Count number of ways to reach destination in a Maze
Given a maze with obstacles, count number of paths to reach rightmost-bottommost cell from topmost-leftmost cell. A cell in given maze has value -1 if it is a blockage or dead end, else 0.
From a given cell, we are allowed to move to cells (i+1, j) and (i, j+1) only.
Examples:
Input: maze[R][C] = {{0, 0, 0, 0},
{0, -1, 0, 0},
{-1, 0, 0, 0},
{0, 0, 0, 0}};
Output: 4
There are four possible paths as shown in
below diagram.
This problem is an extension of below problem.
Backtracking | Set 2 (Rat in a Maze)
In this post a different solution is discussed that can be used to solve the above Rat in a Maze problem also.
The idea is to modify the given grid[][] so that grid[i][j] contains count of paths to reach (i, j) from (0, 0) if (i, j) is not a blockage, else grid[i][j] remains -1.
We can recursively compute grid[i][j] using below
formula and finally return grid[R-1][C-1]
// If current cell is a blockage
if (maze[i][j] == -1)
maze[i][j] = -1; // Do not change
// If we can reach maze[i][j] from maze[i-1][j]
// then increment count.
else if (maze[i-1][j] > 0)
maze[i][j] = (maze[i][j] + maze[i-1][j]);
// If we can reach maze[i][j] from maze[i][j-1]
// then increment count.
else if (maze[i][j-1] > 0)
maze[i][j] = (maze[i][j] + maze[i][j-1]);
Below is the implementation of above idea.
C++
// C++ program to count number of paths in a maze // with obstacles. #include<bits/stdc++.h> using namespace std; #define R 4 #define C 4 // Returns count of possible paths in a maze[R][C] // from (0,0) to (R-1,C-1) int countPaths(int maze[][C]) { // If the initial cell is blocked, there is no // way of moving anywhere if (maze[0][0]==-1) return 0; // Initializing the leftmost column for (int i=0; i<R; i++) { if (maze[i][0] == 0) maze[i][0] = 1; // If we encounter a blocked cell in leftmost // row, there is no way of visiting any cell // directly below it. else break; } // Similarly initialize the topmost row for (int i=1; i<C; i++) { if (maze[0][i] == 0) maze[0][i] = 1; // If we encounter a blocked cell in bottommost // row, there is no way of visiting any cell // directly below it. else break; } // The only difference is that if a cell is -1, // simply ignore it else recursively compute // count value maze[i][j] for (int i=1; i<R; i++) { for (int j=1; j<C; j++) { // If blockage is found, ignore this cell if (maze[i][j] == -1) continue; // If we can reach maze[i][j] from maze[i-1][j] // then increment count. if (maze[i-1][j] > 0) maze[i][j] = (maze[i][j] + maze[i-1][j]); // If we can reach maze[i][j] from maze[i][j-1] // then increment count. if (maze[i][j-1] > 0) maze[i][j] = (maze[i][j] + maze[i][j-1]); } } // If the final cell is blocked, output 0, otherwise // the answer return (maze[R-1][C-1] > 0)? maze[R-1][C-1] : 0; } // Driver code int main() { int maze[R][C] = {{0, 0, 0, 0}, {0, -1, 0, 0}, {-1, 0, 0, 0}, {0, 0, 0, 0}}; cout << countPaths(maze); return 0; } |
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Java
// Java program to count number of paths in a maze // with obstacles. import java.io.*; class GFG { static int R = 4; static int C = 4; // Returns count of possible paths in // a maze[R][C] from (0,0) to (R-1,C-1) static int countPaths(int maze[][]) { // If the initial cell is blocked, // there is no way of moving anywhere if (maze[0][0]==-1) return 0; // Initializing the leftmost column for (int i = 0; i < R; i++) { if (maze[i][0] == 0) maze[i][0] = 1; // If we encounter a blocked cell // in leftmost row, there is no way // of visiting any cell directly below it. else break; } // Similarly initialize the topmost row for (int i =1 ; i< C ; i++) { if (maze[0][i] == 0) maze[0][i] = 1; // If we encounter a blocked cell in // bottommost row, there is no way of // visiting any cell directly below it. else break; } // The only difference is that if a cell // is -1, simply ignore it else recursively // compute count value maze[i][j] for (int i = 1; i < R; i++) { for (int j = 1; j <C ; j++) { // If blockage is found, // ignore this cell if (maze[i][j] == -1) continue; // If we can reach maze[i][j] from // maze[i-1][j] then increment count. if (maze[i - 1][j] > 0) maze[i][j] = (maze[i][j] + maze[i - 1][j]); // If we can reach maze[i][j] from // maze[i][j-1] then increment count. if (maze[i][j - 1] > 0) maze[i][j] = (maze[i][j] + maze[i][j - 1]); } } // If the final cell is blocked, // output 0, otherwise the answer return (maze[R - 1][C - 1] > 0) ? maze[R - 1][C - 1] : 0; } // Driver code public static void main (String[] args) { int maze[][] = {{0, 0, 0, 0}, {0, -1, 0, 0}, {-1, 0, 0, 0}, {0, 0, 0, 0}}; System.out.println (countPaths(maze)); } } // This code is contributed by vt_m |
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Python3
