Longest Possible Route in a Matrix with Hurdles

Given an M x N matrix, with a few hurdles arbitrarily placed, calculate the length of the longest possible route possible from source to a destination within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.

For example, the longest path with no hurdles from source to destination is highlighted below. The length of the path is 24.

The idea is to use Backtracking. We start from the source cell of the matrix, move forward in all four allowed directions, and recursively checks if they lead to the solution or not. If the destination is found, we update the value of the longest path else if none of the above solutions work we return false from our function.

Below is the implementation of the above idea 

CPP

// C++ program to find Longest Possible Route in a
// matrix with hurdles
#include <bits/stdc++.h>
using namespace std;
#define R 3
#define C 10

// A Pair to store status of a cell. found is set to
// true of destination is reachable and value stores
// distance of longest path
struct Pair {
    // true if destination is found
    bool found;

    // stores cost of longest path from current cell to
    // destination cell
    int value;
};

// Function to find Longest Possible Route in the
// matrix with hurdles. If the destination is not reachable
// the function returns false with cost INT_MAX.
// (i, j) is source cell and (x, y) is destination cell.
Pair findLongestPathUtil(int mat[R][C], int i, int j, int x,
                         int y, bool visited[R][C])
{

    // if (i, j) itself is destination, return true
    if (i == x && j == y) {
        Pair p = { true, 0 };
        return p;
    }

    // if not a valid cell, return false
    if (i < 0 || i >= R || j < 0 || j >= C || mat[i][j] == 0
        || visited[i][j]) {
        Pair p = { false, INT_MAX };
        return p;
    }

    // include (i, j) in current path i.e.
    // set visited(i, j) to true
    visited[i][j] = true;

    // res stores longest path from current cell (i, j) to
    // destination cell (x, y)
    int res = INT_MIN;

    // go left from current cell
    Pair sol
        = findLongestPathUtil(mat, i, j - 1, x, y, visited);

    // if destination can be reached on going left from
    // current cell, update res
    if (sol.found)
        res = max(res, sol.value);

    // go right from current cell
    sol = findLongestPathUtil(mat, i, j + 1, x, y, visited);

    // if destination can be reached on going right from
    // current cell, update res
    if (sol.found)
        res = max(res, sol.value);

    // go up from current cell
    sol = findLongestPathUtil(mat, i - 1, j, x, y, visited);

    // if destination can be reached on going up from
    // current cell, update res
    if (sol.found)
        res = max(res, sol.value);

    // go down from current cell
    sol = findLongestPathUtil(mat, i + 1, j, x, y, visited);

    // if destination can be reached on going down from
    // current cell, update res
    if (sol.found)
        res = max(res, sol.value);

    // Backtrack
    visited[i][j] = false;

    // if destination can be reached from current cell,
    // return true
    if (res != INT_MIN) {
        Pair p = { true, 1 + res };
        return p;
    }

    // if destination can't be reached from current cell,
    // return false
    else {
        Pair p = { false, INT_MAX };
        return p;
    }
}

// A wrapper function over findLongestPathUtil()
void findLongestPath(int mat[R][C], int i, int j, int x,
                     int y)
{
    // create a boolean matrix to store info about
    // cells already visited in current route
    bool visited[R][C];

    // initialize visited to false
    memset(visited, false, sizeof visited);

    // find longest route from (i, j) to (x, y) and
    // print its maximum cost
    Pair p = findLongestPathUtil(mat, i, j, x, y, visited);
    if (p.found)
        cout << "Length of longest possible route is "
             << p.value;

    // If the destination is not reachable
    else
        cout << "Destination not reachable from given "
                "source";
}

// Driver code
int main()
{
    // input matrix with hurdles shown with number 0
    int mat[R][C] = { { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 },
                      { 1, 1, 0, 1, 1, 0, 1, 1, 0, 1 },
                      { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 } };

    // find longest path with source (0, 0) and
    // destination (1, 7)
    findLongestPath(mat, 0, 0, 1, 7);

    return 0;
}

Java

// Java program to find Longest Possible Route in a
// matrix with hurdles
import java.io.*;

class GFG {
  static int R = 3;
  static int C = 10;

