Problem Description

You have a 2D grid of size m x n and an integer k. The grid contains:

• Exactly k cells with values from 1 to k (each value appears exactly once)
• All other cells have value 0

Your task is to find a path through the grid that satisfies two conditions:

1. Visit every cell exactly once - The path must cover all m * n cells in the grid
2. Visit numbered cells in order - When encountering cells with values 1 through k, you must visit them in ascending order (1, then 2, then 3, and so on)

Movement rules:

• You can start from any cell in the grid
• From any cell, you can move to its adjacent cells (up, down, left, or right)

The output should be a 2D array of size (m * n) x 2, where each element result[i] = [xi, yi] represents the coordinates of the i-th cell visited in your path.

If multiple valid paths exist, you can return any one of them. If no valid path exists, return an empty array.

For example, if you have a 3x3 grid with k=3:

• Cell (0,0) has value 1
• Cell (1,1) has value 2
• Cell (2,2) has value 3
• All other cells have value 0

A valid path must visit all 9 cells exactly once, and when it encounters the numbered cells, it must visit cell with value 1 first, then the cell with value 2, and finally the cell with value 3.

Intuition

The key insight is that we need to explore all possible paths from every valid starting position to find one that satisfies both constraints. This naturally leads us to think about backtracking with DFS.

Let's think about the constraints:

1. We must visit all cells exactly once - this is a Hamiltonian path problem
2. We must visit numbered cells 1 to k in order

The second constraint actually helps us prune the search space. At any point in our path, we know which number we should encounter next. If we're currently looking for number v, we can only move to cells that are either 0 (empty) or have value v.

Since the grid is small (at most 6 x 6 = 36 cells), we can use bit manipulation to track visited cells efficiently. Each cell can be mapped to a unique bit position using the formula i * n + j, where i is the row and j is the column. This gives us a compact way to represent our visited state with a single integer st.

The strategy becomes:

• Try starting from any cell that has value 0 or 1 (we need to start such that we can visit 1 first)
• Use DFS to explore paths, keeping track of:
  • Current position (i, j)
  • Next required number v to visit in sequence
  • Visited cells using bit mask st
  • Current path being built
• When we move to a cell with value v, increment v for the next required number
• Only move to adjacent cells that are unvisited and have acceptable values (0 or v)
• If we successfully build a path of length m * n, we've found our answer
• If we get stuck, backtrack by removing the last cell from path and unmarking it as visited

The backtracking ensures we explore all possibilities, while the constraints help us avoid invalid paths early, making the search tractable despite the exponential nature of the problem.

Learn more about Recursion patterns.

Solution Approach

The solution uses state compression with DFS and backtracking to find a valid path through the grid.

State Representation: We use bit manipulation to track visited cells. Each cell (i, j) is mapped to a unique bit position using the function f(i, j) = i * n + j. The variable st is an integer where the k-th bit being 1 means the cell at position k has been visited.

Main Algorithm Structure:

1. Helper Function f(i, j): Maps a 2D coordinate to a unique bit position in our state integer.

2. DFS Function dfs(i, j, v):

   • Parameters:
     • (i, j): current cell position
     • v: the next numbered cell we need to visit (starts at 1)
   • Returns True if a valid complete path is found, False otherwise

3. DFS Logic:

   • Add current cell [i, j] to the path
   • Check if we've visited all cells: len(path) == m * n. If yes, we found a solution
   • Mark current cell as visited: st |= 1 << f(i, j) (set the corresponding bit to 1)
   • If current cell value equals v, increment v for the next required number
   • Try moving in 4 directions using dirs = (-1, 0, 1, 0, -1) with pairwise iteration
   • For each adjacent cell (x, y):
     • Check if it's within bounds: 0 <= x < m and 0 <= y < n
     • Check if unvisited: (st & 1 << f(x, y)) == 0
     • Check if value is acceptable: grid[x][y] in (0, v)
     • If all conditions met, recursively call dfs(x, y, v)
   • If no valid path found from current state:
     • Backtrack by removing current cell from path: path.pop()
     • Unmark as visited: st ^= 1 << f(i, j) (flip the bit back to 0)
     • Return False

