How to Generate All Permutations of an Array Recursively in Python?

How can I generate all permutations of an array using recursion in Python?

def generate_permutations(nums):
    if len(nums) == 0:
        return []
    if len(nums) == 1:
        return [nums]

    permutations = []
    for i in range(len(nums)):
        current_num = nums[i]
        remaining_nums = nums[:i] + nums[i+1:]
        for p in generate_permutations(remaining_nums):
            permutations.append([current_num] + p)
    return permutations

Generating all permutations of an array recursively involves selecting each element and recursively permuting the remaining elements. This method will yield every possible arrangement of the array's elements.

def generate_permutations(nums):
    if len(nums) == 0:
        return []
    if len(nums) == 1:
        return [nums]

    permutations = []
    for i in range(len(nums)):
        current_num = nums[i]
        remaining_nums = nums[:i] + nums[i+1:]
        for p in generate_permutations(remaining_nums):
            permutations.append([current_num] + p)
    return permutations

# Example usage:
arr = [1, 2, 3]
print(generate_permutations(arr))

Causes

• Misunderstanding of the recursion mechanism
• Incorrect handling of base cases
• Failure to build permutations correctly from remaining elements

Solutions

• Utilize a base case to terminate recursion when the input list is empty or contains one element.
• Make sure to concatenate the selected element with permutations of the remaining elements in each recursive call.

Mistake: Not managing the base case properly causing infinite recursion.

Solution: Ensure the base case checks for both empty and single-element lists.

Mistake: Updating the same list instead of creating a copy when passing to recursive calls.

Solution: Use slicing (e.g., nums[:i] + nums[i+1:]) to get a new list without modifying the original.