# Find Shortest distance from a guard in a Bank

Given a matrix that is filled with ‘O’, ‘G’, and ‘W’ where ‘O’ represents open space, ‘G’ represents guards and ‘W’ represents walls in a Bank. Replace all of the O’s in the matrix with their shortest distance from a guard, without being able to go through any walls. Also, replace the guards with 0 and walls with -1 in output matrix.

Expected Time complexity is O(MN) for a M x N matrix.

Examples:

```O ==> Open Space
G ==> Guard
W ==> Wall

Input:
O  O  O  O  G
O  W  W  O  O
O  O  O  W  O
G  W  W  W  O
O  O  O  O  G

Output:
3  3  2  1  0
2 -1 -1  2  1
1  2  3 -1  2
0 -1 -1 -1  1
1  2  2  1  0
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to do BFS. We first enqueue all cells containing the guards and loop till queue is not empty. For each iteration of the loop, we dequeue the front cell from the queue and for each of its four adjacent cells, if cell is an open area and its distance from guard is not calculated yet, we update its distance and enqueue it. Finally after BFS procedure is over, we print the distance matrix.

Below is C++ implementation of above idea –

 `// C++ program to replace all of the O's in the matrix ` `// with their shortest distance from a guard ` `#include ` `using` `namespace` `std; ` ` `  `// store dimensions of the matrix ` `#define M 5 ` `#define N 5 ` ` `  `// An Data Structure for queue used in BFS ` `struct` `queueNode ` `{ ` `    ``// i, j and distance stores x and y-coordinates ` `    ``// of a matrix cell and its distance from guard ` `    ``// respectively ` `    ``int` `i, j, distance; ` `}; ` ` `  `// These arrays are used to get row and column ` `// numbers of 4 neighbors of a given cell ` `int` `row[] = { -1, 0, 1, 0}; ` `int` `col[] = { 0, 1, 0, -1 }; ` ` `  `// return true if row number and column number ` `// is in range ` `bool` `isValid(``int` `i, ``int` `j) ` `{ ` `    ``if` `((i < 0 || i > M - 1) || (j < 0 || j > N - 1)) ` `        ``return` `false``; ` ` `  `    ``return` `true``; ` `} ` ` `  `// return true if current cell is an open area and its ` `// distance from guard is not calculated yet ` `bool` `isSafe(``int` `i, ``int` `j, ``char` `matrix[][N], ``int` `output[][N]) ` `{ ` `    ``if` `(matrix[i][j] != ``'O'` `|| output[i][j] != -1) ` `        ``return` `false``; ` ` `  `    ``return` `true``; ` `} ` ` `  `// Function to replace all of the O's in the matrix ` `// with their shortest distance from a guard ` `void` `findDistance(``char` `matrix[][N]) ` `{ ` `    ``int` `output[M][N]; ` `    ``queue q; ` ` `  `    ``// finding Guards location and adding into queue ` `    ``for` `(``int` `i = 0; i < M; i++) ` `    ``{ ` `        ``for` `(``int` `j = 0; j < N; j++) ` `        ``{ ` `            ``// initialize each cell as -1 ` `            ``output[i][j] = -1; ` `            ``if` `(matrix[i][j] == ``'G'``) ` `            ``{ ` `                ``queueNode pos = {i, j, 0}; ` `                ``q.push(pos); ` `                ``// guard has 0 distance ` `                ``output[i][j] = 0; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// do till queue is empty ` `    ``while` `(!q.empty()) ` `    ``{ ` `        ``// get the front cell in the queue and update ` `        ``// its adjacent cells ` `        ``queueNode curr = q.front(); ` `        ``int` `x = curr.i, y = curr.j, dist = curr.distance; ` ` `  `        ``// do for each adjacent cell ` `        ``for` `(``int` `i = 0; i < 4; i++) ` `        ``{ ` `            ``// if adjacent cell is valid, has path and ` `            ``// not visited yet, en-queue it. ` `            ``if` `(isSafe(x + row[i], y + col[i], matrix, output) ` `                ``&& isValid(x + row[i], y + col[i])) ` `            ``{ ` `                ``output[x + row[i]][y + col[i]] = dist + 1; ` ` `  `                ``queueNode pos = {x + row[i], y + col[i], dist + 1}; ` `                ``q.push(pos); ` `            ``} ` `        ``} ` ` `  `        ``// dequeue the front cell as its distance is found ` `        ``q.pop(); ` `    ``} ` ` `  `    ``// print output matrix ` `    ``for` `(``int` `i = 0; i < M; i++) ` `    ``{ ` `        ``for` `(``int` `j = 0; j < N; j++) ` `            ``cout << std::setw(3) << output[i][j]; ` `        ``cout << endl; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``char` `matrix[][N] = ` `    ``{ ` `        ``{``'O'``, ``'O'``, ``'O'``, ``'O'``, ``'G'``}, ` `        ``{``'O'``, ``'W'``, ``'W'``, ``'O'``, ``'O'``}, ` `        ``{``'O'``, ``'O'``, ``'O'``, ``'W'``, ``'O'``}, ` `        ``{``'G'``, ``'W'``, ``'W'``, ``'W'``, ``'O'``}, ` `        ``{``'O'``, ``'O'``, ``'O'``, ``'O'``, ``'G'``} ` `    ``}; ` ` `  `    ``findDistance(matrix); ` ` `  `    ``return` `0; ` `} `

Output:

```  3  3  2  1  0
2 -1 -1  2  1
1  2  3 -1  2
0 -1 -1 -1  1
1  2  2  1  0
```

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