[Leetcode] JS 30 Filter Elements from Arrays

Question

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

Solution

/**
 * @param {number[]} arr
 * @param {Function} fn
 * @return {number[]}
 */
var filter = function(arr, fn) {
    let number = []
    for ( i = 0; i < arr.length; i++ ){
        if(fn(arr[i], i)) {
            number.push(arr[i])
        }
    }
    return number
};

Takeway

Initially, I wrote fn(arr[i], i) == true but found that it couldn't pass certain values. In fact, an if statement like if(fn(arr[i], i)) checks whether the return value is truthy. However, when I write fn(arr[i], i) == true, it strictly checks if the return value equals true, ignoring other truthy values such as non-zero numbers and non-empty strings.