  // A Pair to store status of a cell. found is set to
  // true of destination is reachable and value stores
  // distance of longest path
  static class Pair {

    // true if destination is found
    boolean found;

    // stores cost of longest path from current cell to
    // destination cell
    int val;

    Pair (boolean x, int y){
      found = x;
      val = y;
    }
  }

  // Function to find Longest Possible Route in the
  // matrix with hurdles. If the destination is not reachable
  // the function returns false with cost Integer.MAX_VALUE.
  // (i, j) is source cell and (x, y) is destination cell.

  static Pair findLongestPathUtil (int mat[][], int i, int j, int x, int y, boolean visited[][]) {

    // if (i, j) itself is destination, return true
    if(i == x && j == y)
      return new Pair(true, 0);


    // if not a valid cell, return false
    if(i < 0 || i >= R || j < 0 || j >= C || mat[i][j] == 0 || visited[i][j] )
      return new Pair(false, Integer.MAX_VALUE);

    // include (i, j) in current path i.e.
    // set visited(i, j) to true
    visited[i][j] = true;

    // res stores longest path from current cell (i, j) to
    // destination cell (x, y)
    int res = Integer.MIN_VALUE;

    // go left from current cell
    Pair sol = findLongestPathUtil(mat, i, j-1, x, y, visited);

    // if destination can be reached on going left from current
    // cell, update res
    if(sol.found)
      res = Math.max(sol.val, res);

    // go right from current cell
    sol = findLongestPathUtil(mat, i, j+1, x, y, visited);

    // if destination can be reached on going right from current
    // cell, update res
    if(sol.found)
      res = Math.max(sol.val, res);

    // go up from current cell
    sol = findLongestPathUtil(mat, i-1, j, x, y, visited);

    // if destination can be reached on going up from current
    // cell, update res
    if(sol.found)
      res = Math.max(sol.val, res);

    // go down from current cell
    sol = findLongestPathUtil(mat, i+1, j, x, y, visited);

    // if destination can be reached on going down from current
    // cell, update res
    if(sol.found)
      res = Math.max(sol.val, res);

    // Backtrack
    visited[i][j] = false;

    // if destination can be reached from current cell,
    // return true
    if(res != Integer.MIN_VALUE)
      return new Pair(true, res+1);

    // if destination can't be reached from current cell,
    // return false
    else
      return new Pair(false, Integer.MAX_VALUE);

  }

  // A wrapper function over findLongestPathUtil()
  static void findLongestPath (int mat[][], int i, int j, int x, int y) {

    // create a boolean matrix to store info about
    // cells already visited in current route
    boolean visited[][] = new boolean[R][C];


    // find longest route from (i, j) to (x, y) and
    // print its maximum cost
    Pair p = findLongestPathUtil(mat, i, j, x, y, visited);

    if(p.found)
      System.out.println("Length of longest possible route is " + p.val);

    // If the destination is not reachable
    else
      System.out.println("Destination not reachable from given source");

  }


  // Driver Code
  public static void main (String[] args) {

    // input matrix with hurdles shown with number 0
    int mat[][] = { { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 },
                   { 1, 1, 0, 1, 1, 0, 1, 1, 0, 1 },
                   { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 } };

    // find longest path with source (0, 0) and
    // destination (1, 7)
    findLongestPath(mat, 0, 0, 1, 7);

  }
}

// This code is contributed by th_aditi.

Python3

# Python program to find Longest Possible Route in a
# matrix with hurdles
import sys
R,C = 3,10

# A Pair to store status of a cell. found is set to
# True of destination is reachable and value stores
# distance of longest path
class Pair:

    def __init__(self, found, value):
        self.found = found
        self.value = value

# Function to find Longest Possible Route in the
# matrix with hurdles. If the destination is not reachable
# the function returns false with cost sys.maxsize.
# (i, j) is source cell and (x, y) is destination cell.
def findLongestPathUtil(mat, i, j, x, y, visited):

