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
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 <bits/stdc++.h> 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<queueNode> 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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