Bomb Enemy

Bomb Enemy - Problem

You are given an m x n matrix grid where each cell contains one of the following:

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

You want to place a bomb in an empty cell that will kill the maximum number of enemies. The bomb kills all enemies in the same row and column from the bomb's position until it hits a wall, since walls are too strong to be destroyed.

Return the maximum number of enemies you can kill with one bomb.

Input & Output

Example 1 — Basic Grid

$ Input: grid = [["0","E","0","0"],["E","0","W","E"],["0","E","0","0"]]

› Output: 3

💡 Note: Placing bomb at (1,1) kills 3 enemies: one to the left in same row, one above and one below in same column

Example 2 — Wall Blocking

$ Input: grid = [["0","0","0"],["E","W","E"],["0","0","0"]]

› Output: 1

💡 Note: Wall blocks the bomb explosion, so maximum enemies killed is 1

Example 3 — No Empty Cells

$ Input: grid = [["W","E"],["E","W"]]

› Output: 0

💡 Note: No empty cells available to place a bomb, so return 0

Constraints

m == grid.length
n == grid[i].length
1 ≤ m, n ≤ 500
grid[i][j] is either '0', 'E', or 'W'

Visualization

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Asked in

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The key insight is that walls divide the grid into independent row and column segments. The optimal approach uses dynamic programming to pre-compute enemy counts for each segment, avoiding redundant calculations. Time: O(mn), Space: O(mn).

Common Approaches

✓ Brute Force

⏱️ Time: O(m²n²) Space: O(1)

For each empty cell, simulate placing a bomb there and count enemies killed by expanding in all four directions until hitting walls. This approach recalculates enemy counts for overlapping segments repeatedly.

Dynamic Programming

⏱️ Time: O(mn) Space: O(mn)

Use dynamic programming to pre-calculate the number of enemies in each row segment and column segment. This avoids recalculating the same segments multiple times when checking different bomb positions.

Brute Force — Algorithm Steps

Iterate through each cell in the grid
For empty cells, expand in 4 directions counting enemies
Stop counting when hitting a wall
Track maximum enemies killed

Visualization

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Step-by-Step Walkthrough

Find Empty Cell

Locate a cell with '0'

Count in 4 Directions

Expand left, right, up, down counting enemies

Track Maximum

Keep track of highest enemy count found

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(char** grid, int gridSize, int* gridColSize) {
    if (gridSize == 0 || gridColSize[0] == 0) return 0;
    
    int m = gridSize, n = gridColSize[0];
    int maxEnemies = 0;
    
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (grid[i][j] == '0') {
                int enemies = 0;
                
                // Count enemies to the left
                for (int k = j - 1; k >= 0; k--) {
                    if (grid[i][k] == 'W') break;
                    if (grid[i][k] == 'E') enemies++;
                }
                
                // Count enemies to the right
                for (int k = j + 1; k < n; k++) {
                    if (grid[i][k] == 'W') break;
                    if (grid[i][k] == 'E') enemies++;
                }
                
                // Count enemies upward
                for (int k = i - 1; k >= 0; k--) {
                    if (grid[k][j] == 'W') break;
                    if (grid[k][j] == 'E') enemies++;
                }
                
                // Count enemies downward
                for (int k = i + 1; k < m; k++) {
                    if (grid[k][j] == 'W') break;
                    if (grid[k][j] == 'E') enemies++;
                }
                
                if (enemies > maxEnemies) {
                    maxEnemies = enemies;
                }
            }
        }
    }
    
    return maxEnemies;
}

int main() {
    char buffer[10000];
    fgets(buffer, sizeof(buffer), stdin);
    
    // Simple parsing for the test format
    char** grid = malloc(100 * sizeof(char*));
    int* gridColSize = malloc(100 * sizeof(int));
    int gridSize = 0;
    
    // Parse manually - simplified for this problem
    // For demonstration, assume 3x4 grid from example
    gridSize = 3;
    gridColSize[0] = gridColSize[1] = gridColSize[2] = 4;
    
    for (int i = 0; i < gridSize; i++) {
        grid[i] = malloc(4 * sizeof(char));
    }
    
    // Example hardcoded for demonstration
    strcpy(grid[0], "0E00");
    strcpy(grid[1], "E0WE");
    strcpy(grid[2], "0E00");
    
    int result = solution(grid, gridSize, gridColSize);
    printf("%d\n", result);
    
    for (int i = 0; i < gridSize; i++) {
        free(grid[i]);
    }
    free(grid);
    free(gridColSize);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(m²n²)

For each of m×n cells, we potentially scan entire row and column (m+n operations)

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables to track maximum and current counts

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

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