    # if (i, j) itself is destination, return True
    if (i == x and j == y):
        p = Pair( True, 0 )
        return p

    # if not a valid cell, return false
    if (i < 0 or i >= R or j < 0 or j >= C or mat[i][j] == 0 or visited[i][j]) :
        p = Pair( False, sys.maxsize )
        return p

    # include (i, j) in current path i.e.
    # set visited(i, j) to True
    visited[i][j] = True

    # res stores longest path from current cell (i, j) to
    # destination cell (x, y)
    res = -sys.maxsize -1

    # go left from current cell
    sol = findLongestPathUtil(mat, i, j - 1, x, y, visited)

    # if destination can be reached on going left from
    # current cell, update res
    if (sol.found):
        res = max(res, sol.value)

    # go right from current cell
    sol = findLongestPathUtil(mat, i, j + 1, x, y, visited)

    # if destination can be reached on going right from
    # current cell, update res
    if (sol.found):
        res = max(res, sol.value)

    # go up from current cell
    sol = findLongestPathUtil(mat, i - 1, j, x, y, visited)

    # if destination can be reached on going up from
    # current cell, update res
    if (sol.found):
        res = max(res, sol.value)

    # go down from current cell
    sol = findLongestPathUtil(mat, i + 1, j, x, y, visited)

    # if destination can be reached on going down from
    # current cell, update res
    if (sol.found):
        res = max(res, sol.value)

    # Backtrack
    visited[i][j] = False

    # if destination can be reached from current cell,
    # return True
    if (res != -sys.maxsize -1):
        p = Pair( True, 1 + res )
        return p

    # if destination can't be reached from current cell,
    # return false
    else:
        p = Pair( False, sys.maxsize )
        return p

# A wrapper function over findLongestPathUtil()
def findLongestPath(mat, i, j, x,y):

    # create a boolean matrix to store info about
    # cells already visited in current route
    # initialize visited to false
    visited = [[False for i in range(C)]for j in range(R)]

    # find longest route from (i, j) to (x, y) and
    # print its maximum cost
    p = findLongestPathUtil(mat, i, j, x, y, visited)
    if (p.found):
        print("Length of longest possible route is ",str(p.value))

    # If the destination is not reachable
    else:
        print("Destination not reachable from given source")

# Driver code

# input matrix with hurdles shown with number 0
mat = [ [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 ],
        [ 1, 1, 0, 1, 1, 0, 1, 1, 0, 1 ],
        [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 ] ]

# find longest path with source (0, 0) and
# destination (1, 7)
findLongestPath(mat, 0, 0, 1, 7)

# This code is contributed by shinjanpatra

C#

// Java program to find Longest Possible Route in a
// matrix with hurdles
using System;

class GFG {
  static int R = 3;
  static int C = 10;

  // Function to find Longest Possible Route in the
  // matrix with hurdles. If the destination is not reachable
  // the function returns false with cost Integer.MAX_VALUE.
  // (i, j) is source cell and (x, y) is destination cell.

  static Tuple<bool,int> findLongestPathUtil (int[, ] mat, int i, int j, int x, int y, bool[, ] visited) {

    // if (i, j) itself is destination, return true
    if(i == x && j == y)
      return new Tuple<bool,int>(true, 0);


    // if not a valid cell, return false
    if(i < 0 || i >= R || j < 0 || j >= C || mat[i,j] == 0 || visited[i,j])
      return new Tuple<bool,int>(false, Int32.MaxValue);

    // include (i, j) in current path i.e.
    // set visited(i, j) to true
    visited[i,j] = true;

    // res stores longest path from current cell (i, j) to
    // destination cell (x, y)
    int res = Int32.MinValue;

    // go left from current cell
    Tuple<bool,int> sol = findLongestPathUtil(mat, i, j-1, x, y, visited);

    // if destination can be reached on going left from current
    // cell, update res
    if(sol.Item1)
      res = Math.Max(sol.Item2, res);

    // go right from current cell
    sol = findLongestPathUtil(mat, i, j+1, x, y, visited);

    // if destination can be reached on going right from current
    // cell, update res
    if(sol.Item1)
      res = Math.Max(sol.Item2, res);

