# Generating Permutations of a String in Lexicographic Order Without Recursion

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💡 Problem Formulation: The task is to generate all permutations of a given string in lexicographic (dictionary) order without using a recursive approach. For example, given the string “BAC”, the desired output would be a list of permutations: [“ABC”, “ACB”, “BAC”, “BCA”, “CAB”, “CBA”]. This task can be particularly challenging as recursion is a common method to tackle permutation problems, and avoiding it may require more intricate use of loops and data structures.

## Method 1: Using the itertools module

The itertools module in Python provides a method called `permutations()`, which can be used to generate permutations of a given iterable. After generating the permutations, they can be sorted to achieve the lexicographic order.

Here’s an example:

```from itertools import permutations

def lexicographic_permutations(s):
return sorted([''.join(p) for p in permutations(s)])

# Example use
print(lexicographic_permutations('ABC'))```

Output:

`['ABC', 'ACB', 'BAC', 'BCA', 'CAB', 'CBA']`

This code snippet defines a function `lexicographic_permutations()` which takes a string as input, uses `permutations()` to create all possible permutations, and then sorts them to ensure lexicographic order. The sorted list of permutations is then returned.

## Method 2: Using the next_permutation approach

This method implements the algorithm to find the next lexicographic permutation manually. It involves iterating through the sequence from the end, finding elements to swap, and reversing a segment of the sequence after the swap.

Here’s an example:

```def next_permutation(arr):
i = j = len(arr)-1
while i > 0 and arr[i-1] >= arr[i]:
i -= 1
if i == 0:   # all permutations have been generated
return False
while arr[j] <= arr[i-1]:
j -= 1
arr[i-1], arr[j] = arr[j], arr[i-1]
arr[i:] = reversed(arr[i:])
return True

def lexicographic_permutations(s):
perms = [''.join(s)]
s = list(s)
while next_permutation(s):
perms.append(''.join(s))
return perms

# Example use
print(lexicographic_permutations('ABC'))```

Output:

`['ABC', 'ACB', 'BAC', 'BCA', 'CAB', 'CBA']`

In the provided code, `next_permutation()` is a function that transforms the passed array into its next permutation in place, and returns `False` if there are no more permutations. The function `lexicographic_permutations()` continually applies this function to generate all the permutations and collect them.

## Method 3: Using the Steinhaus–Johnson–Trotter algorithm

The Steinhaus–Johnson–Trotter (SJT) algorithm generates each permutation from the previous one by interchanging a single pair of adjacent elements. To use this algorithm without recursion, we must keep track of each element’s direction and find the right elements to swap at each step.

Here’s an example:

```# This example is a simplified demonstration for readability.
# Full implementation details of SJT are more complex.

def lexicographic_permutations(s):
# SJT implementation is omitted for brevity.
pass

# Example use (this code will not run as-is)
# print(lexicographic_permutations('ABC'))```

This code snippet illustrates that an SJT algorithm can be used; however, due to its complexity, it is not fully implemented here. If implemented, it would generate permutations without using recursion, by moving elements up or down as per their directional flags.

## Method 4: Using a custom function with iterative loops

This method involves creating a custom function that iterates through the string, selects each element as the starting point, and then iterates through the remaining elements to generate permutations. This continues until all permutations are covered.

Here’s an example:

```# A simplified example - full iterative loop implementation is more extensive.
def lexicographic_permutations(s):
# Iterative loop permutation logic is omitted for brevity.
pass

# Example use (this code will not run as-is)
# print(lexicographic_permutations('ABC'))```

While this code snippet indicates the approach of using iterative loops, the extensive logic involved in making sure that all permutations are covered without repeating and in lexicographic order is not displayed due to complexity.

## Bonus One-Liner Method 5: Using a generator expression with sorted()

A one-liner approach to generate permutations in the desired order could involve using a generator expression together with the sorted function.

Here’s an example:

```from itertools import permutations

# Example use
print(sorted(''.join(chars) for chars in permutations('ABC')))```

Output:

`['ABC', 'ACB', 'BAC', 'BCA', 'CAB', 'CBA']`

The single line of code here uses a generator expression inside the `sorted()` function to generate and sort permutations in lexicographic order. This is an elegant and concise way to solve the problem but ultimately relies on the `itertools` module’s `permutations()` function.

## Summary/Discussion

• Method 1: itertools.permutations(): This method is straightforward and relies on Python’s built-in libraries. It is simple to use but may not be optimal for large strings because it has to generate all permutations before sorting them.
• Method 2: Next_permutation approach: This method is efficient as it generates each permutation from the previous one without the need for sorting. However, the implementation is a bit more complex than using itertools.
• Method 3: Steinhaus–Johnson–Trotter algorithm: The SJT algorithm is efficient and generates permutations in lexicographic order by design. However, it is complex and more difficult to implement without recursion.
• Method 4: Custom iterative loops: This approach offers full control over the generation process and can be optimized in various ways. Yet, the implementation can become quite intricate and hard to debug.
• Bonus Method 5: One-liner with generator expression: The simplicity and readability of this method make it attractive for short strings or quick tasks, but it does not stray away from using built-in functionalities like those in itertools.