Python - Permutation Coefficient Problem

Python – Permutation Coefficient Problem

The permutation coefficient, commonly represented as P(n, k) or ⁿP_k, calculates the number of ways to choose and arrange k items from n distinct items. It is mathematically defined as:

P(n, k) = n! / (n – k)!

This tutorial explains how to solve the Permutation Coefficient Problem using Dynamic Programming in Python. We will break down the problem, present sample inputs and outputs, and provide two dynamic programming approaches with detailed explanations for each example.

Problem Statement

Given two integers n and k (with 0 ≤ k ≤ n), compute the permutation coefficient P(n, k). This coefficient represents the number of ways to select and arrange k items out of n distinct items.

If k is greater than n, the permutation coefficient is defined as 0, since it is not possible to arrange more items than available.

Sample Input and Output

Example 1:

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# Input:
n = 5, k = 3

# Calculation:
# P(5, 3) = 5! / (5-3)! = 5! / 2! = (5×4×3×2×1) / (2×1) = 60

# Output:
60

Explanation: There are 60 distinct ways to arrange 3 items from a set of 5. This is calculated by multiplying the top 3 numbers in the factorial sequence: 5 × 4 × 3 = 60.

Example 2:

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# Input:
n = 10, k = 2

# Calculation:
# P(10, 2) = 10! / (10-2)! = 10! / 8! = 10×9 = 90

# Output:
90

Explanation: There are 90 distinct ways to arrange 2 items from a set of 10. This is obtained by multiplying the first two terms: 10 × 9 = 90.

Solution Approach

We can solve the permutation coefficient problem using dynamic programming by breaking it into smaller subproblems. The key recurrence relation is:

P(n, k) = n × P(n-1, k-1)

The base case for this recurrence is P(n, 0) = 1 for any n (choosing 0 items results in exactly one arrangement).

Below, we present two methods using dynamic programming: one using recursion with memoization, and the other using an iterative bottom-up approach.

Python Program

Method 1: Recursive Approach with Memoization

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def permutation_recursive(n, k, memo=None):
    """
    Compute P(n, k) using recursion with memoization.
    
    Parameters:
    n (int): Total number of items.
    k (int): Number of items to arrange.
    memo (dict): Dictionary to store computed values.
    
    Returns:
    int: Permutation coefficient P(n, k).
    """
    if memo is None:
        memo = {}
        
    # Base case: selecting 0 items yields 1 arrangement
    if k == 0:
        return 1
    # If k > n, it's not possible to select k items from n
    if k > n:
        return 0
    # Return cached result if available
    if (n, k) in memo:
        return memo[(n, k)]
        
    # Recurrence relation: P(n, k) = n * P(n-1, k-1)
    memo[(n, k)] = n * permutation_recursive(n - 1, k - 1, memo)
    return memo[(n, k)]

# Test examples
print("Method 1 - Recursive with Memoization")
print("P(5, 3):", permutation_recursive(5, 3))   # Expected output: 60
print("P(10, 2):", permutation_recursive(10, 2))  # Expected output: 90

Explanation: This recursive method breaks down the problem into smaller subproblems, caching the results in a dictionary (memo) to avoid redundant calculations. This dynamic programming technique improves efficiency, especially for larger values of n and k.

Method 2: Iterative Dynamic Programming

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def permutation_iterative(n, k):
    """
    Compute P(n, k) using an iterative dynamic programming approach.
    
    Parameters:
    n (int): Total number of items.
    k (int): Number of items to arrange.
    
    Returns:
    int: Permutation coefficient P(n, k).
    """
    # If k > n, it's impossible to choose k items from n
    if k > n:
        return 0

    # Create a DP table with dimensions (n+1) x (k+1)
    dp = [[0] * (k + 1) for _ in range(n + 1)]
    
    # Base case: For any i, P(i, 0) = 1
    for i in range(n + 1):
        dp[i][0] = 1

    # Build the table using the recurrence relation
    for i in range(1, n + 1):
        # j can be at most i and k
        for j in range(1, min(i, k) + 1):
            dp[i][j] = i * dp[i - 1][j - 1]
            
    return dp[n][k]

# Test examples
print("\nMethod 2 - Iterative Dynamic Programming")
print("P(5, 3):", permutation_iterative(5, 3))   # Expected output: 60
print("P(10, 2):", permutation_iterative(10, 2))  # Expected output: 90

Explanation: This approach builds a two-dimensional DP table where each entry dp[i][j] represents the permutation coefficient for selecting j items from i items. We initialize the base case (dp[i][0] = 1) and then iteratively fill the table using the relation dp[i][j] = i * dp[i-1][j-1]. The final answer is stored in dp[n][k].

Conclusion

In this tutorial, we explored two dynamic programming approaches to solve the Permutation Coefficient Problem:

Recursive with Memoization: Utilizes recursion and caches intermediate results for efficiency.
Iterative Dynamic Programming: Constructs a DP table iteratively to compute the result in a bottom-up manner.

Both methods efficiently compute the number of ways to arrange k items from n distinct items. Choose the approach that best suits your problem constraints and interview requirements.

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Python – Permutation Coefficient Problem