Question

2634. Filter Elements from Array

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Description

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

Solutions

Solution 1: Traversal

We traverse the array $arr$ and for each element $arr[i]$, if $fn(arr[i], i)$ is true, we add it to the answer array. Finally, we return the answer array.

The time complexity is $O(n)$, where $n$ is the length of the array $arr$. Ignoring the space consumption of the answer, the space complexity is $O(1)$.

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;
}