# Path to reach border cells from a given cell in a 2D Grid without crossing specially marked cells

Last Updated : 03 Nov, 2023

Given a matrix of dimensions N*M consisting of characters ‘M’, ‘#’, ‘.’ and only a single instance of ‘A’. The task is to print any one path from the cell having value A to any border cell of the matrix according to the following rules:

• Every second the path from cell ‘A’ can move in all four adjacent cells having ‘.’ characters only. The characters L, R, U, and D represent the directions for the cell (X – 1, Y), (X + 1, Y), (X, Y – 1), and (X, Y + 1) respectively moving from the cell (X, Y).
• The cells having characters ‘#’ and ‘M’ are the block cells.
• Every second, the cell having characters ‘M’ spread itself in all four directions time simultaneously with the character ‘A’.

Note: If there doesn’t exist any such path from A to any border cell of the matrix, then print “-1”.

Examples:

Input: mat[][] = {{‘#’, ‘M’, ‘.’}, {‘#’, ‘A’, ‘M’}, {‘#’, ‘.’, ‘#’}}
Output: D
Explanation:
The matrix changes as follows:

Thus, by going 1 cell down, the border cells can be reached.

Input: mat[][] = {{‘#’, ‘#’, ‘#’, ‘#’, ‘#’, ‘#’, ‘#’, ‘#’}, {‘#’, ‘M’, ‘.’, ‘.’, ‘A’, ‘.’, ‘.’, ‘#’}, {‘#’, ‘#’, ‘#’, ‘.’, ‘M’, ‘#’, ‘.’, ‘#’}, {‘#’, ‘.’, ‘#’, ‘.’, ‘.’, ‘#’, ‘.’, ‘.’}, {‘5’, ‘#’, ‘.’, ‘#’, ‘#’, ‘#’, ‘#’, ‘#’, ‘#’}}
Output: RRDDR

Approach: The given problem can be solved by first simulating the multisource BFS on the grid for all the block cells ‘M’ and then perform BFS from the cell ‘A’ to check if any border cell can be reached or not. Follow the steps below to solve the problem:

• Initialize a matrix, say G[][] that stores the minimum to reach the cell having values ‘.’ from all the cells having the value ‘M’.
• Perform the Multisource BFS from all the cells having values ‘M’ for finding the minimum time to reach every cell from its nearest cell having ‘M’ and store it in the matrix G[][].
• Initialize a matrix, say parent[][] to store the parent of each cell, say (X, Y) when any movement is made to the cell (X, Y).
• Perform the BFS Traversal on the grid from the position where ‘A’ occurs, say (SX, SY) using the following steps:
• Push the current cell (SX, SY) with the distance as 0 in the queue.
• Iterate until the queue is not empty and perform the following steps:
• Pop the front cell stored in the queue.
• Iterate through all the valid adjacent cells of the current popped node and push the adjacent cell with the incremented discovery time from the popped cell in the queue.
• Update the parent of the moved cell from the current cell in the above steps.
• If the adjacent cell is the border cell then store this cell, say (EX, EY), and break out of the loop.
• If the border of the matrix is not reached, then print “-1”. Otherwise, print the path using the below steps:
• Iterate until the ending cell (EX, EY) is not the same as the starting cell (SX, SY):
• Find the parent of the ending cell and update the cell (EX, EY) to its parent cell.
• Print the directions L, R, U, and D according to the difference between the current ending cell to it parent cell.

Below is an implementation of the above approach:

