Python – Binomial Coefficient Problem

Python – Binomial Coefficient Problem

The binomial coefficient, often represented as C(n, k) or “n choose k”, represents the number of ways to choose k items from n items without repetition. In this tutorial, we’ll solve the Binomial Coefficient Problem using a Dynamic Programming approach, which is efficient and avoids redundant calculations.

Problem Statement

Given two integers, n and k, compute the binomial coefficient C(n, k) defined as:

C(n, k) = n! / (k! * (n-k)!)

However, calculating factorials directly can be computationally expensive and may lead to overflow. Instead, we use the following recurrence relation, which is ideal for a dynamic programming solution:

C(n, k) = C(n-1, k-1) + C(n-1, k)

with the base conditions:

1. C(n, 0) = 1 for all n
2. C(n, n) = 1 for all n

Sample Input and Output

Example 1:

Input: n = 5, k = 2
Output: 10

Explanation: There are 10 ways to choose 2 items from a set of 5.

Example 2:

Input: n = 6, k = 3
Output: 20

Explanation: There are 20 ways to choose 3 items from a set of 6.

Solution Approach

We use a bottom-up dynamic programming approach to solve this problem. The steps are as follows:

1. Initialize a DP Table: Create a 2D list dp with dimensions (n+1) x (k+1) to store the intermediate results.
2. Set Base Conditions: For all i, set dp[i][0] = 1 and dp[i][i] = 1 (when k == i).
3. Fill the Table: For each i from 1 to n and for each j from 1 to min(i, k), compute:
   
   dp[i][j] = dp[i-1][j-1] + dp[i-1][j]
   
4. Return the Result: The value at dp[n][k] is the binomial coefficient C(n, k).

Python Program

Method: Dynamic Programming Approach

def binomial_coefficient(n, k):
    """
    Compute the binomial coefficient C(n, k) using dynamic programming.
    
    Parameters:
    n (int): Total number of items.
    k (int): Number of items to choose.
    
    Returns:
    int: The binomial coefficient C(n, k).
    """
    # Create a DP table with (n+1) rows and (k+1) columns initialized to 0
    dp = [[0 for _ in range(k+1)] for _ in range(n+1)]
    
    # Base Cases:
    # For every i, C(i, 0) is 1.
    for i in range(n+1):
        dp[i][0] = 1
        
    # Fill the DP table using the recurrence relation
    for i in range(1, n+1):
        # j cannot be more than i (and k)
        # min(i, k) ensures we don't go out of bounds when k > i.
        for j in range(1, min(i, k) + 1):
            # When i equals j, C(i, i) is 1
            if i == j:
                dp[i][j] = 1
            else:
                dp[i][j] = dp[i-1][j-1] + dp[i-1][j]
    return dp[n][k]

# Example 1:
n1, k1 = 5, 2
result1 = binomial_coefficient(n1, k1)
print(f"C({n1}, {k1}) = {result1}")  # Expected Output: 10

# Example 2:
n2, k2 = 6, 3
result2 = binomial_coefficient(n2, k2)
print(f"C({n2}, {k2}) = {result2}")  # Expected Output: 20

Output for Example 1 and 2:

C(5, 2) = 10
C(6, 3) = 20

Explanation: In the program, we initialize a 2D list dp to store binomial coefficients for subproblems. The nested loops fill this table based on the recurrence relation. The base case dp[i][0] = 1 is set for all i, and when i == j, we set dp[i][j] = 1 because there is only one way to choose all items. Finally, the value dp[n][k] gives us the desired result.

Conclusion

Dynamic Programming provides an efficient way to compute the binomial coefficient by breaking down the problem into smaller overlapping subproblems. The bottom-up approach fills a DP table using a simple recurrence relation, ensuring that each subproblem is solved only once. This method is both time-efficient (O(n*k)) and space-efficient when optimized.