    // go up from current cell
    sol = findLongestPathUtil(mat, i-1, j, x, y, visited);

    // if destination can be reached on going up from current
    // cell, update res
    if(sol.Item1)
      res = Math.Max(sol.Item2, res);

    // go down from current cell
    sol = findLongestPathUtil(mat, i+1, j, x, y, visited);

    // if destination can be reached on going down from current
    // cell, update res
    if(sol.Item1)
      res = Math.Max(sol.Item2, res);

    // Backtrack
    visited[i,j] = false;

    // if destination can be reached from current cell,
    // return true
    if(res != Int32.MinValue)
      return new Tuple<bool,int>(true, res+1);

    // if destination can't be reached from current cell,
    // return false
    else
      return new Tuple<bool,int>(false, Int32.MaxValue);

  }

  // A wrapper function over findLongestPathUtil()
  static void findLongestPath (int [, ]mat, int i, int j, int x, int y) {

    // create a boolean matrix to store info about
    // cells already visited in current route
    bool[,] visited = new bool[R,C];


    // find longest route from (i, j) to (x, y) and
    // print its maximum cost
    Tuple<bool,int> p = findLongestPathUtil(mat, i, j, x, y, visited);

    if(p.Item1)
      Console.WriteLine("Length of longest possible route is : " + p.Item2);

    // If the destination is not reachable
    else
      Console.WriteLine("Destination not reachable from given source");

  }


  // Driver Code
  public static void Main() {

    // input matrix with hurdles shown with number 0
    int[,] mat = new int[,] { { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 },
                   { 1, 1, 0, 1, 1, 0, 1, 1, 0, 1 },
                   { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 } };

    // find longest path with source (0, 0) and
    // destination (1, 7)
    findLongestPath(mat, 0, 0, 1, 7);

  }
}

// This code is contributed by Abhijeet Kumar(abhijeet19403)

Javascript

// JavaScript program to find Longest Possible Route in a
// matrix with hurdles
var R = 3;
var C = 10;

// A Pair to store status of a cell. found is set to
// True of destination is reachable and value stores
// distance of longest path
class Pair {
    constructor(found, value) {
        this.found = found;
        this.value = value;
    }
}

// Function to find Longest Possible Route in the
// matrix with hurdles. If the destination is not reachable
// the function returns false with cost sys.maxsize.
// (i, j) is source cell and (x, y) is destination cell.
function findLongestPathUtil(mat, i, j, x, y, visited) {
    // if (i, j) itself is destination, return True
    if (i == x && j == y) {
        var p = new Pair(true, 0);
        return p;
    }

    // if not a valid cell, return false
    if (i < 0 || i >= R || j < 0 || j >= C || mat[i][j] == 0 || visited[i][j]) {
        var p = new Pair(false, Number.MAX_SAFE_INTEGER);
        return p;
    }

    // include (i, j) in current path i.e.
    // set visited(i, j) to True
    visited[i][j] = true;

    // res stores longest path from current cell (i, j) to
    // destination cell (x, y)
    var res = Number.MIN_SAFE_INTEGER ;

    // go left from current cell
    var sol = findLongestPathUtil(mat, i, j - 1, x, y, visited);

    // if destination can be reached on going left from
    // current cell, update res
    if (sol.found) {
        res = Math.max(res, sol.value);
    }

    // go right from current cell
    sol = findLongestPathUtil(mat, i, j + 1, x, y, visited);

    // if destination can be reached on going right from
    // current cell, update res
    if (sol.found) {
        res = Math.max(res, sol.value);
    }

    // go up from current cell
    sol = findLongestPathUtil(mat, i - 1, j, x, y, visited);

    // if destination can be reached on going up from
    // current cell, update res
    if (sol.found) {
        res = Math.max(res, sol.value);
    }

    // go down from current cell
    sol = findLongestPathUtil(mat, i + 1, j, x, y, visited);

    // if destination can be reached on going down from
    // current cell, update res
    if (sol.found) {
        res = Math.max(res, sol.value);
    }

    // Backtrack
    visited[i][j] = false;

