# 5 Best Ways to Find the N-th Lexicographically Permutation of a String in Python

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π‘ Problem Formulation: Finding the n-th lexicographically permutation of a string involves generating the string sequence that would appear at position n if all permutations were ordered alphabetically. For example, given the string “ABC” and n = 2, the desired output is “ACB”, which is the second permutation when listed in lexicographic order.

## Method 1: Using the itertools Library

The `itertools` library in Python includes a function called `permutations()`, which returns successive r-length permutations of elements in the iterable. By specifying the length of permutations and converting the iterable to a sorted list, we can directly index the n-th permutation (zero-indexed).

Here’s an example:

```from itertools import permutations

def nth_permutation(string, n):
return ''.join(sorted(permutations(string))[n - 1])

# Example usage
print(nth_permutation("ABC", 2))```

Output: “ACB”

This code first imports the permutations function from itertools. It then defines the nth_permutation function that computes all permutations of a given string, sorts them, and retrieves the (n-1)-th item (to account for zero-indexing) as the desired permutation. Finally, it outputs the second permutation of the string “ABC”.

## Method 2: Using the Sorted Function and List Indexing

With the sorted function and list indexing, you can compute permutations of a string, sort them, and access the desired n-th position directly without itertools. This is effective for short strings, as it might not be efficient for strings with a large number of permutations due to the space complexity.

Here’s an example:

```def nth_permutation(string, n):
from math import factorial
result = ""
s_list = sorted(string)
while s_list:
f = factorial(len(s_list) - 1)
pos = (n - 1) // f
result += s_list.pop(pos)
n -= pos * f
return result

# Example usage
print(nth_permutation("ABC", 2))```

Output: “ACB”

This snippet first calculates the number of permutations that can be made with the remaining letters. It then finds the index of the next character to be added to the result by dividing the desired permutation number by the number of permutations possible with the remaining characters. The selected character is removed from the list, and the process is repeated until no characters are left.

## Method 3: Custom Algorithm Without Libraries

Creating a custom function to find the n-th lexicographic permutation without using any extra libraries can be achieved with a bit of algorithmic work. Utilizing mathematical principles of permutation, this function calculates the position of each character and constructs the desired permutation iteratively.

Here’s an example:

```def nth_permutation(string, n):
result, s_list = '', sorted(string)
k, fact = n - 1, [1] * len(s_list)
for i in range(1, len(s_list)):
fact[i] = i * fact[i - 1]
for i in reversed(range(len(s_list))):
div, k = divmod(k, fact[i])
result += s_list.pop(div)
return result

# Example usage
print(nth_permutation("ABC", 2))```

Output: “ACB”

This function computes an array of factorial values to determine the sequence. By looping through the string’s indices in reverse, it calculates which characters should be added based on the remainder (modulus) and quotient of the current k value and the factorial of the current index. Then it constructs the result by popping the characters off the sorted list.

## Method 4: Utilizing Recursion

Recursion can also be a powerful approach in generating permutations by reducing the problem size at each step. The algorithm recursively builds permutations by swapping characters and keeps track of the current permutation count.

Here’s an example:

```def nth_permutation_helper(s_list, n, count):
if len(s_list) == 1:
count[0] += 1
if count[0] == n:
return True, s_list
return False, None
for i in range(len(s_list)):
s_list[0], s_list[i] = s_list[i], s_list[0]
found, perm = nth_permutation_helper(s_list[1:], n, count)
if found:
return True, [s_list[0]] + perm
s_list[0], s_list[i] = s_list[i], s_list[0]
return False, None

def nth_permutation(string, n):
count = [0]
_, result = nth_permutation_helper(list(string), n, count)
return ''.join(result) if result else ''

# Example usage
print(nth_permutation("ABC", 2))```

Output: “ACB”

This code snippet recursively generates permutations by swapping elements at each level of the function calls and counting until it reaches the desired permutation count. Once reached, it propagates the permutation back up the recursion stack as the result.

## Bonus One-Liner Method 5: Using Next and Islice

Python’s itertools library has powerful utilities like `next()` and `islice()` which can be combined to find permutations concisely. This one-liner uses next to skip to the n-th element of the permutations iterator, sliced by islice.

Here’s an example:

```from itertools import permutations, islice

def nth_permutation(string, n):
return ''.join(next(islice(permutations(sorted(string)), n - 1, None)))

# Example usage
print(nth_permutation("ABC", 2))```

Output: “ACB”

By feeding the permutations generator into `islice()` with the appropriate starting index (n-1 for zero-indexing), and stepping through the generator with `next()`, Python will automatically handle the iteration internally and return the n-th permutation efficiently.

## Summary/Discussion

• Method 1: itertools Library. Convenient and readable. Might be inefficient for large strings due to the space complexity.
• Method 2: Sorted Function and List Indexing. Intuitive method without extra libraries. Not as efficient as itertools for large datasets.
• Method 3: Custom Algorithm Without Libraries. Offers fine control over the process. More complex to understand and implement correctly.
• Method 4: Utilizing Recursion. Good for understanding permutation mechanics. Can be inefficient and risk stack overflow for large n.
• Method 5: Using Next and Islice. A concise one-liner. Efficient for large n but might be less readable to those unfamiliar with itertools.