Check for possible path in 2D matrix

Given a 2D array(m x n). The task is to check if there is any path from top left to bottom right. In the matrix, -1 is considered as blockage (can’t go through this cell) and 0 is considered path cell (can go through it).

Note: Top left cell always contains 0

Examples:

Input : arr[][] = {{ 0, 0, 0, -1, 0},
{-1, 0, 0, -1, -1},
{ 0, 0, 0, -1, 0},
{-1, 0, 0, 0, 0},
{ 0, 0, -1, 0, 0}}
Output : Yes
Explanation:

The red cells are blocked, white cell denotes the path and the green cells are not blocked cells.

Input : arr[][] = {{ 0, 0, 0, -1, 0},
{-1, 0, 0, -1, -1},
{ 0, 0, 0, -1, 0},
{-1, 0, -1, 0, 0},
{ 0, 0, -1, 0, 0}}
Output : No
Explanation: There exists no path from start to end.

The red cells are blocked, white cell denotes the path and the green cells are not blocked cells.

Method 1

• Approach: The solution is to perform BFS or DFS to find whether there is a path or not. The graph needs not to be created to perform the bfs, but the matrix itself will be used as a graph. Start the traversal from the top right corner and if there is a way to reach the bottom right corner then there is a path.
• Algorithm:
1. Create a queue that stores pairs (i,j) and insert the (0,0) in the queue.
2. Run a loop till the queue is empty.
3. In each iteration dequeue the queue (a,b), if the front element is the destination (row-1,col-1) then return 1, i,e there is a path and change the value of mat[a][b] to -1, i.e. visited.
4. Else insert the adjacent indices where the value of matrix[i][j] is not -1.

Implementation:

C++

 // C++ program to find if there is path // from top left to right bottom #include using namespace std;   #define row 5 #define col 5   // to find the path from // top left to bottom right bool isPath(int arr[row][col]) {     // directions     int dir[4][2]         = { { 0, 1 }, { 0, -1 }, { 1, 0 }, { -1, 0 } };       // queue     queue > q;       // insert the top right corner.     q.push(make_pair(0, 0));       // until queue is empty     while (q.size() > 0) {         pair p = q.front();         q.pop();           // mark as visited         arr[p.first][p.second] = -1;           // destination is reached.         if (p == make_pair(row - 1, col - 1))             return true;           // check all four directions         for (int i = 0; i < 4; i++) {             // using the direction array             int a = p.first + dir[i][0];             int b = p.second + dir[i][1];               // not blocked and valid             if (arr[a][b] != -1 && a >= 0 && b >= 0                 && a < row && b < col) {                 q.push(make_pair(a, b));             }         }     }     return false; }   // Driver Code int main() {     // Given array     int arr[row][col] = { { 0, 0, 0, -1, 0 },                           { -1, 0, 0, -1, -1 },                           { 0, 0, 0, -1, 0 },                           { -1, 0, 0, 0, 0 },                           { 0, 0, -1, 0, 0 } };       // path from arr[0][0] to arr[row][col]     if (isPath(arr))         cout << "Yes";     else         cout << "No";       return 0; }

Java

 // Java program to find if there is path // from top left to right bottom import java.io.*; import java.util.*;   class pair {     int Item1, Item2;     pair(int f, int s)     {         Item1 = f;         Item2 = s;     } }   class GFG {       static int row = 5;     static int col = 5;       // To find the path from     // top left to bottom right     static boolean isPath(int[][] arr)     {           // Directions         int[][] dir             = { { 0, 1 }, { 0, -1 }, { 1, 0 }, { -1, 0 } };           // Queue         Queue q = new LinkedList<>();           // Insert the top right corner.         q.add(new pair(0, 0));           // Until queue is empty         while (q.size() > 0) {             pair p = (q.peek());             q.remove();               // Mark as visited             arr[p.Item1][p.Item2] = -1;               // Destination is reached.             if (p.Item1 == row - 1 && p.Item2 == col - 1)                 return true;               // Check all four directions             for (int i = 0; i < 4; i++) {                   // Using the direction array                 int a = p.Item1 + dir[i][0];                 int b = p.Item2 + dir[i][1];                   // Not blocked and valid                 if (a >= 0 && b >= 0 && a < row && b < col                     && arr[a][b] != -1) {                     if (a == row - 1 && b == col - 1)                         return true;                       q.add(new pair(a, b));                 }             }         }         return false;     }       // Driver Code     public static void main(String[] args)     {           // Given array         int[][] arr = { { 0, 0, 0, -1, 0 },                         { -1, 0, 0, -1, -1 },                         { 0, 0, 0, -1, 0 },                         { -1, 0, 0, 0, 0 },                         { 0, 0, -1, 0, 0 } };           // Path from arr[0][0] to arr[row][col]         if (isPath(arr))             System.out.println("Yes");         else             System.out.println("No");     } }   // This code is contributed by avanitrachhadiya2155