4. Main Search Loop:

   • Iterate through all cells (i, j) in the grid
   • If grid[i][j] is 0 or 1, try starting DFS from this cell
     • We can only start from cells with value 0 or 1 because we need to visit 1 first
   • If DFS succeeds, return the path
   • If DFS fails, clear the path and reset state st = 0, then try next starting position
   • If no starting position yields a valid path, return empty array

Time Complexity: O(m * n * 2^(m*n)) in worst case, as we might explore all possible paths from each starting position.

Space Complexity: O(m * n) for the recursion stack and path storage.

The elegance of this solution lies in combining bit manipulation for efficient state tracking with constrained DFS that prunes invalid paths early by checking the numbered cell ordering requirement.

Example Walkthrough

Let's walk through a small example with a 3x3 grid where k=2:

Grid:
[1, 0, 0]
[0, 0, 2]
[0, 0, 0]

Goal: Find a path that visits all 9 cells exactly once, visiting cell with value 1 before cell with value 2.

Step 1: Initialize

• m = 3, n = 3, k = 2
• path = [], st = 0 (no cells visited yet)
• Helper function f(i,j) = i * 3 + j maps each cell to a bit position

Step 2: Try Starting Positions We can start from cells with value 0 or 1. Let's try starting from (0,0) which has value 1.

Step 3: DFS from (0,0)

• Add [0,0] to path: path = [[0,0]]
• Mark as visited: st = 000000001 (bit 0 set)
• Since grid[0][0] = 1 and v = 1, increment v to 2
• Now we need to find cell with value 2 next

Step 4: Explore from (0,0) Try moving to adjacent cells. Valid moves must be:

• Unvisited
• Have value 0 or 2 (since v=2)

From (0,0), we can move right to (0,1) or down to (1,0). Both have value 0, so let's try (0,1).

Step 5: DFS from (0,1)

• Add [0,1] to path: path = [[0,0], [0,1]]
• Mark as visited: st = 000000011 (bits 0 and 1 set)
• grid[0,1] = 0, so v stays 2

Step 6: Continue Building Path Let's say we continue: (0,1) → (0,2) → (1,2) → (1,1) → (1,0) → (2,0) → (2,1) → (2,2)

At (1,2), we encounter value 2:

• When we reach (1,2), grid[1,2] = 2 and v = 2, so increment v to 3
• Now all numbered cells have been visited in order

Step 7: Complete the Path After visiting (2,2), we have path.length = 9 = m*n, so we found a valid solution!

Final Path:

[[0,0], [0,1], [0,2], [1,2], [1,1], [1,0], [2,0], [2,1], [2,2]]

Verification:

• ✓ All 9 cells visited exactly once
• ✓ Cell with value 1 at (0,0) visited before cell with value 2 at (1,2)

If Path Fails: If at any point we get stuck (no valid moves), we backtrack:

• Remove last cell from path using path.pop()
• Unmark it as visited using st ^= 1 << f(i,j)
• Try a different direction or backtrack further

The algorithm continues this process, trying different paths and starting positions until a valid path is found or all possibilities are exhausted.

Solution Implementation

Time and Space Complexity

Time Complexity: O(m^2 × n^2)

The analysis breaks down as follows:

• The outer loops iterate through all m × n cells in the grid to find valid starting positions, contributing O(m × n) iterations.
• For each starting position, the DFS explores paths through the grid. In the worst case, the DFS might need to explore all possible paths before finding a valid one or determining none exists.
• At each cell during DFS, we check up to 4 neighboring cells (from the dirs array used with pairwise).
• The maximum depth of recursion is m × n (visiting all cells), and at each level, we might need to backtrack and try different paths.
• In the worst-case scenario, for each starting position, the DFS could potentially explore a significant portion of all possible paths through the grid, which can be approximated as O(m × n) operations per starting position.
• Therefore, the total time complexity is O(m × n) × O(m × n) = O(m^2 × n^2).

