Permutation Sequence

The problem asks for the k-th permutation of the set containing numbers from 1 to n. You can use a mathematical approach to solve this problem without generating all permutations.

The key insight is that there are (n-1)! permutations that start with each digit in the sequence, so you can determine the first digit by finding the (k-1) // (n-1)!-th number, then repeat the process for the remaining numbers.

Python solution:

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from math import factorial

class Solution:
    def getPermutation(self, n: int, k: int) -> str:
        numbers = [str(i) for i in range(1, n+1)]
        result = []
        k -= 1 # Convert to 0-based index

        for i in range(n, 0, -1):
            fact = factorial(i - 1)
            index = k // fact
            result.append(numbers.pop(index))
            k %= fact

        return ''.join(result)

Explanation:

• Create a list numbers of available digits (from 1 to n).
• Initialize an empty list result to build the k-th permutation.
• Convert k to 0-based indexing by subtracting 1.
• Iterate from n down to 1, calculating the factorial of the remaining digits (fact).
• Determine the index of the next digit in the permutation by using integer division of k by fact.
• Add the selected digit to the result and remove it from numbers.
• Update k using the remainder of k divided by fact.
• Concatenate the result list into a string and return it.

This code will correctly output the k-th permutation for any valid input of n and k.

Identifying Problem Isomorphism

“Permutation Sequence” asks for the kth permutation of a sequence of the first n numbers. It requires understanding of permutations and their ordering.

A simpler problem is “Permutations”, where you need to return all possible permutations of a given list of numbers. This problem is simpler as it doesn’t require identifying a specific permutation in the sequence.

A problem with approximately the same level of difficulty is “Next Permutation”, which requires modifying the input sequence to the next lexicographically greater permutation. It’s similar because it also involves knowledge of the order of permutations, but differs as it’s looking for the next permutation rather than a specific one.

A more complex isomorphic problem could be “Minimum Adjacent Swaps to Reach the Kth Smallest Number”. In this problem, you have a number given as a string and are asked to find the minimum number of adjacent swaps required to reach the kth smallest number possible. It increases the complexity by requiring you to track the number of swaps, and having the additional constraint of only being able to swap adjacent digits.

If we rank these problems in terms of increasing complexity:

1. “Permutations” - Generate all permutations of a list of numbers.
2. “Next Permutation” - Find the next lexicographically greater permutation of a sequence.
3. “Permutation Sequence” - Identify a specific permutation in the sequence of all permutations.
4. “Minimum Adjacent Swaps to Reach the Kth Smallest Number” - Find the kth smallest permutation and the number of swaps to reach it.

10 Prerequisite LeetCode Problems

“60. Permutation Sequence” is a combinatorics and backtracking problem where you need to find the kth permutation sequence of n digits. Here are some problems as preparation for it:

1. 46. Permutations: Before you can understand the kth permutation, it might be helpful to solve a problem that requires you to find all permutations.

2. 47. Permutations II: This problem is a slightly more complicated version of the previous one, as it deals with repeated numbers.

3. 77. Combinations: This problem requires understanding of combinatorics, which is essential to solve your target problem.

4. 78. Subsets: Understanding how to generate subsets is a stepping stone to understanding permutations.

5. 79. Word Search: This problem will help you understand backtracking, a key aspect of generating permutations.

6. 90. Subsets II: This problem is a harder version of the Subsets problem, and is good practice for generating permutations.

7. 93. Restore IP Addresses: This problem also uses backtracking to generate all possible solutions, which is similar to your target problem.

8. 131. Palindrome Partitioning: This problem is another good practice for backtracking and permutations.

9. 216. Combination Sum III: This problem also uses combinatorics and is a good practice before tackling more complex permutation problems.

10. 526. Beautiful Arrangement: This problem is about generating permutations under certain constraints.

These cover combinatorics and backtracking, which will be helpful for solving the “60. Permutation Sequence” problem.

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class Solution:
    def getPermutation(self, n: int, k: int) -> str:
        nums = [i for i in range(1, n+1)] 
        factorial = [1] * n
        for i in range(1, n):
            factorial[i] = factorial[i-1] * i

        k -= 1
        result = []
        for i in range(n-1, -1, -1):
            index = k // factorial[i]
            result.append(str(nums[index]))
            nums.pop(index)
            k = k % factorial[i]

        return ''.join(result)

Problem Classification

This problem involves determining the nth permutation of a sequence. It involves understanding the order of permutations and how to generate them, hence it falls under Combinatorial Analysis / Permutations.

This problem asks for the nth arrangement of a sequence. It’s about understanding the order of arrangements and how to create them, so it’s Arrangement Analysis.

Language Agnostic Coding Drills

For the final solution, you need to combine all of these concepts. The problem is solved using a mathematical approach, where for each number from n to 1, we calculate the index of the number that should be added to the result, then we update k and repeat the process until we get the kth permutation.

Here is the breakdown:

• Create a list of numbers and a list of factorial values.
• Decrement k by one because the index is zero-based.
• Then for each number from n to 1, calculate the index of the number that should be added to the result (index = k // factorial[i]).
• Add the number at the calculated index to the result and remove it from the list of numbers.
• Update k to be the remainder of the division (k = k % factorial[i]).
• Repeat the process until you get the kth permutation.
• Finally, join the result list into a string and return it.

Targeted Drills in Python

1. List comprehension in Python: List comprehension is a concise way to create lists. It consists of brackets containing an expression followed by a for statement, then zero or more for or if clauses.

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   # Create a list of numbers from 1 to n n = 5 nums = [i for i in range(1, n+1)] print(nums) # Should print [1, 2, 3, 4, 5]

2. Creating and using factorial list: In this problem, a list of factorial values is created. Here is a simple drill to create and manipulate such a list.

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   n = 5 factorial = [1] * n for i in range(1, n): factorial[i] = factorial[i-1] * i print(factorial) # Should print [1, 1, 2, 6, 24]

3. Integer division and modulus operations: Integer division (//) and modulus (%) are frequently used operations in programming problems.

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   # Integer division print(10 // 3) # Should print 3 # Modulus operation print(10 % 3) # Should print 1

4. List pop operation: Python lists have a built-in pop(index) method that removes the item at the specified index and returns it.

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   nums = [1, 2, 3, 4, 5] element = nums.pop(2) print(element) # Should print 3 print(nums) # Should print [1, 2, 4, 5]

5. Joining a list of strings: In Python, you can join a list of strings into a single string using the join() method.

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   words = ['This', 'is', 'a', 'sentence.'] sentence = ' '.join(words) print(sentence) # Should print "This is a sentence."