What is the time complexity of Permutation Sequence?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Permutation Sequence?

This problem uses Math, Recursion. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Permutation Sequence?

Insight 1: We don't need to generate all n! permutations. That would be very inefficient, especially for larger values of n. Insight 2: We can determine the digits of the k-th permutation one by one. By dividing k by the factorial of (n-1), we can figure out which number should be the first digit.

Which companies ask LeetCode 60. Permutation Sequence in interviews?

This problem is frequently asked at Adobe, Amazon, Jump Trading, X.

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Permutation Sequence - LeetCode 60 Solution

Q: How to solve Permutation Sequence?

The solution uses a mathematical approach to find the k-th permutation without generating all permutations. It iteratively determines each digit of the permutation by dividing k by factorials of decreasing numbers. It maintains a list of available numbers and removes the chosen number at each step. The algorithm cleverly uses the factorials to compute the correct index of each element in the kth permutation.

Permutation Sequence - Complete Solution Guide

Permutation Sequence is LeetCode problem 60, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 k th 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!

Detailed Explanation

The problem asks us to find the k-th permutation in a sequence of all possible permutations of the numbers 1 to n. The permutations are listed in lexicographical order. Given n and k, we need to return the string representation of the k-th permutation. For example, if n = 3, the permutations are "123", "132", "213", "231", "312", "321". If k = 3, the answer is "213". The constraints are 1 <= n <= 9 and 1 <= k <= n!.

Solution Approach

The solution uses a mathematical approach to find the k-th permutation without generating all permutations. It iteratively determines each digit of the permutation by dividing k by factorials of decreasing numbers. It maintains a list of available numbers and removes the chosen number at each step. The algorithm cleverly uses the factorials to compute the correct index of each element in the kth permutation.

Step-by-Step Algorithm

Step 1: Create a list of numbers from 1 to n.
Step 2: Decrement k by 1 because the sequence is 0-indexed in our calculation.
Step 3: Iterate from n-1 down to 0.
Step 4: Calculate the factorial of the current iterator i (i.e., (n-1)!, (n-2)!, ..., 0!).
Step 5: Calculate the index of the digit to select by dividing k by the factorial calculated in step 4 (index = k // factorial).
Step 6: Append the digit at the calculated index to the permutation string, and remove it from the list of available numbers.
Step 7: Update k to be the remainder of the division in step 5 (k = k % factorial).
Step 8: Repeat steps 3-7 until all digits are selected.
Step 9: Return the generated permutation string.

Key Insights

Insight 1: We don't need to generate all n! permutations. That would be very inefficient, especially for larger values of n.
Insight 2: We can determine the digits of the k-th permutation one by one. By dividing k by the factorial of (n-1), we can figure out which number should be the first digit.
Insight 3: After choosing the first digit, we update k to the remainder of the division and continue with the remaining digits.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Math, Recursion.

Companies

Asked at: Adobe, Amazon, Jump Trading, X.

Permutation Sequence | LeetCode 60

Hard

Math Recursion

Time: O(n^2)Space: O(n)

Solution Code

Available in:

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

class Solution {
    public String getPermutation(int n, int k) {
        List<Integer> numbers = new ArrayList<>();
        for (int i = 1; i <= n; i++) {
            numbers.add(i);
        }
        k -= 1;

        List<String> permutation = new ArrayList<>();

        for (int i = n - 1; i >= 0; i--) {
            int fact = factorial(i);
            int index = k / fact;

            permutation.add(String.valueOf(numbers.remove(index)));

            k %= fact;
        }

        return String.join("", permutation);
    }

    private int factorial(int n) {
        int fact = 1;
        for (int i = 1; i <= n; i++) {
            fact *= i;
        }
        return fact;
    }
}

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Problem Info

Description

Problem 60: Permutation Sequence - Detailed explanation and solution approaches.

Topics

Math Recursion

Companies

AdobeAmazonJump TradingX

Stats

Difficulty:Hard

Languages:4

Companies:4

Problem #:60

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

9 steps

1
Step 1: Create a list of numbers from 1 to n.
2
Step 2: Decrement k by 1 because the sequence is 0-indexed in our calculation.
3
Step 3: Iterate from n-1 down to 0.
4
Step 4: Calculate the factorial of the current iterator i (i.e., (n-1)!, (n-2)!, ..., 0!).
5
Step 5: Calculate the index of the digit to select by dividing k by the factorial calculated in step 4 (index = k // factorial).
6
Step 6: Append the digit at the calculated index to the permutation string, and remove it from the list of available numbers.
7
Step 7: Update k to be the remainder of the division in step 5 (k = k % factorial).
8
Step 8: Repeat steps 3-7 until all digits are selected.
9
Step 9: Return the generated permutation string.

💡

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(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the nested loops. The outer loop iterates 'n' times and the pop operation on list takes 'n' in worst case. The space complexity is determined by the list of numbers of size 'n' and permutation list of size 'n' to store the digits during the calculation.

📖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

Recursion

A recursive approach can be used to generate permutations, but is highly inefficient. It will exceed time limit.

Time: O(n!)Space: O(n)

Trade-offs: Recursion is not suitable for this problem because the problem requires finding a specific permutation and not generating them all.

✨

Key Insights

1Insight 1: We don't need to generate all n! permutations. That would be very inefficient, especially for larger values of n.
2Insight 2: We can determine the digits of the k-th permutation one by one. By dividing k by the factorial of (n-1), we can figure out which number should be the first digit.
3Insight 3: After choosing the first digit, we update k to the remainder of the division and continue with the remaining digits.

📚

Key Terms Explained

For Students

Recursion

A function that calls itself to solve smaller versions of the same problem.

Edge Cases

Edge case 1: n = 1, k = 1. The code correctly returns "1".
Edge case 2: n = 9, k = n!. This is handled correctly because k is decremented by 1 at the beginning, so the code will use the 0-indexed value.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to decrement k by 1. The problem gives k as a 1-based index, while our calculations are based on a 0-based index.

✗Common mistake 2: Not handling the removal of elements from the list of available numbers correctly. The index needs to be adjusted after each element is removed.

✗Common mistake 3: Incorrect factorial calculation potentially leading to an infinite loop.

✅

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: This algorithm is applicable in any situation where you need to enumerate permutations in a specific order without needing to generate all of them, such as scheduling tasks.

🏦Real-world scenario 2: Generating unique identifiers or keys based on a specific sequence.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

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

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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 31, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

❓

Frequently Asked Questions

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 "Permutation Sequence"?

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

🤖 Need help with this problem?