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2634. Filter Elements from Array

Description

Given an integer array arr and a filtering function fn, return a new array with a fewer or equal number of elements.

The returned array should only contain elements where fn(arr[i], i) evaluated to a truthy value.

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;
    }
    
    

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