Day 5: Apply Transform Over Each Element in Array — 30 days of LeetCode challenge.

Problem

Given an integer array arr and a mapping function fn, return a new array with a transformation applied to each element.

The returned array should be created such that returnedArray[i] = fn(arr[i], i).

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

Example 1

Input: arr = [1,2,3], fn = function plusone(n) { return n + 1; }
Output: [2,3,4]
Explanation:
const newArray = map(arr, plusone); // [2,3,4]
The function increases each value in the array by one.

Example 2

Input: arr = [1,2,3], fn = function plusI(n, i) { return n + i; }
Output: [1,3,5]
Explanation: The function increases each value by the index it resides in.

Example 3

Input: arr = [10,20,30], fn = function constant() { return 42; }
Output: [42,42,42]
Explanation: The function always returns 42.

Constraints

• 0 <= arr.length <= 1000

• -10<sup>9</sup> <= arr[i] <= 10<sup>9</sup>

• fn returns a number

Solution

* @param {number[]} arr
 * @param {Function} fn
 * @return {number[]}
 */
var map = function(arr, fn) {

 let newArray  = []
for(i= 0; i< arr.length; i++){
    newArray[i] = fn(arr[i] ,i)
}

return newArray
};

Explanation
In the code above, we have a function map that takes two argument arr and fn. fn is a callback function that is applied to each element in the array.

We created an empty array newArray

The for loop is used to iterate over each element in the array and increment it by 1. i stands for the index variable.

newArray[i] = fn(arr[i], i )In this logic, we assigned a callback function fn to the newArray[i]. The callback function gets invoked using the current element arr[i] and the current index i.

Each element in the array gets transformed by the callback function fn and is stored in the corresponding index of newArray.

Conclusion

In this article, we have solved the Apply Transform Over Each Element in Array Leettcode challenge.

Looking forward to solving the next challenge