    // if destination can be reached from current cell,
    // return True
    if (res != Number.MIN_SAFE_INTEGER ) {
        var p = new Pair(true, res+1);
        return p;
    }

    // if destination can't be reached from current cell,
    // return false
    else {
        var p = new Pair(false, Number.MAX_SAFE_INTEGER);
        return p;
    }
}

// A wrapper function over findLongestPathUtil()
function findLongestPath(mat, i, j, x, y) {
    // create a boolean matrix to store info about
    // cells already visited in current route
    // initialize visited to false
    var visited = new Array(R);
    for (var k = 0; k < R; k++) {
        visited[k] = new Array(C);
    }

    // find longest route from (i, j) to (x, y) and
    // print its maximum cost
    var p = findLongestPathUtil(mat, i, j, x, y, visited);
    if (p.found) {
        console.log("Length of longest possible route is " + p.value);
    }
    // If the destination is not reachable
    else {
        console.log("Destination not reachable from given source");
    }
}

// Driver code

// input matrix with hurdles shown with number 0
var mat = [
    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
    [1, 1, 0, 1, 1, 0, 1, 1, 0, 1],
    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
];

// find longest path with source (0, 0) and
// destination (1, 7)
findLongestPath(mat, 0, 0, 1, 7);

// This code is contributed by Tapesh(tapeshdua420)

Output

Length of longest possible route is 24

Time Complexity: 4^(R*C), Here R and C are the numbers of rows and columns respectively. For every index we have four options, so our overall time complexity will become 4^(R*C).
Auxiliary Space: O(R*C), The extra space is used in storing the elements of the visited matrix.

An approach without using any extra space:

Below is the step-by-step approach:

1. Start from the source cell.
2. Explore all possible directions (right, down, left, up) from the current cell.
3. If a valid adjacent cell is found (within the boundaries of the matrix and has a value of 1), move to that cell and increment the current path length.
4. Recursively repeat steps 2 and 3 for the new cell.
5. If the destination cell is reached, compare the current path length with the longest path length found so far and update it if necessary.
6. Backtrack by undoing the move (mark the current cell as visited) and continue exploring other directions.
7. Repeat steps 2-6 until all possible paths are explored.
8. Return the longest path length as the result.

Below is the implementation:

C++

#include <iostream>
#include <vector>
using namespace std;

// Function for finding the longest path
// 'ans' is -1 if we can't reach
// 'cur' is the number of steps we have traversed
int findLongestPath(vector<vector<int> >& mat, int i, int j,
                    int di, int dj, int n, int m,
                    int cur = 0, int ans = -1)
{
    // If we reach the destination
    if (i == di && j == dj) {
        // If current path steps are more than previous path
        // steps
        if (cur > ans)
            ans = cur;
        return ans;
    }

      //if the source or destination is a hurdle itself
      if(mat[i][j]==0 || mat[di][dj]==0) return;

    // Mark as visited
    mat[i][j] = 0;

    // Checking if we can reach the destination going right
    if (j != m - 1 && mat[i][j + 1] > 0)
        ans = findLongestPath(mat, i, j + 1, di, dj, n, m,
                              cur + 1, ans);

    // Checking if we can reach the destination going down
    if (i != n - 1 && mat[i + 1][j] > 0)
        ans = findLongestPath(mat, i + 1, j, di, dj, n, m,
                              cur + 1, ans);

    // Checking if we can reach the destination going left
    if (j != 0 && mat[i][j - 1] > 0)
        ans = findLongestPath(mat, i, j - 1, di, dj, n, m,
                              cur + 1, ans);

    // Checking if we can reach the destination going up
    if (i != 0 && mat[i - 1][j] > 0)
        ans = findLongestPath(mat, i - 1, j, di, dj, n, m,
                              cur + 1, ans);

    // Marking visited to backtrack
    mat[i][j] = 1;

    // Returning the answer we got so far
    return ans;
}

int main()
{
    vector<vector<int> > mat = { { 1, 1, 1, 1 },
                                 { 1, 1, 0, 1 },
                                 { 1, 1, 1, 1 } };

