# Solving Leetcode Interviews in Seconds with AI: Permutation Sequence


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "60" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> The set [1, 2, 3, ..., n] contains a total of n! unique permutations. By listing and labeling all of the permutations in order, we get the following sequence for n = 3:  "123" "132" "213" "231" "312" "321"  Given n and k, return the kth permutation sequence.   Example 1: Input: n = 3, k = 3 Output: "213" Example 2: Input: n = 4, k = 9 Output: "2314" Example 3: Input: n = 3, k = 1 Output: "123"    Constraints:  1 <= n <= 9 1 <= k <= n!  

	# Explanation
	Here's the breakdown of the approach, complexities, and the Python code:

*   **Approach:**
    *   Calculate factorials up to `n`. Determine which number should be at each position by dividing `k` by the factorial of the remaining digits.
    *   Maintain a list of available numbers. Update `k` after selecting each digit to reflect the remaining rank within the smaller set of permutations.

*   **Complexity:**
    *   Runtime: O(n^2) due to list removals. Storage: O(n) for factorial list and candidates list.

	
	# Code
	```python
	def getPermutation(n: int, k: int) -> str:
    """
    Returns the kth permutation sequence of numbers from 1 to n.
    """

    factorials = [1] * (n + 1)
    for i in range(2, n + 1):
        factorials[i] = factorials[i - 1] * i

    numbers = list(range(1, n + 1))
    k -= 1  # Adjust k to be 0-indexed
    result = ""

    for i in range(n, 0, -1):
        index = k // factorials[i - 1]
        result += str(numbers[index])
        numbers.pop(index)
        k %= factorials[i - 1]

    return result
	```