Day-09-Rotting Oranges.cpp

In a given grid, each cell can have one of three values:

the value 0 representing an empty cell;
the value 1 representing a fresh orange;
the value 2 representing a rotten orange.
Every minute, any fresh orange that is adjacent (4-directionally) to a rotten orange becomes rotten.

Return the minimum number of minutes that must elapse until no cell has a fresh orange.  If this is impossible, return -1 instead.



Example 1:



Input: [[2,1,1],[1,1,0],[0,1,1]]
Output: 4
Example 2:

Input: [[2,1,1],[0,1,1],[1,0,1]]
Output: -1
Explanation:  The orange in the bottom left corner (row 2, column 0) is never rotten, because rotting only happens 4-directionally.
Example 3:

Input: [[0,2]]
Output: 0
Explanation:  Since there are already no fresh oranges at minute 0, the answer is just 0.


Note:

1 <= grid.length <= 10
1 <= grid[0].length <= 10
grid[i][j] is only 0, 1, or 2.











class Solution {
public:

    bool isValid(int x, int y, int m, int n){
        return (x>=0 && x<m && y>=0 && y<n);
    }

    int orangesRotting(vector<vector<int>>& grid) {
        int m=grid.size();
        if(!m) return -1;
        int n=grid[0].size();

        queue<pair<int, pair<int, int> > > q;  //  < time, < coord x, coord y > >

        for(int i=0; i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]==2){
                    q.push({0, {i, j}});
                }
            }
        }

        pair<int, pair<int, int> > tmp;

        while(!q.empty()){
            tmp=q.front();
            q.pop();
            int i=tmp.second.first;
            int j=tmp.second.second;

            if(isValid(i-1, j, m, n) && grid[i-1][j]==1){ // top
                grid[i-1][j]=2;
                q.push({tmp.first+1, {i-1, j}});
            }

            if(isValid(i, j+1, m, n) && grid[i][j+1]==1){  // right
                grid[i][j+1]=2;
                q.push({tmp.first+1, {i, j+1}});
            }

            if(isValid(i+1, j, m, n) && grid[i+1][j]==1){  // down
                grid[i+1][j]=2;
                q.push({tmp.first+1, {i+1, j}});
            }

            if(isValid(i, j-1, m, n) && grid[i][j-1]==1){  // left
                grid[i][j-1]=2;
                q.push({tmp.first+1, {i, j-1}});
            }
        }

        int resultTime=tmp.first;

        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]==1){
                    resultTime=-1; break;
                }
            }
        }
        return resultTime;
    }
};