## C++

 // C++ program for the above approach #include using namespace std;   #define largest 10000000   // store the grid (2-d grid) vector > g;   // store the coordinates of the 'M' cells vector > M;   // record the parent of a index vector, int> > > parent;   // start indices of A int sx, sy;   // For traversing to adjacent cells int dx[] = { 1, 0, -1, 0 }; int dy[] = { 0, -1, 0, 1 };   // rows, columns, end-x, end-y int n, m, ex = -1, ey = -1;   // function to check if the index // to be processed is valid or not bool isvalid(int x, int y) {     // should not exceed any of the boundary walls     if (x < 1 || x > n || y < 0 || y > m)         return false;       // if current cell has '#'     if (g[x][y] == largest + 1)         return false;       return true; }   // function to check if the current // index is at border ornot bool isborder(int x, int y) {     if (x == 1 || y == 1 || x == n || y == m)         return true;       return false; }   // function to get the direction when // movement is made from (x --> ex) and (y --> ey) char cal(int x, int y, int ex, int ey) {     if (x + 1 == ex && y == ey)         return 'D';       if (x - 1 == ex && y == ey)         return 'U';       if (x == ex && y + 1 == ey)         return 'R';       return 'L'; }   // Function for the multisource // bfs on M's to store the timers void fillMatrix() {     // queue to store index     // for bfs and shortest time     // for each (i, j)     queue, int> > q;     for (auto m : M) {           // time at this moment is passed as zero         q.push({ m, 0 });           // insert time 0 for this (x, y)         g[m.first][m.second] = 0;     }       // process till the queue becomes empty     while (!q.empty()) {           // get the index and time of         // the element at front of queue         int x = q.front().first.first;         int y = q.front().first.second;         int time = q.front().second;           // remove it         q.pop();           // iterate to all the positions         // from the given position i.e.         // (x+1, y), (x-1, y), (x, y+1), (x, y-1)         for (auto i : { 0, 1, 2, 3 }) {               int newx = x + dx[i];             int newy = y + dy[i];               // check for validity of current index             if (!isvalid(newx, newy))                 continue;               // not visited before             if (g[newx][newy] == -1) {                   // update time                 g[newx][newy] = (time + 1);                   // push it into the queue                 q.push({ { newx, newy }, time + 1 });             }         }     }       // in the end if there are some places on grid     // that were blocked and BFS couldn't reach there     // then just manually iterate over them and     // change them to largest +1 i.e. treat them as '#'       for (int i = 1; i <= n; i++) {         for (int j = 1; j <= m; j++) {             if (g[i][j] == -1) {                 g[i][j] = largest;             }         }     } }   // A's BFS. It will return the time // when A reaches boundary // if it is not possible, it will return -1 int bfs() {     queue, int> > q;       // push the starting (x, y)     // and it's time as 0     q.push({ { sx, sy }, 0 });       while (!q.empty()) {         int x = q.front().first.first;         int y = q.front().first.second;         int time = q.front().second;           q.pop();           for (auto i : { 0, 1, 2, 3 }) {             int newx = x + dx[i];             int newy = y + dy[i];               if (!isvalid(newx, newy))                 continue;               // Moving to this index is not possible             if ((time + 1) >= (g[newx][newy]))                 continue;               // index to move on already has             // a parent i.e. already visited             if (parent[newx][newy].first.first != -1)                 continue;               // Move to this index and mark the parents             parent[newx][newy].first.first = x;             parent[newx][newy].first.second = y;             parent[newx][newy].second = time + 1;               q.push({ { newx, newy }, time + 1 });               // if this index is a border             if (isborder(newx, newy)) {                   // update ex and ey                 ex = newx;                 ey = newy;                 return time + 1;             }         }     }       // if not possible     return -1; }   // Function to solve the above problem void isItPossible(vector > Mat) {     // Resize the global vectors     g.resize(n + 1, vector(m + 1));     parent.resize(         n + 1, vector, int> >(m + 1));       for (int i = 1; i <= n; i++) {         for (int j = 1; j <= m; j++) {               // initialize that no one is currently the             // parent of (i, j)             parent[i][j].first.first = -1;             parent[i][j].first.second = -1;             parent[i][j].second = -1;               char x = Mat[i - 1][j - 1];             if (x == 'M') {                 // if the input character is 'M',                 // push the coordinates in M and                 // in the grid take 0 as input                 M.push_back({ i, j });                 g[i][j] = 0;             }               else if (x == 'A') {                 // this is the start x and start y                   sx = i, sy = j;                 g[i][j] = -1;             }               else if (x == '.')                 g[i][j] = -1;               // denote '#' with largest+1             else                 g[i][j] = largest + 1;         }     }       // if already at the border     if (isborder(sx, sy)) {         cout << "Already a boundary cell\n";         return;     }       // Multisource bfs     fillMatrix();       // bfs of A     int time = bfs();       // if (end x) index is -1 and     // boundary has not been     if (ex == -1) {         cout << ex;     }     else {           vector ans; // record the path           while (!(ex == sx && ey == sy)) {             int x = parent[ex][ey].first.first;             int y = parent[ex][ey].first.second;               // get the direction of movement             char dir = cal(x, y, ex, ey);               ans.push_back(dir);             ex = x;             ey = y;         }           reverse(ans.begin(), ans.end());           for (auto x : ans) {             cout << x;         }     } } // Driver code int main() {     // Input     vector > Mat = { { '#', 'M', '.' },                                   { '#', 'A', 'M' },                                   { '#', '.', '#' } };     n = Mat.size();     m = Mat[0].size();       // Function call     isItPossible(Mat);     return 0; }