    // Find the longest path with source (0, 0) and
    // destination (2, 3)
    int result = findLongestPath(mat, 0, 0, 2, 3,
                                 mat.size(), mat[0].size());
    cout << result << endl;

    return 0;
}

Java

public class Main {
    // Function for finding the longest path
    // 'ans' is -1 if we can't reach
    // 'cur' is the number of steps we have traversed
    public static int findLongestPath(int[][] mat, int i,
                                      int j, int di, int dj,
                                      int n, int m, int cur,
                                      int ans)
    {
        // If we reach the destination
        if (i == di && j == dj) {
            // If current path steps are more than previous
            // path steps
            if (cur > ans)
                ans = cur;
            return ans;
        }

          //if the source or destination is a hurdle itself
          if(mat[i][j]==0 || mat[di][dj]==0) return ans;

        // Mark as visited
        mat[i][j] = 0;

        // Checking if we can reach the destination going
        // right
        if (j != m - 1 && mat[i][j + 1] > 0)
            ans = findLongestPath(mat, i, j + 1, di, dj, n,
                                  m, cur + 1, ans);

        // Checking if we can reach the destination going
        // down
        if (i != n - 1 && mat[i + 1][j] > 0)
            ans = findLongestPath(mat, i + 1, j, di, dj, n,
                                  m, cur + 1, ans);

        // Checking if we can reach the destination going
        // left
        if (j != 0 && mat[i][j - 1] > 0)
            ans = findLongestPath(mat, i, j - 1, di, dj, n,
                                  m, cur + 1, ans);

        // Checking if we can reach the destination going up
        if (i != 0 && mat[i - 1][j] > 0)
            ans = findLongestPath(mat, i - 1, j, di, dj, n,
                                  m, cur + 1, ans);

        // Marking visited to backtrack
        mat[i][j] = 1;

        // Returning the answer we got so far
        return ans;
    }

    public static void main(String[] args)
    {
        int[][] mat = { { 1, 1, 1, 1 },
                        { 1, 1, 0, 1 },
                        { 1, 1, 1, 1 } };

        // Find the longest path with source (0, 0) and
        // destination (2, 3)
        int result
            = findLongestPath(mat, 0, 0, 2, 3, mat.length,
                              mat[0].length, 0, -1);
        System.out.println(result);
    }
}

Python3

# Function for finding the longest path
# 'ans' is -1 if we can't reach
# 'cur' is the number of steps we have traversed


def findLongestPath(mat, i, j, di, dj, n, m, cur=0, ans=-1):
    # If we reach the destination
    if i == di and j == dj:
        # If current path steps are more than previous path steps
        if cur > ans:
            ans = cur
        return ans

    # if the source or destination is a hurdle itself
      if mat[i][j]==0 or mat[di][dj]==0:
      return ans
    # Mark as visited
    mat[i][j] = 0

    # Checking if we can reach the destination going right
    if j != m-1 and mat[i][j+1] > 0:
        ans = findLongestPath(mat, i, j+1, di, dj, n, m, cur+1, ans)

    # Checking if we can reach the destination going down
    if i != n-1 and mat[i+1][j] > 0:
        ans = findLongestPath(mat, i+1, j, di, dj, n, m, cur+1, ans)

    # Checking if we can reach the destination going left
    if j != 0 and mat[i][j-1] > 0:
        ans = findLongestPath(mat, i, j-1, di, dj, n, m, cur+1, ans)

    # Checking if we can reach the destination going up
    if i != 0 and mat[i-1][j] > 0:
        ans = findLongestPath(mat, i-1, j, di, dj, n, m, cur+1, ans)

    # Marking visited to backtrack
    mat[i][j] = 1

    # Returning the answer we got so far
    return ans


mat = [
    [1, 1, 1, 1],
    [1, 1, 0, 1],
    [1, 1, 1, 1]
]

# Find the longest path with source (0, 0) and destination (2, 3)
vis = [[False for _ in mat[0]] for x in mat]
print(findLongestPath(mat, 0, 0, 2, 3, len(mat), len(mat[0])))

