# HackerEarth Escape from grid problem solution

In this HackerEarth Escape from grid problem solution Assume that you are given a two-dimensional grid G that contains N rows and M columns. The grid  consists of only three integers (0, 1, and 2).
•  0 denotes an empty cell
•  1 denotes that a cell contains a plant
•  2 denotes a cell where you are standing initially
You can move into an adjacent cell if that adjacent cell is empty. Two cells are adjacent if they share a side. In other words, if a cell is empty, then you can move in one of the four directions, Up, Down, Left, and Right.
You cannot move out of the grid G.

Your task is to find the length of the shortest path to reach one of the boundary edges of the grid without stepping on a plant. The length of a path is equal to the number of moves you make.

## HackerEarth Escape from grid problem solution.

`#include <iostream>#include <queue>using namespace std;const int MAX = 1e3 + 5;int grid[MAX][MAX], dist[MAX][MAX], n, m;bool vis[MAX][MAX];int dr[] = {0, 0, -1, 1};int dc[] = {-1, 1, 0, 0};int bfs(int fromX, int fromY){    queue <pair<int, int> > q;    pair <int, int> p;    dist[fromX][fromY] = 0;    q.push({fromX, fromY});    while(!q.empty())    {        p = q.front();        q.pop();        fromX = p.first;        fromY = p.second;        if(fromX == 0 or fromX == n-1 or fromY == 0 or fromY == m-1)            return dist[fromX][fromY];        for(int i = 0;i < 4;++i)        {            int x = fromX + dr[i];            int y = fromY + dc[i];            if(x >= 0 and x < n and y >= 0 and y < m and grid[x][y] != 1 and vis[x][y] == false)            {                vis[x][y] = true;                dist[x][y] = dist[fromX][fromY] + 1;                q.push({x, y});            }        }    }    return -1;}int main(){    int ans, x, y;    cin >> n >> m;    for(int i = 0;i < n;++i) for(int j = 0;j < m;++j)    {        cin >> grid[i][j];        vis[i][j] = false;        dist[i][j] = 0;        if(grid[i][j] == 2)            x = i, y = j;    }    ans = bfs(x, y);    cout << ans << endl;    return 0;}`