Solution 1: Dynamic Programming

This problem assesses your ability to use dynamic programming to efficiently find the number of unique paths in a grid.

Here’s a Python implementation of the unique_paths function using dynamic programming:

def unique_paths(m: int, n: int) -> int:
    # Create a 2D array to store the number of unique paths for each cell
    dp = [[0] * n for _ in range(m)]

    # Initialize the top row and leftmost column to 1
    for i in range(m):
        dp[i][0] = 1
    for j in range(n):
        dp[0][j] = 1

    # Fill in the rest of the array using dynamic programming
    for i in range(1, m):
        for j in range(1, n):
            dp[i][j] = dp[i - 1][j] + dp[i][j - 1]

    return dp[-1][-1]  # Number of unique paths for the bottom-right corner

# Example usage:
result1 = unique_paths(3, 7)
print(result1)  # Output: 28

result2 = unique_paths(3, 2)
print(result2)  # Output: 3

result3 = unique_paths(2, 2)
print(result3)  # Output: 2

result4 = unique_paths(1, 1)
print(result4)  # Output: 1

This solution uses dynamic programming to fill in a 2D array dp where dp[i][j] represents the number of unique paths to reach the cell at row i and column j.

Time Complexity:

The time complexity is O(m * n), where m is the number of rows and n is the number of columns in the grid.

Space Complexity:

The space complexity is O(m * n) due to the 2D array.

Solution 2: Backtracking

Using backtracking for this problem is not the most efficient approach because it leads to exponential time complexity. However, for the sake of completeness, here’s an implementation using backtracking:

def unique_paths(m: int, n: int) -> int:
    def backtrack(row: int, col: int) -> int:
        # Base case: reached the bottom-right corner
        if row == m - 1 and col == n - 1:
            return 1

        # Check if moving down is within bounds
        down_paths = 0
        if row + 1 < m:
            down_paths = backtrack(row + 1, col)

        # Check if moving right is within bounds
        right_paths = 0
        if col + 1 < n:
            right_paths = backtrack(row, col + 1)

        return down_paths + right_paths

    # Start backtracking from the top-left corner
    return backtrack(0, 0)

# Example usage:
print(unique_paths(3, 7))  # Output: 28
print(unique_paths(3, 2))  # Output: 3
print(unique_paths(2, 2))  # Output: 2
print(unique_paths(1, 1))  # Output: 1

Explanation:

  • We define a recursive function backtrack that explores all possible paths from the top-left corner to the bottom-right corner.
  • At each step, we have two choices: move down or move right. We explore both choices recursively until we reach the bottom-right corner.
  • The base case of the recursion is when we reach the bottom-right corner, in which case we return 1 to indicate that we have found a valid path.
  • We keep track of the number of paths found by summing the paths obtained from moving down and moving right.

Time Complexity:

The time complexity of this solution is exponential, as it explores all possible paths in the worst case. It’s roughly O(2^(m+n)), where m is the number of rows and n is the number of columns. Let’s analyze why it is so:

  1. Number of Recursive Calls:
    • At each step of the recursion, we have two choices: to move down or to move right.
    • Therefore, each recursive call results in branching into two recursive calls.
    • As a result, the number of recursive calls grows exponentially with the size of the input grid.
  2. Recursion Depth:
    • The recursion depth corresponds to the number of steps needed to reach the bottom-right corner of the grid.
    • In the worst case, the robot needs to move m - 1 steps downwards and n - 1 steps to the right to reach the destination.
    • Thus, the recursion depth is m + n - 2.
  3. Combination of Factors:
    • Since each recursive call branches into two calls, and the recursion depth is m + n - 2, the total number of recursive calls made is roughly 2^(m+n-2).
    • Therefore, the time complexity of the backtracking solution is exponential, approximately O(2^(m+n)) where m is the number of rows and n is the number of columns in the grid.

In summary, the backtracking solution explores all possible paths from the top-left corner to the bottom-right corner of the grid, leading to an exponential time complexity. As a result, it is not suitable for large grids due to its inefficiency.

Space Complexity:

The space complexity is O(m + n) due to the recursion depth, as the maximum depth of the recursion tree is bounded by the sum of m and n.

Solution 3: Combinatorics

Another solution to find the number of unique paths is to use combinatorics. We can observe that to reach the bottom-right corner from the top-left corner, the robot must move m-1 steps downwards and n-1 steps to the right, for a total of m-1 + n-1 = m + n - 2 steps.

Out of these m + n - 2 steps, we need to choose m - 1 steps to move downwards, and the remaining steps will be to move to the right. Therefore, the number of unique paths is the number of combinations of choosing m - 1 out of m + n - 2 steps.

We can use the formula for combinations (n choose k) to compute this:

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

import math

def unique_paths(m: int, n: int) -> int:
    # Calculate the number of unique paths using combinatorics formula
    return math.comb(m + n - 2, m - 1)

# Example usage:
print(unique_paths(3, 7))  # Output: 28
print(unique_paths(3, 2))  # Output: 3
print(unique_paths(2, 2))  # Output: 2
print(unique_paths(1, 1))  # Output: 1

Explanation:

  • We use the math.comb function from the math module to compute the number of combinations of choosing m - 1 out of m + n - 2 steps.
  • This directly gives us the number of unique paths from the top-left corner to the bottom-right corner of the grid.

Applying the same logic, we can consider it as selecting n - 1 steps to move to the right, leaving the rest of the steps to move downwards:

import math

def unique_paths(m: int, n: int) -> int:
    # Calculate the number of unique paths using combinatorics formula
    return math.comb(m + n - 2, n - 1)

# Example usage:
print(unique_paths(3, 7))  # Output: 28
print(unique_paths(3, 2))  # Output: 3
print(unique_paths(2, 2))  # Output: 2
print(unique_paths(1, 1))  # Output: 1

Time Complexity:

The time complexity of this solution is O(1), as it involves simple arithmetic operations to compute the result.

Space Complexity:

The space complexity is also O(1), as we are not using any additional data structures that depend on the input size.