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 oranges So 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 cannot be rotten. Below is algorithm. 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. Note: USE C++ LANGUAGE
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 oranges So 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 cannot be rotten.
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Note:
USE C++ LANGUAGE
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