Problem Description

You have a 2D grid of size m x n that represents a box with an open top and bottom. You need to drop n balls from the top, one ball from each column, and determine where each ball exits at the bottom (or if it gets stuck).

Each cell in the grid contains a diagonal board that redirects balls:

• A value of 1 represents a board spanning from top-left to bottom-right (redirects ball to the right)
• A value of -1 represents a board spanning from top-right to bottom-left (redirects ball to the left)

A ball gets stuck in two scenarios:

1. V-shaped pattern: When two adjacent cells form a "V" shape - this happens when a cell with value 1 (redirects right) is next to a cell with value -1 (redirects left)
2. Wall collision: When a board redirects the ball into either the left or right wall of the box

The ball movement works as follows:

• When a ball encounters a cell with value 1, it moves diagonally down-right to the next row and next column
• When a ball encounters a cell with value -1, it moves diagonally down-left to the next row and previous column
• If the ball successfully passes through all rows without getting stuck, it falls out of the bottom

Your task is to return an array answer of size n where answer[i] represents:

• The column index where the ball dropped from column i exits at the bottom, or
• -1 if the ball gets stuck somewhere in the box

For example, if you drop a ball from column 2 and it exits from column 3 at the bottom, then answer[2] = 3. If the ball gets stuck, then answer[2] = -1.

Intuition

The key insight is to simulate the path of each ball as it falls through the grid. We can think of this as tracing the ball's journey from top to bottom, following the diagonal boards at each step.

When we drop a ball at column j, it encounters a series of diagonal boards as it falls. At each cell (i, j), the ball needs to move to the next row based on the board's direction. If the board value is 1, the ball tries to move to (i+1, j+1). If the board value is -1, the ball tries to move to (i+1, j-1).

The critical observation is identifying when a ball gets stuck. Let's visualize what happens:

• If we're at the leftmost column (j = 0) and the board directs left (grid[i][j] = -1), the ball hits the left wall
• If we're at the rightmost column (j = n-1) and the board directs right (grid[i][j] = 1), the ball hits the right wall
• The "V" pattern occurs when adjacent cells have opposite directions. If grid[i][j] = 1 (directing right) and grid[i][j+1] = -1 (directing left), the ball gets trapped between them
• Similarly, if grid[i][j] = -1 (directing left) and grid[i][j-1] = 1 (directing right), the ball is trapped

This naturally leads to a recursive approach: for each ball starting at column j, we can use depth-first search (DFS) to trace its path. At each step, we check if the ball gets stuck according to the conditions above. If not, we recursively move to the next position. When the ball reaches row m (past the last row), we know it has successfully fallen through and can return its final column position.

The beauty of this approach is that each ball's path is independent, so we can simply run the DFS simulation for each starting column and collect all the results.

Solution Approach

We implement the solution using Depth-First Search (DFS) to simulate each ball's path through the grid.

DFS Function Design: We create a recursive function dfs(i, j) that returns the exit column for a ball currently at position (i, j). The function returns -1 if the ball gets stuck.

Base Case: When i == m (the ball has passed through all rows), we return the current column j as the exit position.

Checking for Stuck Conditions: Before moving the ball, we check four conditions where the ball gets stuck:

1. Left wall collision: j == 0 and grid[i][j] == -1

   • Ball is at the leftmost column and the board directs it left into the wall

2. Right wall collision: j == n-1 and grid[i][j] == 1

   • Ball is at the rightmost column and the board directs it right into the wall

3. V-pattern (right-left): grid[i][j] == 1 and grid[i][j+1] == -1

   • Current cell directs right while the adjacent right cell directs left, forming a "V"

4. V-pattern (left-right): grid[i][j] == -1 and grid[i][j-1] == 1

   • Current cell directs left while the adjacent left cell directs right, forming a "V"

If any of these conditions are met, we immediately return -1.

Recursive Movement: If the ball doesn't get stuck, we move it to the next row:

• If grid[i][j] == 1: Move diagonally down-right by calling dfs(i+1, j+1)
• If grid[i][j] == -1: Move diagonally down-left by calling dfs(i+1, j-1)

Main Function: We initialize the grid dimensions m and n, then use list comprehension to call dfs(0, j) for each starting column j from 0 to n-1. This gives us the final position (or -1) for each ball dropped from the top.

The time complexity is O(m × n) as each ball can potentially traverse through all rows, and we have n balls. The space complexity is O(m) for the recursion stack depth in the worst case.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Consider this 3×3 grid:

grid = [[1,  1, -1],
        [1, -1,  1],
        [1,  1,  1]]

Let's trace what happens when we drop balls from each column:

Ball from Column 0:

• Start at position (0, 0), value is 1 (directs right)
• Check stuck conditions:
  • Not at right wall (j=0, not n-1)
  • Check right neighbor: grid[0][1] = 1 (not -1), so no V-pattern
• Move diagonally down-right to (1, 1)
• At (1, 1), value is -1 (directs left)
• Check stuck conditions:
  • Not at left wall (j=1, not 0)
  • Check left neighbor: grid[1][0] = 1, forms V-pattern!
  • Ball gets stuck
• Return -1

Ball from Column 1:

• Start at position (0, 1), value is 1 (directs right)
• Check stuck conditions:
  • Not at right wall (j=1, not n-1)
  • Check right neighbor: grid[0][2] = -1, forms V-pattern!
  • Ball gets stuck
• Return -1

Ball from Column 2:

• Start at position (0, 2), value is -1 (directs left)
• Check stuck conditions:
  • Not at left wall (j=2, not 0)
  • Check left neighbor: grid[0][1] = 1, forms V-pattern!
  • Ball gets stuck
• Return -1

Final result: [-1, -1, -1] - all balls get stuck!

