How do you solve Permutation Sequence on LeetCode?

Permutation Sequence (LeetCode #60) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding factorial growth helps in optimizing the solution.

What is the brute force solution for Permutation Sequence?

The brute force approach for Permutation Sequence is Brute Force, with O(n!) time complexity and O(n) space complexity. The brute force approach generates all possible permutations of the numbers from 1 to n and then sorts them to find the k-th permutation. This is straightforward but inefficient for larger n due to the factorial growth of permutations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Permutation Sequence?

Permutation Sequence uses the following concepts: Math, Recursion. The recommended patterns to study are: Math, Backtracking.

What are the interview tips for Permutation Sequence?

Explain your thought process clearly while coding. Discuss the trade-offs between brute force and optimal solutions. Practice writing clean and efficient code.

What are common mistakes when solving Permutation Sequence?

Forgetting to convert k to 0-indexed. Not handling the removal of elements correctly in the list.

#60

Permutation Sequence

Hard

MathRecursionMathBacktracking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a mathematical approach to directly compute the k-th permutation without generating all permutations. It leverages factorials to determine the correct digits in the sequence based on the value of k.

⚙️

Algorithm

4 steps

1Step 1: Create a list of numbers from 1 to n.
2Step 2: Calculate the factorial values for numbers from 1 to n-1.
3Step 3: Determine the correct digit for each position by using k and the factorial values to find the index in the list of available numbers.
4Step 4: Remove the used number from the list and repeat until the permutation is complete.

solution.py14 lines

1# Full working Python code
2import math
3
4def get_permutation(n, k):
5    nums = list(range(1, n + 1))
6    k -= 1  # Convert to 0-indexed
7    permutation = []
8    factorial = [math.factorial(i) for i in range(n)]
9    for i in range(n):
10        index = k // factorial[n - 1 - i]
11        permutation.append(str(nums[index]))
12        nums.pop(index)
13        k %= factorial[n - 1 - i]
14    return ''.join(permutation)

ℹ

Complexity note: The time complexity is O(n) since we perform a constant amount of work for each of the n digits. The space complexity is O(n) for storing the list of numbers and the resulting permutation.

1Understanding factorial growth helps in optimizing the solution.
2Directly calculating the k-th permutation avoids unnecessary computations.

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