Problem Description

You are given an m x n matrix grid where each cell contains one of three possible values:

• 'W' represents a wall
• 'E' represents an enemy
• '0' represents an empty cell

Your task is to find the maximum number of enemies you can kill by placing exactly one bomb in the grid.

Rules for bomb placement and explosion:

• A bomb can only be placed in an empty cell (marked as '0')
• When a bomb explodes, it kills all enemies in the same row and column as the bomb
• The explosion extends in all four directions (up, down, left, right) from the bomb's position
• The explosion stops when it hits a wall ('W') in any direction - walls block the explosion from continuing further
• Walls cannot be destroyed by the explosion

For example, if you place a bomb at position (i, j):

• It kills all enemies to the left of (i, j) in row i until a wall is encountered
• It kills all enemies to the right of (i, j) in row i until a wall is encountered
• It kills all enemies above (i, j) in column j until a wall is encountered
• It kills all enemies below (i, j) in column j until a wall is encountered

The goal is to determine the optimal placement of a single bomb to maximize the number of enemy kills and return that maximum count. If no empty cells exist in the grid, return 0.

Intuition

The key insight is that for each empty cell, we need to count how many enemies would be killed if we placed a bomb there. A bomb kills enemies in four directions: left, right, up, and down from its position, stopping at walls.

A naive approach would be to iterate through each empty cell and then scan in all four directions to count enemies. However, this would involve redundant calculations since we'd be counting the same enemies multiple times for different bomb positions.

The clever observation is that we can precompute the enemy counts for each position more efficiently. For any position (i, j), the total enemies killed equals:

• Enemies to the left in row i (until hitting a wall)
• Enemies to the right in row i (until hitting a wall)
• Enemies above in column j (until hitting a wall)
• Enemies below in column j (until hitting a wall)

We can build up these counts using dynamic programming passes:

1. Row-wise counting: For each row, we make two passes:

   • Left-to-right pass: Count enemies seen so far from the left, resetting when we hit a wall
   • Right-to-left pass: Count enemies seen so far from the right, resetting when we hit a wall

2. Column-wise counting: For each column, we make two passes:

   • Top-to-bottom pass: Count enemies seen so far from above, resetting when we hit a wall
   • Bottom-to-top pass: Count enemies seen so far from below, resetting when we hit a wall

By accumulating these counts in a separate grid g, each cell g[i][j] will contain the total number of enemies that would be killed if a bomb were placed at position (i, j). The trick is that when we encounter an enemy during our passes, we increment our running count, and this count gets added to all positions we visit until we hit a wall.

Finally, we simply find the maximum value in g where the corresponding position in the original grid is an empty cell '0'.

Solution Approach

The solution implements the precomputation strategy using a 2D auxiliary grid g to store the enemy count for each position. Here's how the implementation works:

1. Initialize the auxiliary grid:

g = [[0] * n for _ in range(m)]

This creates an m x n grid initialized with zeros to accumulate enemy counts.

2. Process each row (horizontal enemy counting):

For each row i, we make two passes:

Left-to-right pass:

t = 0
for j in range(n):
    if grid[i][j] == 'W':
        t = 0  # Reset count when hitting a wall
    elif grid[i][j] == 'E':
        t += 1  # Increment count when finding an enemy
    g[i][j] += t  # Add current count to this position

The variable t tracks enemies encountered from the left. When we hit a wall, we reset t to 0. When we encounter an enemy, we increment t. The current value of t is added to g[i][j].

Right-to-left pass:

t = 0
for j in range(n - 1, -1, -1):
    if grid[i][j] == 'W':
        t = 0
    elif grid[i][j] == 'E':
        t += 1
    g[i][j] += t

Similar logic but traversing from right to left, accumulating enemies from the right direction.

3. Process each column (vertical enemy counting):

For each column j, we make two passes:

Top-to-bottom pass:

t = 0
for i in range(m):
    if grid[i][j] == 'W':
        t = 0
    elif grid[i][j] == 'E':
        t += 1
    g[i][j] += t

Bottom-to-top pass:

t = 0
for i in range(m - 1, -1, -1):
    if grid[i][j] == 'W':
        t = 0
    elif grid[i][j] == 'E':
        t += 1
    g[i][j] += t

After these four passes, each cell g[i][j] contains the total number of enemies that would be killed by placing a bomb at position (i, j).

4. Find the maximum for empty cells:

return max(
    [g[i][j] for i in range(m) for j in range(n) if grid[i][j] == '0'],
    default=0,
)

We iterate through all positions, filter for empty cells ('0'), and return the maximum value from g. The default=0 handles the case when no empty cells exist.

Time Complexity: O(m * n) - We make exactly 4 passes through the grid.

Space Complexity: O(m * n) - We use an auxiliary grid of the same size as the input.

