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:
- Create a queue that stores pairs (i,j) and insert the (0,0) in the queue.
- Run a loop till the queue is empty.
- 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.
- 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 <bits/stdc++.h>using namespace std;#define row 5#define col 5// to find the path from// top left to bottom rightbool isPath(int arr[row][col]){ // directions int dir[4][2] = { { 0, 1 }, { 0, -1 }, { 1, 0 }, { -1, 0 } }; // queue queue<pair<int, int> > q; // insert the top right corner. q.push(make_pair(0, 0)); // until queue is empty while (q.size() > 0) { pair<int, int> 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 Codeint 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 bottomimport 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<pair> 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 bottomrow = 5col = 5# to find the path from# top left to bottom rightdef 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 arrayarr = [[ 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 bottomusing 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<pair> q = new LinkedList<pair>(); // 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
<script>// JavaScript program to find if there is path // from top left to right bottom var row = 5; var col = 5;// To find the path from // top left to bottom right function isPath(arr) { // Directions var dir = [ [ 0, 1 ], [ 0, -1 ], [ 1, 0 ], [ -1, 0 ] ]; // Queue var q = []; // Insert the top right corner. q.push([0, 0]); // Until queue is empty while (q.length > 0) { var p = q[0]; q.shift(); // Mark as visited arr[p[0]][p[1]] = -1; // Destination is reached. if (p[0]==row-1 && p[1]==col-1) return true; // Check all four directions for(var i = 0; i < 4; i++) { // Using the direction array var a = p[0] + dir[i][0]; var b = p[1] + 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.push([a,b]); } } } return false; } // Driver Code// Given array var 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)) document.write("Yes"); else document.write("No"); </script> |
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.
- Time Complexity: O(R*C).
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:
- Mark the cell 0,0 as 1.
- 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.
- 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.
- 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 (check only the cell above and the cell to the left). is marked 1 or not. If marked 1 then mark the cell 1.
- 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 <iostream>using namespace std;#define row 5#define col 5// to find the path from// top left to bottom rightbool 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 Codeint 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 bottomimport 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 = 5col = 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 bottomusing 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 |
PHP
<?php// PHP 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 rightfunction isPath($arr){ global $row, $col; $arr[0][0] = 1; // Mark reachable (from // top left) nodes in // first row and first column. for ($i = 1; $i < $row; $i++) if ($arr[$i][0] != -1) $arr[$i][0] = $arr[$i - 1][0]; for ($j = 1; $j < $col; $j++) if ($arr[0][$j] != -1) $arr[0][$j] = $arr[0][$j - 1]; // Mark reachable nodes // in remaining matrix. for ($i = 1; $i < $row; $i++) for ($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// Given array$arr = array(array(0, 0, 0, 1, 0), array(-1, 0, 0, -1, -1), array(0, 0, 0, -1, 0), array(-1, 0, -1, 0, -1), array(0, 0, -1, 0, 0));// path from arr[0][0]// to arr[row][col]if (isPath($arr))echo "Yes";elseecho "No"; // This code is contributed by anuj_67.?> |
Javascript
<script>// JavaScript program to find if there is path // from top left to right bottomvar arr = [[5], [5]]// to find the path from// top left to bottom rightfunction isPath(arr){ // set arr[0][0] = 1 arr[0][0] = 1; // Mark reachable (from top left) nodes // in first row and first column. for (var i = 1; i < 5; i++) if (arr[i][0] != -1) arr[i][0] = arr[i - 1][0]; for (var j = 1; j < 5; j++) if (arr[0][j] != -1) arr[0][j] = arr[0][j - 1]; // Mark reachable nodes in remaining // matrix. for (var i = 1; i < 5; i++) for (var 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 // Given array var 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)) document.write("Yes"); else document.write("No"); // This code is contributed by Mayank Tyagi</script> |
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.
- Time Complexity: O(R*C).
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