Let's try another example where some balls make it through:

grid = [[1,  1,  1],
        [1,  1,  1]]

Ball from Column 0:

• Start at (0, 0), value is 1 (directs right)
• No stuck conditions (grid[0][1] = 1, same direction)
• Move to (1, 1)
• At (1, 1), value is 1 (directs right)
• No stuck conditions (grid[1][2] = 1, same direction)
• Move to (2, 2) - past the last row!
• Return 2 (exit column)

Ball from Column 1:

• Start at (0, 1), value is 1 (directs right)
• No stuck conditions (grid[0][2] = 1, same direction)
• Move to (1, 2)
• At (1, 2), value is 1 (directs right)
• Stuck condition: at right wall (j=2 is n-1) and directing right
• Return -1

Ball from Column 2:

• Start at (0, 2), value is 1 (directs right)
• Stuck condition: at right wall (j=2 is n-1) and directing right
• Return -1

Final result: [2, -1, -1]

This walkthrough demonstrates how the DFS function checks for stuck conditions at each step and recursively traces the ball's path until it either exits or gets stuck.

Solution Implementation

Time and Space Complexity

Time Complexity: O(m × n)

Each ball starting from the top row follows a unique path through the grid. In the worst case, a ball traverses from row 0 to row m-1, making exactly m recursive calls. Since we simulate the path for all n balls (one starting from each column), the total time complexity is O(m × n).

Space Complexity: O(m)

The space complexity is determined by the maximum depth of the recursive call stack. For each ball, the DFS function makes at most m recursive calls (one for each row traversed), resulting in a maximum recursion depth of m. Although we process n balls, we process them sequentially (not simultaneously), so only one recursive call stack exists at any given time. Therefore, the space complexity is O(m).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect V-Pattern Detection Logic

One of the most common mistakes is incorrectly identifying when a ball gets stuck in a V-shaped pattern. Developers often check the wrong adjacent cell or forget to verify both types of V-patterns.

Incorrect Implementation:

# Wrong: Only checking one type of V-pattern
if grid[i][j] == 1 and grid[i][j+1] == -1:
    return -1
# Missing the check for grid[i][j] == -1 and grid[i][j-1] == 1

Correct Implementation:

# Check both V-pattern scenarios
if grid[i][j] == 1 and j < n-1 and grid[i][j+1] == -1:
    return -1
if grid[i][j] == -1 and j > 0 and grid[i][j-1] == 1:
    return -1

2. Array Index Out of Bounds

Another frequent error is accessing array indices without proper boundary checks, especially when checking adjacent cells for V-patterns.

Incorrect Implementation:

# Wrong: No boundary check before accessing grid[i][j+1]
if grid[i][j] == 1 and grid[i][j+1] == -1:  # Can crash if j == n-1
    return -1

Correct Implementation:

# First check if we're at the boundary
if j == n-1 and grid[i][j] == 1:  # Wall collision check
    return -1
# Then safe to check V-pattern since we know j < n-1
if grid[i][j] == 1 and grid[i][j+1] == -1:
    return -1

3. Confusing Ball Movement Direction

A subtle but critical mistake is misunderstanding how the ball moves through the grid. The ball doesn't simply move down one row; it moves diagonally based on the board's direction.

Incorrect Implementation:

# Wrong: Ball only moves down, not diagonally
if grid[i][j] == 1:
    return dfs(i+1, j)  # Should be j+1
else:
    return dfs(i+1, j)  # Should be j-1

Correct Implementation:

# Ball moves diagonally based on board direction
if grid[i][j] == 1:
    return dfs(i+1, j+1)  # Move down-right
else:
    return dfs(i+1, j-1)  # Move down-left

4. Forgetting Base Case or Using Wrong Termination Condition

Some implementations forget to handle when the ball successfully exits the bottom, or use the wrong row index for termination.

Incorrect Implementation:

# Wrong: Using m-1 instead of m
if i == m-1:  # This means we're at the last row, not past it
    return j

Correct Implementation:

# Ball has passed through all rows (row index equals total rows)
if i == m:
    return j

Complete Corrected Solution Pattern:

def dfs(i, j):
    # Base case: successfully reached bottom
    if i == m:
        return j
  
    # Check boundaries first to avoid index errors
    if j == 0 and grid[i][j] == -1:
        return -1
    if j == n-1 and grid[i][j] == 1:
        return -1
  
    # Check V-patterns (boundaries already handled above)
    if grid[i][j] == 1 and grid[i][j+1] == -1:
        return -1
    if grid[i][j] == -1 and grid[i][j-1] == 1:
        return -1
  
    # Move diagonally based on board direction
    if grid[i][j] == 1:
        return dfs(i+1, j+1)
    else:
        return dfs(i+1, j-1)