Space Complexity: O(m × n)

The space usage consists of:

• The path list stores at most m × n coordinate pairs (when a complete path is found).
• The recursion stack depth can go up to m × n in the worst case (visiting all cells).
• The st variable uses bit manipulation to track visited cells, requiring O(1) space (as it's a single integer, though it represents m × n bits).
• Other variables (dirs, loop variables) use O(1) space.
• The dominant factor is the recursion stack and the path storage, both of which are O(m × n).

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

Common Pitfalls

1. Incorrect State Management During Backtracking

One of the most common pitfalls in this solution is forgetting to properly restore the state when backtracking. The code correctly handles this, but many implementations fail in one of these areas:

Pitfall Example:

# WRONG: Forgetting to restore the visited state
def dfs(row, col, current_value):
    visited_bitmask |= 1 << cell_to_bitmask_index(row, col)
    path.append([row, col])
  
    # ... recursive calls ...
  
    path.pop()  # Only removing from path
    # Missing: visited_bitmask ^= 1 << cell_to_bitmask_index(row, col)

Solution: Always ensure both the path AND the visited state are restored:

# Backtrack: remove current position from path and mark as unvisited
path.pop()
visited_bitmask ^= 1 << cell_to_bitmask_index(row, col)

2. Starting Position Logic Error

Pitfall: Starting the DFS from any cell without considering the constraint that numbered cells must be visited in order:

# WRONG: Starting from any cell
for start_row in range(rows):
    for start_col in range(cols):
        if depth_first_search(start_row, start_col, 1):
            return path

This would fail if you start from a cell with value 2 or higher, as you'd never be able to visit cell 1 first.

Solution: Only start from cells that allow visiting numbered cells in order:

# CORRECT: Only start from cells with value 0 or 1
if grid[start_row][start_col] in (0, 1):
    if depth_first_search(start_row, start_col, 1):
        return path

3. Bit Manipulation Errors

Pitfall: Using incorrect bit operations for marking/unmarking visited cells:

# WRONG: Using OR to unmark (doesn't work)
visited_bitmask |= 1 << cell_to_bitmask_index(row, col)  # Mark as visited
visited_bitmask |= 1 << cell_to_bitmask_index(row, col)  # Still marked!

# WRONG: Using AND to unmark
visited_bitmask &= 1 << cell_to_bitmask_index(row, col)  # This clears everything except the bit

Solution: Use XOR (^) to toggle bits or AND with complement to clear:

# CORRECT: Mark as visited
visited_bitmask |= 1 << cell_to_bitmask_index(row, col)

# CORRECT: Unmark as visited (toggle back)
visited_bitmask ^= 1 << cell_to_bitmask_index(row, col)

# Alternative: AND with complement
visited_bitmask &= ~(1 << cell_to_bitmask_index(row, col))

4. Value Checking Logic

Pitfall: Not properly checking if a cell can be visited based on its value:

# WRONG: Only checking if the cell equals current_value
if grid[next_row][next_col] == current_value:
    # This prevents visiting cells with value 0

Solution: Allow visiting cells with value 0 (empty cells) OR the current expected numbered value:

# CORRECT: Can visit empty cells (0) or the next required number
if grid[next_row][next_col] in (0, current_value):
    # Proceed with DFS

5. Global State Not Reset Between Starting Positions

Pitfall: Forgetting to reset the global state when trying different starting positions:

# WRONG: Not resetting state between attempts
for start_row in range(rows):
    for start_col in range(cols):
        if grid[start_row][start_col] in (0, 1):
            if depth_first_search(start_row, start_col, 1):
                return path
            # Missing reset here!

Solution: Always clear the path and reset the visited bitmask:

# CORRECT: Reset state for next attempt
if depth_first_search(start_row, start_col, 1):
    return path

# Reset for next starting position
path.clear()
visited_bitmask = 0