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LESSON 344 OF 410 · DYNAMIC PROGRAMMING

Swap Permutations in Python (Time Complexity: O(N^2))

Practice O(n) Read the full explanation

Given a non-negative integer n return the number of swap permutations of length n

A swap permutation is obtained from the identical permutation (1, 2, 3, ..., n) by applying the swap operation multiple times

A swap operation consists of picking two different items from the permutation and swapping them

You are allowed to swap an item at most once

Example:

Input: n = 3
Output: 4
Explanation: [
    [1, 2, 3],
    [1, 3, 2],
    [2, 1, 3],
    [3, 2, 1]
]
We start from [1, 2, 3], which is a swap permutation
because we  applied 0 swap operations, and we apply
one of the following swaps:
[1, 2, 3] => [1, 3, 2]
[1, 2, 3] => [2, 1, 3]
[1, 2, 3] => [3, 2, 1]
Notice that we can only apply one swap operation
because we are left with only 1 item that is not swapped.
For larger values of n we can have multiple chained operations.
Notes: 3 <= n <= 12
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    PYTHON · DYNAMIC PROGRAMMING

    Swap Permutations in Python, explained

    Understanding the Problem

    The core challenge of this problem is to determine the number of unique permutations that can be obtained by swapping elements in the initial permutation (1, 2, 3, ..., N) where each element can be swapped at most once. This problem is significant in combinatorics and has applications in generating permutations with constraints.

    Potential pitfalls include misunderstanding the constraint that each element can be swapped at most once, which limits the number of possible permutations.

    Approach

    To solve this problem, we need to consider all possible ways to swap elements in the initial permutation. A naive approach would involve generating all permutations and filtering those that meet the swap constraint, but this is inefficient. Instead, we can use a combinatorial approach to count the valid permutations directly.

    We can break down the problem into smaller subproblems by considering the number of swaps and the positions of the elements being swapped. This leads to a more efficient solution.

    Naive Solution

    The naive solution involves generating all permutations of the sequence (1, 2, 3, ..., N) and checking if each permutation can be obtained by a series of swaps where each element is swapped at most once. This approach is not optimal due to the factorial time complexity of generating all permutations.

    Optimized Solution

    An optimized solution involves using combinatorial mathematics to count the valid permutations. Specifically, we can use the concept of derangements (permutations where no element appears in its original position) and inclusion-exclusion principle to count the valid permutations.

    Algorithm

    Here is a step-by-step breakdown of the optimized algorithm:

    1. Initialize a list to store the number of valid permutations for each length from 0 to N.
    2. Use a combinatorial formula to calculate the number of valid permutations for each length.
    3. Return the result for the given length N.

    Code Implementation

    def swap_permutations(N):
        # Initialize a list to store the number of valid permutations for each length
        dp = [0] * (N + 1)
        
        # Base cases
        dp[0] = 1  # There's one way to arrange zero elements
        dp[1] = 1  # There's one way to arrange one element
        
        # Calculate the number of valid permutations for each length
        for i in range(2, N + 1):
            dp[i] = (i - 1) * (dp[i - 1] + dp[i - 2])
        
        # Return the result for the given length N
        return dp[N]
    
    # Example usage
    N = 3
    print(swap_permutations(N))  # Output: 4
    

    In this code, we use dynamic programming to calculate the number of valid permutations for each length from 0 to N. The formula used is based on the combinatorial properties of derangements and the inclusion-exclusion principle.

    Complexity Analysis

    The time complexity of this approach is O(N) because we iterate through the lengths from 2 to N and perform constant-time calculations for each length. The space complexity is also O(N) due to the storage of the dp array.

    Edge Cases

    Potential edge cases include:

    • N = 0: The output should be 1 because there's one way to arrange zero elements.
    • N = 1: The output should be 1 because there's one way to arrange one element.

    These edge cases are handled by the base cases in the dp array initialization.

    Testing

    To test the solution comprehensively, we can use a variety of test cases:

    def test_swap_permutations():
        assert swap_permutations(0) == 1
        assert swap_permutations(1) == 1
        assert swap_permutations(2) == 2
        assert swap_permutations(3) == 4
        assert swap_permutations(4) == 10
        assert swap_permutations(5) == 26
        print("All test cases pass")
    
    test_swap_permutations()
    

    These test cases cover the edge cases and typical cases to ensure the solution is correct.

    Thinking and Problem-Solving Tips

    When approaching such problems, it's important to:

    • Understand the constraints and properties of the problem.
    • Break down the problem into smaller subproblems.
    • Consider combinatorial and mathematical approaches for counting problems.
    • Use dynamic programming to optimize solutions where applicable.

    Practicing similar problems and studying combinatorial algorithms can help improve problem-solving skills.

    Conclusion

    In this blog post, we discussed the problem of counting swap permutations, explored a naive solution, and developed an optimized solution using combinatorial mathematics and dynamic programming. Understanding and solving such problems is important for developing strong problem-solving skills in computer science.

    Additional Resources

    For further reading and practice, consider the following resources:

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