What is the time complexity of The kth Factor of n?

The optimal solution has a time complexity of O(sqrt(n)).

What data structures are used to solve The kth Factor of n?

This problem uses Math, Number Theory. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving The kth Factor of n?

Insight 1: We only need to iterate up to the square root of 'n' to find all factors. If 'i' is a factor of 'n', then 'n/i' is also a factor. Insight 2: We can optimize space by not explicitly storing all factors in a single list initially. Instead, we can keep track of factors less than or equal to the square root and factors greater than the square root separately and then combine them. The Python solution avoids explicit list creation.

Which companies ask LeetCode 1492. The kth Factor of n in interviews?

This problem is frequently asked at Expedia, IBM.

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The kth Factor of n - LeetCode 1492 Solution

The kth Factor of n - Complete Solution Guide

The kth Factor of n is LeetCode problem 1492, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given two positive integers n and k . A factor of an integer n is defined as an integer i where n % i == 0 . Consider a list of all factors of n sorted in ascending order , return the k th factor in this list or return -1 if n has less than k factors. Example 1: Input: n = 12, k = 3 Output: 3 Explanation: Factors list is [1, 2, 3, 4, 6, 12], the 3 rd factor is 3. Example 2: Input: n = 7, k = 2 Output: 7 Explanation: Factors list is [1, 7], the 2 nd factor is 7. Example 3: Input: n = 4, k

Detailed Explanation

The problem asks us to find the k-th factor of a given number 'n'. A factor is a number 'i' that divides 'n' without leaving a remainder (n % i == 0). We need to find all factors of 'n', sort them in ascending order, and return the k-th element in this sorted list. If 'n' has fewer than 'k' factors, we should return -1.

Solution Approach

The solution iterates from 1 up to the square root of 'n'. For each number 'i' in this range, it checks if 'i' is a factor of 'n'. If it is, 'i' is added to a list of 'small_factors'. If 'i * i' is not equal to 'n' (to avoid duplicates for perfect squares), 'n / i' is added to a list of 'large_factors'. After the loop, the 'large_factors' list is reversed and appended to the 'small_factors' list to form the complete list of factors in ascending order. Finally, it checks if 'k' is within the bounds of the factor list and returns the k-th factor (k-1 because of 0-based indexing) or -1 if 'k' is out of bounds. The Python solution does this without explicit lists, further optimizing space.

Step-by-Step Algorithm

Step 1: Calculate the square root of 'n'.
Step 2: Iterate from 'i' = 1 to the square root of 'n'.
Step 3: If 'i' is a factor of 'n' (n % i == 0), check if k is one. If it is, return the value of i, and decrement k
Step 4: If 'i * i' is equal to 'n' (meaning n is a perfect square), then skip the next step in the first loop.
Step 5: Second Loop is from sqrt(n) downto 1, check if the current i squared equals n. If so, continue the current loop iteration.
Step 6: check if n is divisible by i. If so, decrement k. If k is 0, return n divided by i
Step 7: If the loop finishes without finding the k-th factor, return -1.

Key Insights

Insight 1: We only need to iterate up to the square root of 'n' to find all factors. If 'i' is a factor of 'n', then 'n/i' is also a factor.
Insight 2: We can optimize space by not explicitly storing all factors in a single list initially. Instead, we can keep track of factors less than or equal to the square root and factors greater than the square root separately and then combine them. The Python solution avoids explicit list creation.
Insight 3: We need to handle the case where 'n' is a perfect square (e.g., n = 9). In this case, the square root is counted only once as a factor.

Complexity Analysis

Time Complexity: O(sqrt(n))

Space Complexity: O(1)

Topics

This problem involves: Math, Number Theory.

Companies

Asked at: Expedia, IBM.

The kth Factor of n | LeetCode 1492

Medium

Math Number Theory

Time: O(sqrt(n))Space: O(1)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

class Solution {
    public int kthFactor(int n, int k) {
        List<Integer> small_factors = new ArrayList<>();
        List<Integer> large_factors = new ArrayList<>();

        for (int i = 1; i <= Math.sqrt(n); i++) {
            if (n % i == 0) {
                small_factors.add(i);
                if (i * i != n) {
                    large_factors.add(n / i);
                }
            }
        }

        List<Integer> all_factors = new ArrayList<>(small_factors);
        Collections.reverse(large_factors);
        all_factors.addAll(large_factors);

        if (k > all_factors.size()) {
            return -1;
        }

        return all_factors.get(k - 1);
    }
}

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Problem Info

Description

Problem 1492: The kth Factor of n - Detailed explanation and solution approaches.

Topics

Math Number Theory

Companies

ExpediaIBM

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1492

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Calculate the square root of 'n'.
2
Step 2: Iterate from 'i' = 1 to the square root of 'n'.
3
Step 3: If 'i' is a factor of 'n' (n % i == 0), check if k is one. If it is, return the value of i, and decrement k
4
Step 4: If 'i * i' is equal to 'n' (meaning n is a perfect square), then skip the next step in the first loop.
5
Step 5: Second Loop is from sqrt(n) downto 1, check if the current i squared equals n. If so, continue the current loop iteration.
6
Step 6: check if n is divisible by i. If so, decrement k. If k is 0, return n divided by i
7
Step 7: If the loop finishes without finding the k-th factor, return -1.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(sqrt(n))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(sqrt(n)) because the loop iterates up to the square root of 'n'. The python solution space complexity is O(1) because it uses constant extra space and doesn't build a factors list. The other solutions' space complexity is O(sqrt(n)) because, in the worst case, they store the small factors and large factors in separate lists, each potentially containing up to sqrt(n) elements.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Iterate from 1 to n, find all factors, store them in a list, and return the k-th element. This is less efficient.

Time: O(n)Space: O(number of factors), which can be O(n) in the worst case

Trade-offs: Simpler to implement but less efficient than the square root approach. Only suitable for small values of 'n'.

✨

Key Insights

1Insight 1: We only need to iterate up to the square root of 'n' to find all factors. If 'i' is a factor of 'n', then 'n/i' is also a factor.
2Insight 2: We can optimize space by not explicitly storing all factors in a single list initially. Instead, we can keep track of factors less than or equal to the square root and factors greater than the square root separately and then combine them. The Python solution avoids explicit list creation.
3Insight 3: We need to handle the case where 'n' is a perfect square (e.g., n = 9). In this case, the square root is counted only once as a factor.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: n = 1, k = 1. The factor list is [1], and the output should be 1. The solution handles this correctly.
Edge case 2: n is a perfect square (e.g., n = 9, k = 2). The factor list is [1, 3, 9], and we need to avoid adding the square root twice. The solution uses the `i * i != n` condition to prevent duplicates. The python solution addresses this in its logic to find kth factor.
Edge case 3: k > number of factors. The solution returns -1, which is the correct behavior.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where 'n' is a perfect square and adding the square root twice to the factor list.

✗Common mistake 2: Incorrectly reversing the order of 'large_factors' and not constructing the final sorted list properly.

✗Common mistake 3: Not checking if 'k' is within the bounds of the factor list and trying to access an invalid index, leading to an error.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Finding divisors is a basic operation in cryptography, especially when dealing with prime numbers and factorization.

🏦Real-world scenario 2: In resource allocation or scheduling, you might need to find factors to divide tasks or resources evenly among different entities.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 23, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "The kth Factor of n"?

For this medium-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether number theory techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?