Minimum time required to rot all oranges - GeeksforGeeks
Given a matrix of dimension m*n where each cell in the matrix can have values 0, 1 or 2 which has the following meaning:
Examples:
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Given a matrix of dimension m*n where each cell in the matrix can have values 0, 1 or 2 which has the following meaning:
0: Empty cell
1: Cells have fresh oranges
2: Cells have rotten orangesSo we have to determine what is the minimum time required so that all the oranges become rotten. A rotten orange at index [i,j] can rot other fresh orange at indexes [i-1,j], [i+1,j], [i,j-1], [i,j+1] (up, down, left and right). If it is impossible to rot every orange then simply return -1.
Examples:
Input: arr[][C] = { {2, 1, 0, 2, 1},
{1, 0, 1, 2, 1},
{1, 0, 0, 2, 1}};
Output:
All oranges can become rotten in 2 time frames.
Input: arr[][C] = { {2, 1, 0, 2, 1},
{0, 0, 1, 2, 1},
{1, 0, 0, 2, 1}};
Output:
All oranges cannot be rotten.
1) Create an empty Q.
2) Find all rotten oranges and enqueue them to Q. Also enqueue
a delimiter to indicate beginning of next time frame.
3) While Q is not empty do following
3.a) While delimiter in Q is not reached
(i) Dequeue an orange from queue, rot all adjacent oranges.
While rotting the adjacents, make sure that time frame
is incremented only once. And time frame is not icnremented
if there are no adjacent oranges.
3.b) Dequeue the old delimiter and enqueue a new delimiter. The
oranges rotten in previous time frame lie between the two
delimiters.
bool isdelim(ele temp){ return (temp.x == -1 && temp.y == -1);}// Function to check whether there is still a fresh// orange remainingbool checkall(int arr[][C]){ for (int i=0; i<R; i++) for (int j=0; j<C; j++) if (arr[i][j] == 1) return true; return false;}// This function finds if it is possible to rot all oranges or not.// If possible, then it returns minimum time required to rot all,// otherwise returns -1int rotOranges(int arr[][C]){ // Create a queue of cells queue<ele> Q; ele temp; int ans = 0; // Store all the cells having rotten orange in first time frame for (int i=0; i<R; i++) { for (int j=0; j<C; j++) { if (arr[i][j] == 2) { temp.x = i; temp.y = j; Q.push(temp); } } } // Separate these rotten oranges from the oranges which will rotten // due the oranges in first time frame using delimiter which is (-1, -1) temp.x = -1; temp.y = -1; Q.push(temp); // Process the grid while there are rotten oranges in the Queue while (!Q.empty()) { // This flag is used to determine whether even a single fresh // orange gets rotten due to rotten oranges in current time // frame so we can increase the count of the required time. bool flag = false; // Process all the rotten oranges in current time frame. while (!isdelim(Q.front())) { temp = Q.front(); // Check right adjacent cell that if it can be rotten if (isvalid(temp.x+1, temp.y) && arr[temp.x+1][temp.y] == 1) { // if this is the first orange to get rotten, increase // count and set the flag. if (!flag) ans++, flag = true; // Make the orange rotten arr[temp.x+1][temp.y] = 2; // push the adjacent orange to Queue temp.x++; Q.push(temp); temp.x--; // Move back to current cell } // Check left adjacent cell that if it can be rotten if (isvalid(temp.x-1, temp.y) && arr[temp.x-1][temp.y] == 1) { if (!flag) ans++, flag = true; arr[temp.x-1][temp.y] = 2; temp.x--; Q.push(temp); // push this cell to Queue temp.x++; } // Check top adjacent cell that if it can be rotten if (isvalid(temp.x, temp.y+1) && arr[temp.x][temp.y+1] == 1) { if (!flag) ans++, flag = true; arr[temp.x][temp.y+1] = 2; temp.y++; Q.push(temp); // Push this cell to Queue temp.y--; } // Check bottom adjacent cell if it can be rotten if (isvalid(temp.x, temp.y-1) && arr[temp.x][temp.y-1] == 1) { if (!flag) ans++, flag = true; arr[temp.x][temp.y-1] = 2; temp.y--; Q.push(temp); // push this cell to Queue } Q.pop(); } // Pop the delimiter Q.pop(); // If oranges were rotten in current frame than separate the // rotten oranges using delimiter for the next frame for processing. if (!Q.empty()) { temp.x = -1; temp.y = -1; Q.push(temp); } // If Queue was empty than no rotten oranges left to process so exit } // Return -1 if all arranges could not rot, otherwise -1. return (checkall(arr))? -1: ans;}