Unique Paths

 

  1. Clarify the problem:

    • The problem requires us to find the number of unique paths from the top-left corner to the bottom-right corner of a grid.
    • We can only move either down or right at any point in time.
    • The grid is represented as a 2D matrix, and we are given its dimensions.

  2. Analyze the problem:

    • Input: The dimensions of the grid (m x n).
    • Output: The number of unique paths from the top-left to the bottom-right corner of the grid.
    • Constraints:
      • The dimensions m and n are positive integers.
      • The grid can be of varying sizes.

  3. Design an algorithm:

    • We can solve this problem using dynamic programming.
    • Create a 2D grid of size m x n to store the number of unique paths for each position.
    • Initialize the first row and first column of the grid to 1 since there is only one way to reach any position in the first row or column.
    • For each cell (i, j) in the grid, the number of unique paths is the sum of the number of paths from the cell above (i-1, j) and the cell to the left (i, j-1).
    • Traverse the grid from left to right and top to bottom, updating each cell with the number of unique paths.
    • Return the value in the bottom-right corner of the grid, which represents the total number of unique paths.

  4. Explain your approach:

    • We will define a function uniquePaths that takes the dimensions m and n as input.
    • Create a 2D grid of size m x n and initialize all cells to 0.
    • Initialize the first row and first column of the grid to 1.
    • Traverse the grid from the second row and second column.
    • For each cell (i, j), update its value as the sum of the value in the cell above (i-1, j) and the value in the cell to the left (i, j-1).
    • Finally, return the value in the bottom-right corner of the grid.

  5. Write clean and readable code:

python
def uniquePaths(m, n):
    # Create a 2D grid of size m x n
    grid = [[0] * n for _ in range(m)]

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

    # Traverse the grid and update each cell with the number of unique paths
    for i in range(1, m):
        for j in range(1, n):
            grid[i][j] = grid[i-1][j] + grid[i][j-1]

    # Return the value in the bottom-right corner
    return grid[m-1][n-1]
  1. Test your code:
python
# Test case 1:
m1, n1 = 3, 7
result1 = uniquePaths(m1, n1)
# Expected output: 28 (3x7 grid has 28 unique paths)

# Test case 2:
m2, n2 = 3, 2
result2 = uniquePaths(m2, n2)
# Expected output: 3 (3x2 grid has 3 unique paths)

# Test case 3:
m3, n3 = 7, 3
result3 = uniquePaths(m3, n3)
# Expected output: 28 (7x3 grid has 28 unique paths)

# Test case 4:
m4, n4 = 1, 1
result4 = uniquePaths(m4, n4)
# Expected output: 1 (1x1 grid has only 1 unique path)

# Test case 5:
m5, n5 = 2, 2
result5 = uniquePaths(m5, n5)
# Expected output: 2 (2x2 grid has 2 unique paths)

# Test case 6:
m6, n6 = 0, 0
result6 = uniquePaths(m6, n6)
# Expected output: 0 (Empty grid has 0 unique paths)

# Print the results
print(result1)
print(result2)
print(result3)
print(result4)
print(result5)
print(result6)

  1. Optimize if necessary:

    • The solution provided already has an optimal time complexity of O(m*n) since we traverse the entire grid once.

  2. Handle error cases:

    • The solution handles the case when either m or n is 0 by returning 0, indicating that there are no unique paths in an empty grid.
    • We assume that the input dimensions m and n are non-negative integers.

  3. Discuss complexity analysis:

    • Time complexity: The time complexity of the solution is O(m*n) since we traverse the entire grid once.
    • Space complexity: The space complexity is O(m*n) as we create a 2D grid to store the number of unique paths for each position.
Next Post Previous Post