Python3

 # Python3 program to find if there is path # from top left to right bottom row = 5 col = 5   # to find the path from # top left to bottom right def isPath(arr) :       # directions     Dir = [ [0, 1], [0, -1], [1, 0], [-1, 0]]           # queue     q = []           # insert the top right corner.     q.append((0, 0))           # until queue is empty     while(len(q) > 0) :         p = q[0]         q.pop(0)                   # mark as visited         arr[p[0]][p[1]] = -1                   # destination is reached.         if(p == (row - 1, col - 1)) :             return True                       # check all four directions         for i in range(4) :                       # using the direction array             a = p[0] + Dir[i][0]             b = p[1] + Dir[i][1]                           # not blocked and valid             if(a >= 0 and b >= 0 and a < row and b < col and arr[a][b] != -1) :                            q.append((a, b))     return False     # Given array arr = [[ 0, 0, 0, -1, 0 ],           [ -1, 0, 0, -1, -1],           [ 0, 0, 0, -1, 0 ],           [ -1, 0, 0, 0, 0 ],           [ 0, 0, -1, 0, 0 ] ]   # path from arr[0][0] to arr[row][col] if (isPath(arr)) :     print("Yes") else :     print("No")       # This code is contributed by divyesh072019

C#

 // C# program to find if there is path // from top left to right bottom using System; using System.Collections.Generic;   class pair {   public int Item1, Item2;   public pair(int f, int s)   {     Item1 = f;     Item2 = s;   } }   class GFG {     static int row = 5;   static int col = 5;     // To find the path from   // top left to bottom right   static bool isPath(int[, ] arr)   {       // Directions     int[, ] dir       = { { 0, 1 }, { 0, -1 }, { 1, 0 }, { -1, 0 } };       // Queue     LinkedList q = new LinkedList();       // Insert the top right corner.     q.AddLast(new pair(0, 0));       // Until queue is empty     while (q.Count > 0) {       pair p = (pair)(q.First.Value);       q.RemoveFirst();         // Mark as visited       arr[p.Item1, p.Item2] = -1;         // Destination is reached.       if (p.Item1 == row - 1 && p.Item2 == col - 1)         return true;         // Check all four directions       for (int i = 0; i < 4; i++) {           // Using the direction array         int a = p.Item1 + dir[i, 0];         int b = p.Item2 + dir[i, 1];           // Not blocked and valid         if (a >= 0 && b >= 0 && a < row && b < col             && arr[a, b] != -1) {           if (a == row - 1 && b == col - 1)             return true;             q.AddLast(new pair(a, b));         }       }     }     return false;   }     // Driver Code   public static void Main(string[] args)   {       // Given array     int[, ] arr = { { 0, 0, 0, -1, 0 },                    { -1, 0, 0, -1, -1 },                    { 0, 0, 0, -1, 0 },                    { -1, 0, 0, 0, 0 },                    { 0, 0, -1, 0, 0 } };       // Path from arr[0, 0] to arr[row, col]     if (isPath(arr))       Console.WriteLine("Yes");     else       Console.WriteLine("No");   } }   // This code is contributed by phasing17

Javascript



Output

Yes
• Complexity Analysis:
• Time Complexity: O(R*C).
Every element of the matrix can be inserted once in the queue, so time Complexity is O(R*C).
• Space Complexity: O(R*C).
To store all the elements in a queue O(R*C) space is needed.

Method 2

• Approach: The only problem with the above solution is it uses extra space. This approach will eliminate the need for extra space. The basic idea is very similar. This algorithm will also perform BFS but the need for extra space will be eliminated by marking the array. So First run a loop and check which elements of the first column and the first row is accessible from 0,0 by using only the first row and column. mark them as 1. Now traverse the matrix from start to the end row-wise in increasing index of rows and columns. If the cell is not blocked then check that any of its adjacent cells is marked 1 or not. If marked 1 then mark the cell 1.
• Algorithm:
1. Mark the cell 0,0 as 1.
2. Run a loop from 0 to row length and if the cell above is marked 1 and the current cell is not blocked then mark the current cell as 1.
3. Run a loop from 0 to column length and if the left cell is marked 1 and the current cell is not blocked then mark the current cell as 1.
4. Traverse the matrix from start to the end row-wise in increasing index of rows and columns.
5. If the cell is not blocked then check that any of its adjacent cells (check only the cell above and the cell to the left). is marked 1 or not. If marked 1 then mark the cell 1.
6. If the cell (row-1, col-1) is marked 1 return true else return false.

