HackerEarth Travel diaries problem solution

In this HackerEarth Travel Diaries problem solution You are given a matrix of size N  X M that contains the digits 0, 1, or 2 only. All the cells that contain 1 and are adjacent to any cell that contains 2 will be converted from 1 to 2, simultaneously in 1 second. Write a program to find the minimum time to convert all the cells having value 1 to 2.

HackerEarth Travel diaries problem solution.

#include<bits/stdc++.h>
using namespace std;
 
// function to check whether a cell is valid / invalid
bool isvalid(int i, int j,int n,int m)
{
    return (i >= 0 && j >= 0 && i < n && j < m);
}
 
// structure for storing coordinates of the cell
struct ele {
    int x, y;
};
 
// Drive program
int main()
{
    int n,m,i,j;
    cin>>n>>m;
    
    short int arr[1001][1001];
    
    for(i=0;i<n;i++){
        for(j=0;j<m;j++)cin>>arr[i][j];
     }
     
    // Create a queue of cells
    queue<ele> Q;
    ele temp;
    int ans = 0;
 
    // Store all the cells having Bad People in first time frame
    for (int i=0; i<n; i++)
    {
       for (int j=0; j<m; j++)
       {
            if (arr[i][j] == 2)
            {
                temp.x = i;
                temp.y = j;
                Q.push(temp);
            }
        }
    }
    temp.x = -1;
    temp.y = -1;
    Q.push(temp);
 
    // Process the grid while there are Bad People in the Queue
    while (!Q.empty())
    {
        bool flag = false;
 
        // Process all the Bad People in current time frame until 
        //it is not a delimiter.
        while (!(Q.front().x ==-1 && Q.front().y == -1))
        {
            temp = Q.front();
 
            // Check right adjacent cell that if it has Good Person
            if (isvalid(temp.x+1, temp.y,n,m) && arr[temp.x+1][temp.y] == 1)
            {
                if (!flag) ans++, flag = true;
                arr[temp.x+1][temp.y] = 2;
 
                // push the adjacent Person to Queue
                temp.x++;
                Q.push(temp);
 
                temp.x--; // Move back to current cell
            }
            if (isvalid(temp.x-1, temp.y,n,m) && 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++;
            }
            if (isvalid(temp.x, temp.y+1,n,m) && 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--;
            }
            if (isvalid(temp.x, temp.y-1,n,m) && 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 (!Q.empty()) {
            temp.x = -1;
            temp.y = -1;
            Q.push(temp);
        }
 
        // If Queue was empty than no Bad People left to process so exit
    }
 
    int f=0;
    for (int i=0; i<n; i++)
        for (int j=0; j<m; j++)
            if(arr[i][j]==1){
                f=1;
                break;
            }
 
    //f=1 if some Good People Left
    if (f == 1)
        cout << -1<<endl;
    else
         cout << ans << endl;
    return 0;
}