Question

2635. Apply Transform Over Each Element in Array

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Description

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
• -109 <= arr[i] <= 109
• fn returns a number

Solutions

Solution 1: traversal

We traverse the array $arr$, for each element $arr[i]$, replace it with $fn(arr[i], i)$. Finally, return the array $arr$.

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

function map(arr: number[], fn: (n: number, i: number) => number): number[] {
  for (let i = 0; i < arr.length; ++i) {
    arr[i] = fn(arr[i], i);
  }
  return arr;
}