Counting the Ways to Win at Most K Consecutive Games in Python

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💡 Problem Formulation: This article delves into the intriguing problem of calculating the number of different ways one can win at most k consecutive games. Given the total number of games n, and an upper limit k for consecutive wins, the goal is to find all possible win/loss sequences that satisfy these conditions. For example, with n=3 games and a maximum of k=2 consecutive wins, the output should be the count of sequences such as WLL, LWL, WLW, LLW, and LLL, where ‘W’ stands for a win and ‘L’ for a loss.

Method 1: Dynamic Programming

This method employs dynamic programming to systematically break down our problem into smaller subproblems. We construct an array to keep track of the count of ways to win for each game i with up to j consecutive wins, taking into account the upper limit k.

Here’s an example:

def count_ways(n, k):
    dp = [[0]*(k+1) for _ in range(n+1)]
    for i in range(n+1):
        dp[i][0] = 1  # One way to have zero consecutive wins
    for i in range(1, n+1):
        for j in range(1, min(i, k)+1):
            dp[i][j] = dp[i-1][j-1] + dp[i-1][j]
    return sum(dp[n])

# Example usage:
print(count_ways(3, 2))

Output: 5

The function count_ways initializes a 2D list dp to store counts of winning ways and iterates through the games, progressively calculating the count. It returns the sum of all possible sequences for the n-th game while adhering to the maximum k consecutive wins.

Method 2: Recursive with Memoization

Another approach uses recursion with memoization. By defining a recursive function that computes the number of ways to win for each game taking into account the limit k, and storing these results, we avoid repeatedly solving the same subproblems.

Here’s an example:

def count_ways_memo(n, k, memo=None):
    if memo is None:
        memo = {}
    if n == 0:
        return 1
    if n in memo:
        return memo[n]
    ways = 0
    for i in range(min(n, k)):
        ways += count_ways_memo(n - i - 1, k, memo)
    memo[n] = ways
    return ways

# Example usage:
print(count_ways_memo(3, 2))

Output: 5

In this snippet, count_ways_memo is a recursive function that calculates the number of ways to win for each game. The results are stored in a dictionary memo, which prevents redundant computations and vastly increases efficiency.

Method 3: Iterative with Space Optimization

The space-optimized iterative method improves upon the dynamic programming approach by reducing space complexity. It uses a single array to keep track of the counts for the ongoing iteration instead of maintaining a full 2D array.

Here’s an example:

def count_ways_iterative(n, k):
    ways = [0] * (k + 1)
    ways[0] = 1
    for _ in range(1, n + 1):
        for i in reversed(range(1, k + 1)):
            ways[i] = ways[i] + ways[i - 1]
    return sum(ways)

# Example usage:
print(count_ways_iterative(3, 2))

Output: 5

The function count_ways_iterative uses a single list, ways, to accumulate the result. This method iteratively computes the solution with a bottom-up approach, updating the list values while traversing them in reverse to avoid overwriting needed values.

Method 4: Mathematical Combinatorics

For a more mathematically inclined solution, we can solve the problem using combinatorics. This involves counting how many sequences of wins and losses can be formed under the constraints provided by k.

Here’s an example:

from math import comb

def count_ways_combinatorics(n, k):
    return sum(comb(n-i, i) for i in range(k+1))

# Example usage:
print(count_ways_combinatorics(3, 2))

Output: 5

The function count_ways_combinatorics takes advantage of the combinatorics function comb, which computes combinations. It iterates up to k to calculate the amount of distinct sequences where the streak of wins does not exceed the given limit.

Bonus One-Liner Method 5: Functional Approach

If you love a compact, functional style of coding, here’s a one-liner that utilizes Python’s itertools alongside list comprehension to solve the problem.

Here’s an example:

from itertools import product

count_ways_oneliner = lambda n, k: sum(1 for sequence in product('WL', repeat=n) if 'W'*(k+1) not in ''.join(sequence))

# Example usage:
print(count_ways_oneliner(3, 2))

Output: 5

This one-liner defines a lambda function count_ways_oneliner that generates all possible sequences of ‘W’ and ‘L’ for n games using product and filters out those with more than k consecutive wins.

Summary/Discussion

  • Method 1: Dynamic Programming. Highly efficient and systematic. May use significant space for larger inputs.
  • Method 2: Recursive with Memoization. Easy to understand. Uses recursion which could lead to stack overflow on large inputs.
  • Method 3: Iterative with Space Optimization. More space-efficient than Method 1. Does not have recursion depth limitations.
  • Method 4: Mathematical Combinatorics. Elegant and fast for small k. The efficiency decreases as k grows larger.
  • Bonus Method 5: Functional Approach. Simple and concise. Less efficient due to generating all sequences.