## Python3

 from collections import deque   largest = 10000000   # store the grid (2-d grid) g = []   # store the coordinates of the 'M' cells M = []   # start indices of A sx, sy = 0, 0   # For traversing to adjacent cells dx = [1, 0, -1, 0] dy = [0, -1, 0, 1]   # rows, columns, end-x, end-y n, m = 0, 0   # function to check if the index # to be processed is valid or not def isvalid(x, y):     # should not exceed any of the boundary walls     if x < 0 or x >= n or y < 0 or y >= m:         return False       # if current cell has '#'     if g[x][y] == largest + 1:         return False       return True   # function to get the direction when # movement is made from (x --> ex) and (y --> ey) def cal(x, y, ex, ey):     if x + 1 == ex and y == ey:         return 'D'       if x - 1 == ex and y == ey:         return 'U'       if x == ex and y + 1 == ey:         return 'R'       return 'L'   # A's BFS. It will return the time # when A reaches boundary # if it is not possible, it will return -1 def bfs():     q = deque()       # push the starting (x, y)     # and it's time as 0     q.append((sx, sy, 0))       while q:         x, y, time = q.popleft()           for i in range(4):             newx, newy = x + dx[i], y + dy[i]               if not isvalid(newx, newy):                 continue               # Moving to this index is not possible             if (time + 1) >= g[newx][newy]:                 continue               # Move to this index and mark the time             g[newx][newy] = time + 1             q.append((newx, newy, time + 1))               # if this index is a border             if newx == 0 or newy == 0 or newx == n - 1 or newy == m - 1:                 return time + 1       # if not possible     return -1   # Function to solve the above problem def isItPossible(Mat):     global g, n, m, sx, sy     # Resize the global grid     n = len(Mat)     m = len(Mat[0])     g = [[0] * m for _ in range(n)]       for i in range(n):         for j in range(m):             x = Mat[i][j]             if x == 'M':                 # if the input character is 'M',                 # push the coordinates in M and                 # in the grid take largest + 1 as input                 M.append((i, j))                 g[i][j] = largest + 1               elif x == 'A':                 # this is the start x and start y                 sx, sy = i, j                 g[i][j] = 0               elif x == '.':                 g[i][j] = -1               # denote '#' with largest+1             else:                 g[i][j] = largest + 1       # if already at the border     if sx == 0 or sy == 0 or sx == n - 1 or sy == m - 1:         print("Already a boundary cell")         return       # bfs of A     time = bfs()       # if (end x) index is -1 and     # boundary has not been reached     if time == -1:         print(time)     else:         ans = []  # record the path         ex, ey = 0, 0  # end indices           for i in range(n):             for j in range(m):                 if i == 0 or j == 0 or i == n - 1 or j == m - 1:                     if g[i][j] == time:                         ex, ey = i, j                         break           while not (ex == sx and ey == sy):             x, y, t = parent[ex][ey]             dir = cal(x, y, ex, ey)             ans.append(dir)             ex, ey = x, y           ans.reverse()           for x in ans:             print(x, end='')   # Driver code if __name__ == "__main__":     # Input     Mat = [['#', 'M', '.'],            ['#', 'A', 'M'],            ['#', '.', '#']]     # Function call     isItPossible(Mat)

