The Find the Factor Challenge on HackerRank is a popular coding problem that tests your problem-solving skills and understanding of mathematical concepts. In this guide, we'll walk you through the solution for the Find the Factor Challenge using JavaScript. By the end of this tutorial, you'll have a solid understanding of the problem and how to solve it efficiently.

## Table of Contents

## Problem Overview

**Problem Link**: Find the Factor Challenge

In this challenge, you are given an integer `n`

and you need to find the two factors of `n`

that have the minimum absolute difference. If multiple pairs have the same minimum absolute difference, return the pair with the smaller first element.

### Input Format

- A single integer
`n`

(1 <= n <= 10^12).

### Output Format

- Two space-separated integers that represent the two factors of
`n`

with the minimum absolute difference.

### Example

**Input:**

```
45
```

**Output:**

```
5 9
```

## Solution Approach

To solve this problem efficiently, we'll be using the square root of the given integer `n`

. The square root is the central point that allows us to find the two factors with the minimum absolute difference.

Here are the steps we'll follow:

- Calculate the square root of
`n`

. - From the square root, iterate downwards to find the first factor of
`n`

. - Divide
`n`

by the first factor to get the second factor. - Return the two factors.

## Step-by-Step Solution

### Step 1: Calculate the square root of `n`

First, we'll calculate the square root of the given integer `n`

. In JavaScript, we can use the `Math.sqrt()`

function to achieve this.

```
const sqrtN = Math.sqrt(n);
```

### Step 2: Iterate downwards to find the first factor of `n`

Now, we'll iterate downwards from the square root of `n`

to find the first factor.

```
for (let i = parseInt(sqrtN); i > 0; i--) {
if (n % i === 0) {
firstFactor = i;
break;
}
}
```

### Step 3: Divide `n`

by the first factor to get the second factor

After finding the first factor, we'll divide `n`

by the first factor to get the second factor.

```
const secondFactor = n / firstFactor;
```

### Step 4: Return the two factors

Finally, we'll return the two factors as the solution.

```
return `${firstFactor} ${secondFactor}`;
```

## Final Solution Code

Here is the complete solution code in JavaScript:

```
function findFactors(n) {
const sqrtN = Math.sqrt(n);
let firstFactor;
for (let i = parseInt(sqrtN); i > 0; i--) {
if (n % i === 0) {
firstFactor = i;
break;
}
}
const secondFactor = n / firstFactor;
return `${firstFactor} ${secondFactor}`;
}
```

## FAQs

### Q1: Can I solve this problem using a different programming language?

**A1**: Yes, you can solve this problem using other programming languages as well. The logic and approach remain the same, but the syntax and functions may vary depending on the language you choose. HackerRank supports multiple programming languages, including Python, Java, C++, and more.

### Q2: Can I use a brute force approach to solve this problem?

**A2**: While a brute force approach might work for smaller values of `n`

, it is not efficient for larger values. Using the square root method, as shown in this guide, provides a more efficient and optimized solution.

### Q3: Why do we need to calculate the square root of `n`

?

**A3**: We calculate the square root of `n`

because it helps us find the two factors with the minimum absolute difference. By starting from the square root and iterating downwards, we can quickly find the closest pair of factors.

### Q4: Is there any other way to optimize this solution?

**A4**: The solution provided in this guide is already optimized for solving the Find the Factor Challenge. The use of the square root method ensures that we find the closest pair of factors efficiently.

### Q5: How can I practice more similar problems on HackerRank?

**A5**: HackerRank has a vast collection of coding challenges that you can practice to improve your problem-solving skills. Visit the HackerRank Practice page and choose a domain or specific challenge to start practicing.