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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 $O(1)$.
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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; }