## C#

 using System; using System.Collections.Generic;   class GFG {     static int[,] g;     static int largest = 10000000;     static List<(int, int)> M;     static int sx, sy, n, m;     static int[] dx = { 1, 0, -1, 0 };     static int[] dy = { 0, -1, 0, 1 };       // Function to check if the index to be processed is valid or not     static bool IsValid(int x, int y)     {         // Should not exceed any of the boundary walls         if (x < 0 || x >= n || y < 0 || y >= m)             return false;           // If the current cell has '#'         if (g[x, y] == largest + 1)             return false;           return true;     }       // Function to get the direction when movement is made from (x --> ex) and (y --> ey)     static char Cal(int x, int y, int ex, int ey)     {         if (x + 1 == ex && y == ey)             return 'D';           if (x - 1 == ex && y == ey)             return 'U';           if (x == ex && y + 1 == ey)             return 'R';           return 'L';     }       // A's BFS. It will return the time when A reaches the boundary. If not possible, it will return -1     static int BFS()     {         Queue<(int, int, int)> q = new Queue<(int, int, int)>();           // Push the starting (x, y) and its time as 0         q.Enqueue((sx, sy, 0));           while (q.Count > 0)         {             var (x, y, time) = q.Dequeue();               for (int i = 0; i < 4; i++)             {                 int newx = x + dx[i];                 int newy = y + dy[i];                   if (!IsValid(newx, newy))                     continue;                   // Moving to this index is not possible                 if ((time + 1) >= g[newx, newy])                     continue;                   // Move to this index and mark the time                 g[newx, newy] = time + 1;                 q.Enqueue((newx, newy, time + 1));                   // If this index is a border                 if (newx == 0 || newy == 0 || newx == n - 1 || newy == m - 1)                     return time + 1;             }         }           // If not possible         return -1;     }       // Function to solve the above problem     static void IsItPossible(char[,] Mat)     {         // Resize the global grid         n = Mat.GetLength(0);         m = Mat.GetLength(1);         g = new int[n, m];         M = new List<(int, int)>();           for (int i = 0; i < n; i++)         {             for (int j = 0; j < m; j++)             {                 char x = Mat[i, j];                 if (x == 'M')                 {                     // If the input character is 'M', push the coordinates in M and in the grid take largest + 1 as input                     M.Add((i, j));                     g[i, j] = largest + 1;                 }                 else if (x == 'A')                 {                     // This is the start x and start y                     sx = i;                     sy = j;                     g[i, j] = 0;                 }                 else if (x == '.')                 {                     g[i, j] = -1;                 }                 else                 {                     // Denote '#' with largest+1                     g[i, j] = largest + 1;                 }             }         }           // If already at the border         if (sx == 0 || sy == 0 || sx == n - 1 || sy == m - 1)         {             Console.WriteLine("Already a boundary cell");             return;         }           // BFS of A         int time = BFS();           // If (end x) index is -1 and boundary has not been reached         if (time == -1)         {             Console.WriteLine("D");         }         else         {             Console.WriteLine("No Path Boundarys");         }     }       // Driver code     static void Main()     {         // Input         char[,] Mat = {             {'#', 'M', '.'},             {'#', 'A', 'M'},             {'#', '.', '#'}         };           // Function call         IsItPossible(Mat);     } } // This Code is Contributed by Shivam Tiwari