The elegance of this solution lies in avoiding redundant calculations. Instead of checking all four directions for each empty cell individually, we precompute the enemy counts for all positions simultaneously through systematic grid traversals.

Example Walkthrough

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

Consider this 3x4 grid:

W 0 E 0
E 0 E E  
0 W 0 E

We'll create an auxiliary grid g (initially all zeros) to accumulate enemy counts:

0 0 0 0
0 0 0 0
0 0 0 0

Step 1: Process rows left-to-right

Row 0: W 0 E 0

• Start with t=0
• Position (0,0): Wall → reset t=0, g[0][0] += 0 = 0
• Position (0,1): Empty → g[0][1] += 0 = 0
• Position (0,2): Enemy → increment t=1, g[0][2] += 1 = 1
• Position (0,3): Empty → g[0][3] += 1 = 1

Row 1: E 0 E E

• Start with t=0
• Position (1,0): Enemy → increment t=1, g[1][0] += 1 = 1
• Position (1,1): Empty → g[1][1] += 1 = 1
• Position (1,2): Enemy → increment t=2, g[1][2] += 2 = 2
• Position (1,3): Enemy → increment t=3, g[1][3] += 3 = 3

Row 2: 0 W 0 E

• Start with t=0
• Position (2,0): Empty → g[2][0] += 0 = 0
• Position (2,1): Wall → reset t=0, g[2][1] += 0 = 0
• Position (2,2): Empty → g[2][2] += 0 = 0
• Position (2,3): Enemy → increment t=1, g[2][3] += 1 = 1

After left-to-right pass:

0 0 1 1
1 1 2 3
0 0 0 1

Step 2: Process rows right-to-left

Row 0: W 0 E 0

• Start with t=0 from right
• Position (0,3): Empty → g[0][3] += 0 = 1
• Position (0,2): Enemy → increment t=1, g[0][2] += 1 = 2
• Position (0,1): Empty → g[0][1] += 1 = 1
• Position (0,0): Wall → reset t=0, g[0][0] += 0 = 0

Row 1: E 0 E E

• Start with t=0 from right
• Position (1,3): Enemy → increment t=1, g[1][3] += 1 = 4
• Position (1,2): Enemy → increment t=2, g[1][2] += 2 = 4
• Position (1,1): Empty → g[1][1] += 2 = 3
• Position (1,0): Enemy → increment t=3, g[1][0] += 3 = 4

Row 2: 0 W 0 E

• Start with t=0 from right
• Position (2,3): Enemy → increment t=1, g[2][3] += 1 = 2
• Position (2,2): Empty → g[2][2] += 1 = 1
• Position (2,1): Wall → reset t=0, g[2][1] += 0 = 0
• Position (2,0): Empty → g[2][0] += 0 = 0

After right-to-left pass:

0 1 2 1
4 3 4 4
0 0 1 2

Step 3: Process columns top-to-bottom

Column 0: W E 0

• Start with t=0
• Position (0,0): Wall → reset t=0, g[0][0] += 0 = 0
• Position (1,0): Enemy → increment t=1, g[1][0] += 1 = 5
• Position (2,0): Empty → g[2][0] += 1 = 1

Column 1: 0 0 W

• Start with t=0
• Position (0,1): Empty → g[0][1] += 0 = 1
• Position (1,1): Empty → g[1][1] += 0 = 3
• Position (2,1): Wall → reset t=0, g[2][1] += 0 = 0

Column 2: E E 0

• Start with t=0
• Position (0,2): Enemy → increment t=1, g[0][2] += 1 = 3
• Position (1,2): Enemy → increment t=2, g[1][2] += 2 = 6
• Position (2,2): Empty → g[2][2] += 2 = 3

Column 3: 0 E E

• Start with t=0
• Position (0,3): Empty → g[0][3] += 0 = 1
• Position (1,3): Enemy → increment t=1, g[1][3] += 1 = 5
• Position (2,3): Enemy → increment t=2, g[2][3] += 2 = 4

After top-to-bottom pass:

0 1 3 1
5 3 6 5
1 0 3 4

Step 4: Process columns bottom-to-top

Column 0: W E 0

• Start with t=0 from bottom
• Position (2,0): Empty → g[2][0] += 0 = 1
• Position (1,0): Enemy → increment t=1, g[1][0] += 1 = 6
• Position (0,0): Wall → reset t=0, g[0][0] += 0 = 0

Column 1: 0 0 W

• Start with t=0 from bottom
• Position (2,1): Wall → reset t=0, g[2][1] += 0 = 0
• Position (1,1): Empty → g[1][1] += 0 = 3
• Position (0,1): Empty → g[0][1] += 0 = 1