Implementation:

C++

 // C++ program to find if there is path // from top left to right bottom #include using namespace std;   #define row 5 #define col 5   // to find the path from // top left to bottom right bool isPath(int arr[row][col]) {     // set arr[0][0] = 1     arr[0][0] = 1;       // Mark reachable (from top left) nodes     // in first row and first column.     for (int i = 1; i < row; i++)         if (arr[i][0] != -1)             arr[i][0] = arr[i - 1][0];         for (int j = 1; j < col; j++)         if (arr[0][j] != -1)             arr[0][j] = arr[0][j - 1];          // Mark reachable nodes in remaining     // matrix.     for (int i = 1; i < row; i++)         for (int j = 1; j < col; j++)           if (arr[i][j] != -1)               arr[i][j] = max(arr[i][j - 1],                             arr[i - 1][j]);                 // return yes if right bottom     // index is 1     return (arr[row - 1][col - 1] == 1); }   // Driver Code int main() {     // Given array     int arr[row][col] = { { 0, 0, 0, -1, 0 },                           { -1, 0, 0, -1, -1 },                           { 0, 0, 0, -1, 0 },                           { -1, 0, -1, 0, -1 },                           { 0, 0, -1, 0, 0 } };       // path from arr[0][0] to arr[row][col]     if (isPath(arr))       cout << "Yes";     else       cout << "No";   return 0; }

Java

 // Java program to find if there is path // from top left to right bottom import java.util.*; import java.io.*;   class GFG {     // to find the path from     // top left to bottom right     static boolean isPath(int arr[][])     {         // set arr[0][0] = 1         arr[0][0] = 1;           // Mark reachable (from top left) nodes         // in first row and first column.         for (int i = 1; i < 5; i++)             if (arr[0][i] != -1)                 arr[0][i] = arr[0][i - 1];         for (int j = 1; j < 5; j++)             if (arr[j][0] != -1)                 arr[j][0] = arr[j - 1][0];           // Mark reachable nodes in         //  remaining matrix.         for (int i = 1; i < 5; i++)             for (int j = 1; j < 5; j++)                 if (arr[i][j] != -1)                     arr[i][j] = Math.max(arr[i][j - 1],                                         arr[i - 1][j]);           // return yes if right         // bottom index is 1         return (arr[5 - 1][5 - 1] == 1);     }            //Driver code     public static void main(String[] args)     {         // Given array         int arr[][] = { { 0, 0, 0, -1, 0 },                         { -1, 0, 0, -1, -1 },                         { 0, 0, 0, -1, 0 },                         { -1, 0, -1, 0, -1 },                         { 0, 0, -1, 0, 0 } };           // path from arr[0][0]         // to arr[row][col]         if (isPath(arr))             System.out.println("Yes");         else             System.out.println("No");     } } // This code is contributed // by prerna saini

Python3

 # Python3 program to find if there # is path from top left to right bottom row = 5 col = 5   # to find the path from # top left to bottom right def isPath(arr):           # set arr[0][0] = 1     arr[0][0] = 1       # Mark reachable (from top left)     # nodes in first row and first column.     for i in range(1, row):         if (arr[i][0] != -1):             arr[i][0] = arr[i-1][0]       for j in range(1, col):         if (arr[0][j] != -1):             arr[0][j] = arr[0][j-1]                   # Mark reachable nodes in     # remaining matrix.     for i in range(1, row):         for j in range(1, col):             if (arr[i][j] != -1):                 arr[i][j] = max(arr[i][j - 1],                                 arr[i - 1][j])                                       # return yes if right     # bottom index is 1     return (arr[row - 1][col - 1] == 1)   # Driver Code   # Given array arr = [[ 0, 0, 0, -1, 0 ],        [-1, 0, 0, -1, -1],        [ 0, 0, 0, -1, 0 ],        [-1, 0, -1, 0, -1],        [ 0, 0, -1, 0, 0 ]]   # path from arr[0][0] to arr[row][col] if (isPath(arr)):     print("Yes") else:     print("No")   # This code is contributed # by sahilshelangia

C#

 // C# program to find if there is path // from top left to right bottom using System;   class GFG {     // to find the path from     // top left to bottom right     static bool isPath(int [,]arr)     {         // set arr[0][0] = 1         arr[0, 0] = 1;           // Mark reachable (from top left) nodes         // in first row and first column.         for (int i = 1; i < 5; i++)             if (arr[i, 0] != -1)                 arr[i, 0] = arr[i - 1, 0];         for (int j = 1; j < 5; j++)             if (arr[0,j] != -1)                 arr[0,j] = arr[0, j - 1];           // Mark reachable nodes in         // remaining matrix.         for (int i = 1; i < 5; i++)             for (int j = 1; j < 5; j++)                 if (arr[i, j] != -1)                     arr[i, j] = Math.Max(arr[i, j - 1],                                         arr[i - 1, j]);           // return yes if right         // bottom index is 1         return (arr[5 - 1, 5 - 1] == 1);     }           //Driver code     public static void Main()     {         // Given array         int [,]arr = { { 0, 0, 0, -1, 0 },                         { -1, 0, 0, -1, -1 },                         { 0, 0, 0, -1, 0 },                         { -1, 0, -1, 0, -1 },                         { 0, 0, -1, 0, 0 } };           // path from arr[0][0]         // to arr[row][col]         if (isPath(arr))             Console.WriteLine("Yes");         else             Console.WriteLine("No");     } }   // This code is contributed // by vt_m



Javascript



Output

No
• Complexity Analysis:
• Time Complexity: O(R*C).
Every element of the matrix is traversed, so time Complexity is O(R*C).
• Space Complexity: O(1).
No extra space is needed.

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