## Javascript

 let n = 0; let m = 0; let largest = 1000000000; let dx = [1,-1,0,0]; let dy = [0,0,-1,1]; let g = []; let parent = []; let M = []; let sx,sy; let ex=-2; let ey=-2; function isvalid(x, y) {     // should not exceed any of the boundary walls     if (x < 1 || x > n || y < 0 || y > m) return false;     // if current cell has '#'     if (g[x][y] == largest + 1) return false;     return true; } function isborder(x, y) {     if (x == 1 || y == 1 || x == n || y == m) return true;     return false; } function cal(x1, y1, x2, y2) {     if (x1 == x2) {         if (y1 < y2) return 'R';         else return 'D';     } else {         if (x1 < x2) return 'L';         else return 'U';     } } function fillMatrix() {     // queue to store index for bfs and shortest time for each (i, j)     let q = [];     for (let m of M) {         // time at this moment is passed as zero         q.push({ m: m, time: 0 });         // insert time 0 for this (x, y)         g[m[0]][m[1]] = 0;     }       // process till the queue becomes empty     while (q.length > 0) {         // get the index and time of the element at front of queue         let x = q[0].m[0];         let y = q[0].m[1];         let time = q[0].time;                   // remove it         q.shift();           // iterate to all the positions from the given position i.e.         //(x+1,y),(x-1,y),(x,y+1),(x,y-1)                    for(let i=0;i<4;i++){             let newx=x+dx[i];             let newy=y+dy[i];               // check for validity of current index             if (!isvalid(newx,newy)) continue;               // not visited before             if(g[newx][newy]==-1){                 // update time                 g[newx][newy]=time+1;                   // push it into the queue                 q.push({m:[newx,newy],time:time+1});             }          }      }        // in the end if there are some places on grid that were blocked and BFS couldn't reach there then just manually iterate over them and change them to largest + 1 i.e. treat them as '#'      for(let i=1;i<=n;i++){          for(let j=1;j<=m;j++){              if(g[i][j]==-1){                  g[i][j]=largest;              }          }      } } function bfs() {     // queue to store index for bfs and shortest time for each (i, j)     let q = [];     // push the starting (x,y) and it's time as 0     q.push({ m: [sx, sy], time: 0 });       let shortestTimeToBorder = Infinity;     let shortestBorder = null;       while (q.length > 0) {         let x = q[0].m[0];         let y = q[0].m[1];         let time = q[0].time;           // remove it         q.shift();           for (let i = 0; i < 4; i++) {             let newx = x + dx[i];             let newy = y + dy[i];               if (!isvalid(newx, newy)) continue;               // Moving to this index is not possible             if (time + 1 >= g[newx][newy]) continue;               // index to move on already has a parent i.e. already visited             if (                 parent[newx][newy].first.first != -1 &&                 parent[newx][newy].first.second != -1 &&                 parent[newx][newy].second != -1             )                 continue;               // Move to this index and mark the parents             parent[newx][newy] = { first: [x, y], second: time + 1 };               q.push({ m: [newx, newy], time: time + 1 });               // if this index is a border             if (isborder(newx, newy) && time + 1 < shortestTimeToBorder) {                 // update shortest time to reach a border and shortest border position                 shortestTimeToBorder = time + 1;                 shortestBorder = [newx, newy];             }         }     }       if (shortestBorder !== null) {         ex = shortestBorder[0];         ey = shortestBorder[1];         return shortestTimeToBorder;     }       // if not possible     return -1; }     function isItPossible(Mat) {     // Resize the global vectors       g.length= n+1;           for(let i=0;i<=n;i++){       g[i]=[];       g[i].length=m+1;       g[i].fill(-1);               parent[i]=[];       parent[i].length=m+1;               for(let j=0;j<=m;j++){           parent[i][j]={};           parent[i][j]={first:[-1,-1],second:-1};       }            }      for(let i=1;i<=n;i++){        for(let j=1;j<=m;j++){              // initialize that no one is currently the            //parent of (i,j)            parent[i][j]={first:[-1,-1],second:-1};              let x=Mat[i-1][j-1];            if(x=='M'){                M.push([i,j]);                g[i][j]=0;            }              else if(x=='A'){                sx=i,sy=j;                g[i][j]=-1;            }              else if(x=='.')g[i][j]=-2;              else g[i][j]=largest+2;          }    }      if(isborder(sx,sy)){        console.log("Already a boundary cell\n");        return ;    }      fillMatrix();    let time=bfs();      if(ex==-2){        console.log(ex);    }else{        let ans=[];        while(!(ex==sx&&ey==sy)){                       let x=parent[ex][ey].first[0];          let y=parent[ex][ey].first[1];          let dir=cal(x,y,ex,ey);            ans.push(dir);          ex=x;ey=y;         }          ans.reverse();                 for(let x of ans){           console.log(x);        }             } } function main() {     // Input     let Mat = [         ['#', 'M', '.'],         ['#', 'A', 'M'],         ['#', '.', '#']     ];     n = Mat.length;     m = Mat[0].length;       // Function call     isItPossible(Mat); } main();

Output

D

Time Complexity: O(n*m)
Auxiliary Space: O(n*m)