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Leetcode #1492: The kth Factor of n

In this guide, we solve Leetcode #1492 The kth Factor of n in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

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.

Quick Facts

  • Difficulty: Medium
  • Premium: No
  • Tags: Math, Number Theory

Intuition

There is a mathematical invariant or formula that directly leads to the result.

Using math avoids unnecessary loops and reduces complexity.

Approach

Derive the formula or update rule, then compute the answer directly.

Handle edge cases like overflow or zero carefully.

Steps:

  • Identify the math relationship.
  • Compute the result with a loop or formula.
  • Handle edge cases.

Example

Input: n = 12, k = 3 Output: 3 Explanation: Factors list is [1, 2, 3, 4, 6, 12], the 3rd factor is 3.

Python Solution

class Solution: def kthFactor(self, n: int, k: int) -> int: for i in range(1, n + 1): if n % i == 0: k -= 1 if k == 0: return i return -1

Complexity

The time complexity is O(n)O(n)O(n), and the space complexity is O(1)O(1)O(1). The space complexity is O(1)O(1)O(1).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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