], fn = function firstIndex(n, i) { return i === 0; }
Output: [1]
Explanation:
fn can also accept the index of each element
In this case, the function removes elements not at index 0

Example 3:

Input: arr = [-2,-1,0,1,2], fn = function plusOne(n) { return n + 1 }
Output: [-2,0,1,2]
Explanation:
Falsey values such as 0 should be filtered out

Constraints:

  • 0 <= arr.length <= 1000
  • -109 <= arr[i] <= 109

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: Given an integer array arr and a filtering function fn, return a filtered array filteredArr. The fn function takes one or two arguments: arr[i] - number from the arr i - index of arr[i] filteredArr should only contain the elements from the arr for which the expression fn(arr[i], i) evaluates to a truthy value. A truthy value is a value where Boolean(value) returns true. Please solve it without the built-in Array.filter method.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

function greaterThan10(n) { return n > 10; }
[0,10,20,30]

Example 2

function firstIndex(n, i) { return i === 0; }
[1,2,3]

Example 3

function plusOne(n) { return n + 1 }
[-2,-1,0,1,2]

Related Problems

  • Group By (group-by)
  • Apply Transform Over Each Element in Array (apply-transform-over-each-element-in-array)
  • Array Reduce Transformation (array-reduce-transformation)
Step 02

Core Insight

What unlocks the optimal approach

  • Start by declaring a new array which will eventually be returned.
  • In Javascript, there is the concept of "truthiness" and "falsiness". Values such as 0, undefined, null, and false are falsy. Most values are truthy: 1, {}, [], true, etc. In Javascript, the contents of if-statements don't need to be booleans. You can say "if ([1,2,3]) {}", and it's equivalent to saying 'if (true) {}".
  • Loop over each element in the array. If fn(arr[i]) is truthy, push it to the array.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2634: Filter Elements from Array
// Auto-generated Java example from ts.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (ts):
// // Accepted solution for LeetCode #2634: Filter Elements from Array
// function filter(arr: number[], fn: (n: number, i: number) => any): number[] {
//     const ans: number[] = [];
//     for (let i = 0; i < arr.length; ++i) {
//         if (fn(arr[i], i)) {
//             ans.push(arr[i]);
//         }
//     }
//     return ans;
// }
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(1)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED
O(n) time
O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.