# Python 3 program to count number of paths # in a maze with obstacles. R = 4C = 4 # Returns count of possible paths in a # maze[R][C] from (0,0) to (R-1,C-1) def countPaths(maze): # If the initial cell is blocked, # there is no way of moving anywhere if (maze[0][0] == -1): return 0 # Initializing the leftmost column for i in range(R): if (maze[i][0] == 0): maze[i][0] = 1 # If we encounter a blocked cell in # leftmost row, there is no way of # visiting any cell directly below it. else: break # Similarly initialize the topmost row for i in range(1, C, 1): if (maze[0][i] == 0): maze[0][i] = 1 # If we encounter a blocked cell in # bottommost row, there is no way of # visiting any cell directly below it. else: break # The only difference is that if a cell is -1, # simply ignore it else recursively compute # count value maze[i][j] for i in range(1, R, 1): for j in range(1, C, 1): # If blockage is found, ignore this cell if (maze[i][j] == -1): continue # If we can reach maze[i][j] from # maze[i-1][j] then increment count. if (maze[i - 1][j] > 0): maze[i][j] = (maze[i][j] + maze[i - 1][j]) # If we can reach maze[i][j] from # maze[i][j-1] then increment count. if (maze[i][j - 1] > 0): maze[i][j] = (maze[i][j] + maze[i][j - 1]) # If the final cell is blocked, # output 0, otherwise the answer if (maze[R - 1][C - 1] > 0): return maze[R - 1][C - 1] else: return 0 # Driver code if __name__ == '__main__': maze = [[0, 0, 0, 0], [0, -1, 0, 0], [-1, 0, 0, 0], [0, 0, 0, 0 ]] print(countPaths(maze)) # This code is contributed by # Surendra_Gangwar |
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C#
// C# program to count number of paths in a maze // with obstacles. using System; class GFG { static int R = 4; static int C = 4; // Returns count of possible paths in // a maze[R][C] from (0,0) to (R-1,C-1) static int countPaths(int [,]maze) { // If the initial cell is blocked, // there is no way of moving anywhere if (maze[0,0]==-1) return 0; // Initializing the leftmost column for (int i = 0; i < R; i++) { if (maze[i,0] == 0) maze[i,0] = 1; // If we encounter a blocked cell // in leftmost row, there is no way // of visiting any cell directly below it. else break; } // Similarly initialize the topmost row for (int i =1 ; i< C ; i++) { if (maze[0,i] == 0) maze[0,i] = 1; // If we encounter a blocked cell in // bottommost row, there is no way of // visiting any cell directly below it. else break; } // The only difference is that if a cell // is -1, simply ignore it else recursively // compute count value maze[i][j] for (int i = 1; i < R; i++) { for (int j = 1; j <C ; j++) { // If blockage is found, // ignore this cell if (maze[i,j] == -1) continue; // If we can reach maze[i][j] from // maze[i-1][j] then increment count. if (maze[i - 1,j] > 0) maze[i,j] = (maze[i,j] + maze[i - 1,j]); // If we can reach maze[i][j] from // maze[i][j-1] then increment count. if (maze[i,j - 1] > 0) maze[i,j] = (maze[i,j] + maze[i,j - 1]); } } // If the final cell is blocked, // output 0, otherwise the answer return (maze[R - 1,C - 1] > 0) ? maze[R - 1,C - 1] : 0; } // Driver code public static void Main () { int [,]maze = { {0, 0, 0, 0}, {0, -1, 0, 0}, {-1, 0, 0, 0}, {0, 0, 0, 0}}; Console.Write (countPaths(maze)); } } // This code is contributed by nitin mittal. |
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PHP
<?php // PHP program to count number // of paths in a maze with obstacles. $R = 4; $C = 4; // Returns count of possible // paths in a maze[R][C] // from (0,0) to (R-1,C-1) function countPaths( $maze) { global $R, $C; // If the initial cell is // blocked, there is no // way of moving anywhere if ($maze[0][0] == - 1) return 0; // Initializing the // leftmost column for ( $i = 0; $i < $R; $i++) { if ($maze[$i][0] == 0) $maze[$i][0] = 1; // If we encounter a blocked // cell in leftmost row, // there is no way of // visiting any cell // directly below it. else break; } // Similarly initialize // the topmost row for($i = 1; $i < $C; $i++) { if ($maze[0][$i] == 0) $maze[0][$i] = 1; // If we encounter a blocked // cell in bottommost row, // there is no way of // visiting any cell // directly below it. else break; } // The only difference is // that if a cell is -1, // simply ignore it else // recursively compute // count value maze[i][j] for($i = 1; $i < $R; $i++) { for($j = 1; $j < $C; $j++) { // If blockage is found, // ignore this cell if ($maze[$i][$j] == -1) continue; // If we can reach maze[i][j] // from maze[i-1][j] // then increment count. if ($maze[$i - 1][$j] > 0) $maze[$i][$j] = ($maze[$i][$j] + $maze[$i - 1][$j]); // If we can reach maze[i][j] // from maze[i][j-1] // then increment count. if ($maze[$i][$j - 1] > 0) $maze[$i][$j] = ($maze[$i][$j] + $maze[$i][$j - 1]); } } // If the final cell is // blocked, output 0, // otherwise the answer return ($maze[$R - 1][$C - 1] > 0) ? $maze[$R - 1][$C - 1] : 0; } // Driver Code $maze = array(array(0, 0, 0, 0), array(0, -1, 0, 0), array(-1, 0, 0, 0), array(0, 0, 0, 0)); echo countPaths($maze); // This code is contributed by anuj_67. ?> |
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Output:
4
Time Complexity : O(R x C)
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