Python Coding challenge - Day 522| What is the output of the following Python Code?

 

Code Explanation:

1. Function Definition

def count_paths(r, c):

This defines a function named count_paths that takes two parameters:

r: number of rows

c: number of columns

2. Base Case: Grid size is 0

    if r == 0 or c == 0:

        return 0

If either r or c is 0, it means there's no grid (invalid), so there are 0 paths.

This is an edge case guard to avoid negative recursion or invalid grids.

3. Base Case: At destination

    if r == 1 and c == 1:

        return 1

This checks if you're at the starting point, which is also the destination in a 1x1 grid.

In this case, there's exactly 1 path — you’re already there.

This acts as the stopping condition for the recursion.

4. Recursive Case: Count from top and left

    return count_paths(r - 1, c) + count_paths(r, c - 1)

This is the heart of the recursive logic.

To reach cell (r, c), you could have come:

from the cell above: (r-1, c)

from the cell to the left: (r, c-1)

So, total paths = paths from top + paths from left.

5. Function Call

print(count_paths(3, 3))

This calls the function with r = 3, c = 3 and prints the result.

It calculates the number of unique paths in a 3×3 grid.

Output Explanation

Let’s trace what count_paths(3, 3) does:

It breaks into:

count_paths(2, 3) + count_paths(3, 2)

Each of these breaks down similarly, and eventually reaches the base case (1,1) multiple times. After full recursion, the number of unique paths = 6.

Final Output:

6