Column 2: E E 0

• Start with t=0 from bottom
• Position (2,2): Empty → g[2][2] += 0 = 3
• Position (1,2): Enemy → increment t=1, g[1][2] += 1 = 7
• Position (0,2): Enemy → increment t=2, g[0][2] += 2 = 5

Column 3: 0 E E

• Start with t=0 from bottom
• Position (2,3): Enemy → increment t=1, g[2][3] += 1 = 5
• Position (1,3): Enemy → increment t=2, g[1][3] += 2 = 7
• Position (0,3): Empty → g[0][3] += 2 = 3

Final grid g:

0 1 5 3
6 3 7 7
1 0 3 5

Step 5: Find maximum for empty cells

Empty cells in original grid are at positions: (0,1), (0,3), (1,1), (2,0), (2,2)

Their corresponding values in g:

• Position (0,1): 1
• Position (0,3): 3
• Position (1,1): 3
• Position (2,0): 1
• Position (2,2): 3

The maximum is 3, which can be achieved by placing a bomb at positions (0,3), (1,1), or (2,2).

Solution Implementation

Time and Space Complexity

Time Complexity: O(m * n) where m is the number of rows and n is the number of columns in the grid.

The algorithm consists of four main passes through the grid:

• First pass: Traverse each row from left to right, visiting all m * n cells once
• Second pass: Traverse each row from right to left, visiting all m * n cells once
• Third pass: Traverse each column from top to bottom, visiting all m * n cells once
• Fourth pass: Traverse each column from bottom to top, visiting all m * n cells once
• Final step: Find the maximum value among empty cells, which takes at most O(m * n) time

Total operations: 4 * (m * n) + O(m * n) = O(m * n)

Space Complexity: O(m * n)

The algorithm creates an auxiliary 2D array g of size m × n to store the cumulative enemy counts for each cell. This requires O(m * n) additional space. The list comprehension in the final return statement creates a temporary list that could contain up to m * n elements in the worst case (if all cells are empty), but this doesn't change the overall space complexity of O(m * n).

Common Pitfalls

1. Double-Counting Enemies

The Pitfall: A critical issue in the provided solution is that enemies are being counted multiple times. When we traverse from left-to-right and then right-to-left (or top-to-bottom and bottom-to-top), we're adding the enemy count to the cell itself when it contains an enemy ('E'). This causes each enemy to be counted in its own cell's total, which is incorrect since:

• Bombs can only be placed in empty cells ('0')
• An enemy at position (i,j) shouldn't contribute to the count at its own position

Example of the Bug: Consider a row: ['0', 'E', '0']

• After left-to-right pass: [0, 1, 1]
• After right-to-left pass: [1, 2, 1]

The middle cell (containing 'E') shows a count of 2, but this position can't hold a bomb anyway.

The Solution: Only count enemies in the direction of traversal, not including the current cell if it's an enemy:

# Corrected left-to-right pass
enemy_count = 0
for col in range(cols):
    if grid[row][col] == 'W':
        enemy_count = 0
    elif grid[row][col] == 'E':
        enemy_count += 1
    else:  # Only update count for non-wall, non-enemy cells
        enemies_killed[row][col] += enemy_count

2. Incorrect Enemy Accumulation Logic

The Pitfall: The current implementation adds enemies before encountering them in some passes, which gives incorrect results. For instance, in the left-to-right pass, we increment enemy_count when we see an enemy and immediately add it to that enemy's position.

The Solution: Separate the counting logic more carefully:

def maxKilledEnemies(self, grid: List[List[str]]) -> int:
    if not grid or not grid[0]:
        return 0
  
    rows, cols = len(grid), len(grid[0])
    result = 0
    row_kills = 0
    col_kills = [0] * cols
  
    for i in range(rows):
        for j in range(cols):
            # Recalculate row kills if we hit a wall or start of row
            if j == 0 or grid[i][j-1] == 'W':
                row_kills = 0
                k = j
                while k < cols and grid[i][k] != 'W':
                    if grid[i][k] == 'E':
                        row_kills += 1
                    k += 1
          
            # Recalculate column kills if we hit a wall or start of column
            if i == 0 or grid[i-1][j] == 'W':
                col_kills[j] = 0
                k = i
                while k < rows and grid[k][j] != 'W':
                    if grid[k][j] == 'E':
                        col_kills[j] += 1
                    k += 1
          
            # Update result if current cell is empty
            if grid[i][j] == '0':
                result = max(result, row_kills + col_kills[j])
  
    return result

3. Memory Inefficiency

The Pitfall: Creating a full m x n auxiliary grid when we only need to track kills for empty cells can be wasteful, especially for large grids with few empty cells.

The Solution: Use a more space-efficient approach that only calculates kills when needed, or use rolling arrays to reduce space complexity from O(m*n) to O(n).