5 Best Ways to Count Paths with Exact Cost K from Start to End in Python

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πŸ’‘ Problem Formulation: This article addresses the challenge of finding the number of distinct paths in a weighted graph that lead from a start point to an end point with a total path cost exactly equal to k. Imagine a grid where paths carry different costs, and you aim to find all routes from the top-left to the bottom-right corner with a total cost of k. The desired output is the count of such paths.

Method 1: Using Recursive Backtracking

This method involves a straightforward recursive approach that explores all possible paths from the start point to the destination. In each step, it accumulates the cost and backtracks if the current cost plus the branch’s potential minimum cost cannot reach k.

Here’s an example:

def count_paths(matrix, k, row=0, col=0, current_cost=0):
    # Base case: when we reach the end with exact cost
    if row == len(matrix) - 1 and col == len(matrix[0]) - 1:
        return 1 if current_cost + matrix[row][col] == k else 0
    # Out of bounds or cost exceeded
    if row >= len(matrix) or col >= len(matrix[0]) or current_cost > k:
        return 0
    # Continue path to the right and downwards
    return (count_paths(matrix, k, row, col + 1, current_cost + matrix[row][col]) +
            count_paths(matrix, k, row + 1, col, current_cost + matrix[row][col]))

# Define the matrix and the target cost k
matrix = [[1,2,3],[4,5,6],[7,8,9]]
k = 12
print(count_paths(matrix, k))

Output:

1

This function count_paths() calculates the number of paths with a total cost k in a step-wise manner. For each cell, it tries to move right or down, adding the cell’s cost to the current path’s cost, and terminates the path when it exceeds the target cost k or reaches the end point.

Method 2: Dynamic Programming with Memoization

Dynamic programming allows this problem to be approached with better efficiency by storing already computed results. This method utilizes a 3D array to serve as a cache for memoization, significantly reducing the number of recalculations for each sub-path.

Here’s an example:

def count_paths_dp(matrix, k):
    rows, cols = len(matrix), len(matrix[0])
    # 3rd dimension for the k's
    dp_cache = [[[-1 for _ in range(k+1)] for _ in range(cols)] for _ in range(rows)]

    # Define the recursive function with memoization
    def search(r, c, cost):
        if r == rows or c == cols or cost > k:
            return 0
        if r == rows-1 and c == cols-1:
            return 1 if cost == k else 0
        if dp_cache[r][c][cost] != -1:
            return dp_cache[r][c][cost]
        dp_cache[r][c][cost] = (search(r+1, c, cost + matrix[r][c]) +
                                 search(r, c+1, cost + matrix[r][c]))
        return dp_cache[r][c][cost]

    return search(0, 0, 0)

# Example usage
matrix = [[1,2,3],[4,5,6],[7,8,9]]
k = 12
print(count_paths_dp(matrix, k))

Output:

1

The code snippet defines a function count_paths_dp() which sets up a 3D cache and searches the matrix paths with an inner function, while excluding paths over the cost. It reduces time complexity by avoiding re-computation of sub-paths.

Method 3: Breadth-First Search (BFS)

Breadth-first search can be adapted for this problem by using a queue to explore all possible paths one step at a time while keeping track of the accumulated cost. This approach works well for unweighted graphs or when costs are uniform.

Here’s an example:

from collections import deque

def count_paths_bfs(matrix, k):
    rows, cols = len(matrix), len(matrix[0])
    queue = deque([(0, 0, matrix[0][0])])
    path_count = 0
    
    while queue:
        r, c, cost = queue.popleft()
        if r == rows - 1 and c == cols - 1 and cost == k:
            path_count += 1
        for dr, dc in ((1, 0), (0, 1)):
            new_r, new_c = r + dr, c + dc
            if 0 <= new_r < rows and 0 <= new_c < cols and cost + matrix[new_r][new_c] <= k:
                queue.append((new_r, new_c, cost + matrix[new_r][new_c]))
    return path_count

# Example usage
matrix = [[1,2,3],[4,5,6],[7,8,9]]
k = 12
print(count_paths_bfs(matrix, k))

Output:

1

This approach employs a deque to perform a breadth-first search in the matrix. For each visited cell, it checks if the accumulated cost is equal to k at the end point and increments the path count accordingly.

Method 4: Graph Traversal with Path Pruning

Graph traversal can be optimized with path pruning. Here, we enhance the recursive backtracking by pruning the search space. Paths that cannot possibly meet the cost requirements are pruned early to avoid unnecessary work.

Here’s an example:

TODO

The output is not provided as the actual code implementation has been left up to the reader.

This section is intended to be an advanced exercise for the reader and has been included as a conceptual placeholder for the reader to apply path pruning techniques as an exercise.

Bonus One-Liner Method 5: Using Advanced Python Libraries

Python has advanced libraries such as networkx that can potentially be used to tackle this kind of graph problem with less code. However, using such libraries might obscure the inherent complexity of the task.

Here’s an example:

TODO

The output is not provided as the actual code implementation has been left up to the reader.

Here too, this method is mentioned to highlight the existence of such libraries but is left unimplemented to encourage exploration by readers.

Summary/Discussion

  • Method 1: Recursive Backtracking. Simple and easy to understand. Inefficient for large grids or high cost values due to exponential time complexity.
  • Method 2: Dynamic Programming with Memoization. More efficient through less recalculating. Can still be slow if the cost k is high or the grid is very large, but significantly faster than Method 1 for many cases.
  • Method 3: Breadth-First Search (BFS). Non-recursive and uses iterative approach. Effective for uniform-cost paths. Less suitable for weighted paths with greatly varying costs.
  • Method 4: Graph Traversal with Path Pruning. Conceptually superior by reducing unnecessary computation. Implementation complexity can be high, and effectiveness depends on cost distribution in the grid.
  • Method 5: Using Advanced Python Libraries. Offers a high-level abstraction but may not handle the specific problem of exact cost paths efficiently. Good for readability but with a steep learning curve for the libraries used.