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Rotten Oranges.cpp
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Rotten Oranges.cpp
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/*
Problem Link: https://practice.geeksforgeeks.org/problems/rotten-oranges2536/1
*/
class Solution
{
public:
//Function to find minimum time required to rot all oranges.
int orangesRotting(vector<vector<int>>& grid) {
// Code here
int n= grid.size();
int m= grid[0].size();
queue<pair<pair<int,int>, int>> q;
// {{r,c},t} will be the queue pattern
vector<vector<int>> vis(n, vector<int>(m, 0));
int cnt= 0, cntIn= 0;
for(int i=0; i<n; i++){
for(int j=0; j<m; j++){
if(grid[i][j]==1) cntIn++;
if(grid[i][j]==2){
q.push({{i,j}, 0});
vis[i][j]=2;
}
}
}
int tmx=0;
while(!q.empty()){
int row= q.front().first.first;
int col= q.front().first.second;
int t= q.front().second;
q.pop();
tmx= max(tmx, t);
int delrow[]={-1,0,1,0};
int delcol[]={0,1,0,-1};
for(int i=0; i<4; i++){
int nrow= row + delrow[i];
int ncol= col + delcol[i];
if(nrow>=0 && nrow<n && ncol>=0 && ncol<m && grid[nrow][ncol]==1 && vis[nrow][ncol] !=2){
q.push({{nrow, ncol}, t+1});
cnt++;
vis[nrow][ncol]=2; // to make sure it is never visited twice
}
}
}
if(cnt != cntIn) return -1;
return tmx;
}
};