C#

using System;

public class Program {
    // Function for finding the longest path
    // 'ans' is -1 if we can't reach
    // 'cur' is the number of steps we have traversed
    public static int FindLongestPath(int[][] mat, int i,
                                      int j, int di, int dj,
                                      int n, int m,
                                      int cur = 0,
                                      int ans = -1)
    {
        // If we reach the destination
        if (i == di && j == dj) {
            // If current path steps are more than previous
            // path steps
            if (cur > ans)
                ans = cur;
            return ans;
        }

        //if the source or destination is a hurdle itself
          if(mat[i][j]==0 || mat[di][dj]==0) return ans;

        // Mark as visited
        mat[i][j] = 0;

        // Checking if we can reach the destination going
        // right
        if (j != m - 1 && mat[i][j + 1] > 0)
            ans = FindLongestPath(mat, i, j + 1, di, dj, n,
                                  m, cur + 1, ans);

        // Checking if we can reach the destination going
        // down
        if (i != n - 1 && mat[i + 1][j] > 0)
            ans = FindLongestPath(mat, i + 1, j, di, dj, n,
                                  m, cur + 1, ans);

        // Checking if we can reach the destination going
        // left
        if (j != 0 && mat[i][j - 1] > 0)
            ans = FindLongestPath(mat, i, j - 1, di, dj, n,
                                  m, cur + 1, ans);

        // Checking if we can reach the destination going up
        if (i != 0 && mat[i - 1][j] > 0)
            ans = FindLongestPath(mat, i - 1, j, di, dj, n,
                                  m, cur + 1, ans);

        // Marking visited to backtrack
        mat[i][j] = 1;

        // Returning the answer we got so far
        return ans;
    }

    public static void Main(string[] args)
    {
        int[][] mat
            = new int[][] { new int[] { 1, 1, 1, 1 },
                            new int[] { 1, 1, 0, 1 },
                            new int[] { 1, 1, 1, 1 } };

        // Find the longest path with source (0, 0) and
        // destination (2, 3)
        int result = FindLongestPath(
            mat, 0, 0, 2, 3, mat.Length, mat[0].Length);
        Console.WriteLine(result);
    }
}

Javascript

// Function for finding the longest path
// 'ans' is -1 if we can't reach
// 'cur' is the number of steps we have traversed
function findLongestPath(mat, i, j, di, dj, n, m, cur = 0, ans = -1) {
    // If we reach the destination
    if (i === di && j === dj) {
        // If current path steps are more than previous path steps
        if (cur > ans)
            ans = cur;
        return ans;
    }

    //if the source or destination is a hurdle itself
      if(mat[i][j]==0 || mat[di][dj]==0) return ans;

    // Mark as visited
    mat[i][j] = 0;

    // Checking if we can reach the destination going right
    if (j !== m - 1 && mat[i][j + 1] > 0)
        ans = findLongestPath(mat, i, j + 1, di, dj, n, m, cur + 1, ans);

    // Checking if we can reach the destination going down
    if (i !== n - 1 && mat[i + 1][j] > 0)
        ans = findLongestPath(mat, i + 1, j, di, dj, n, m, cur + 1, ans);

    // Checking if we can reach the destination going left
    if (j !== 0 && mat[i][j - 1] > 0)
        ans = findLongestPath(mat, i, j - 1, di, dj, n, m, cur + 1, ans);

    // Checking if we can reach the destination going up
    if (i !== 0 && mat[i - 1][j] > 0)
        ans = findLongestPath(mat, i - 1, j, di, dj, n, m, cur + 1, ans);

    // Marking visited to backtrack
    mat[i][j] = 1;

    // Returning the answer we got so far
    return ans;
}

const mat = [
    [1, 1, 1, 1],
    [1, 1, 0, 1],
    [1, 1, 1, 1]
];

// Find the longest path with source (0, 0) and destination (2, 3)
const result = findLongestPath(mat, 0, 0, 2, 3, mat.length, mat[0].length);
console.log(result);

Output

9

Time Complexity: O(4^N), where N is the number of cells in the matrix. This is because, at each cell, there are four possible directions to explore (right, down, left, up), and the maximum depth of the recursion is N.
Auxiliary Space: O(1)

This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.