How do you solve The kth Factor of n on LeetCode?

The kth Factor of n (LeetCode #1492) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Factors are numbers that divide n evenly.

What is the brute force solution for The kth Factor of n?

The brute force approach for The kth Factor of n is Brute Force, with O(n) time complexity and O(n) space complexity. The brute force approach involves checking every number from 1 to n to see if it is a factor of n. If it is, we add it to a list of factors and then return the k-th factor from that list. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve The kth Factor of n?

The kth Factor of n uses the following concepts: Math, Number Theory. The recommended patterns to study are: Brute Force, Counting, Early Exit.

What are the interview tips for The kth Factor of n?

Clarify the problem statement and constraints before coding. Discuss your thought process and any edge cases you consider. Write clean and efficient code, and explain your reasoning.

What are common mistakes when solving The kth Factor of n?

Not checking if k is within the valid range. Confusing 0-based and 1-based indexing when returning the k-th factor.

#1492

The kth Factor of n

Medium

MathNumber TheoryBrute ForceCountingEarly Exit

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(n)
Space	O(n)	O(1)

💡

Intuition

Time O(n)Space O(1)

Instead of storing all factors, we can directly count them until we reach the k-th factor. This way, we avoid using extra space for storing all factors and stop early if we find the k-th factor.

⚙️

Algorithm

5 steps

1Step 1: Initialize a count variable to track the number of factors found.
2Step 2: Loop through all integers from 1 to n.
3Step 3: For each integer, check if it divides n evenly (n % i == 0). If it does, increment the count.
4Step 4: If the count equals k, return the current integer i as the k-th factor.
5Step 5: If the loop ends and the count is less than k, return -1.

solution.py10 lines

1# Full working Python code
2
3def kth_factor(n, k):
4    count = 0
5    for i in range(1, n + 1):
6        if n % i == 0:
7            count += 1
8            if count == k:
9                return i
10    return -1

ℹ

Complexity note: The time complexity remains O(n) since we still check each number from 1 to n. The space complexity is O(1) because we do not use any additional space to store factors.

1Factors are numbers that divide n evenly.
2We can optimize by counting factors instead of storing them.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.