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# Count all possible paths from top left to bottom right of a mXn matrix

The problem is to count all the possible paths from the top left to the bottom right of a M X N matrix with the constraints that from each cell you can either move only to the right or down

Examples:

Input:  M = 2, N = 2
Output: 2
Explanation: There are two paths
(0, 0) -> (0, 1) -> (1, 1)
(0, 0) -> (1, 0) -> (1, 1)

Input:  M = 2, N = 3
Output: 3
Explanation: There are three paths
(0, 0) -> (0, 1) -> (0, 2) -> (1, 2)
(0, 0) -> (0, 1) -> (1, 1) -> (1, 2)
(0, 0) -> (1, 0) -> (1, 1) -> (1, 2)

Recommended Practice

## Count all possible paths from top left to the bottom right of a M X N matrix using Recursion:

To solve the problem follow the below idea:

We can recursively move to right and down from the start until we reach the destination and then add up all valid paths to get the answer.

Follow the below steps to solve the problem:

• Create a recursive function with parameters as row and column index
• Call this recursive function for N-1 and M-1
• In the recursive function
• If N == 1 or M == 1 then return 1
• else call the recursive function with (N-1, M) and (N, M-1) and return the sum of this

Below is the implementation of the above approach:

## C++

 // A C++  program to count all possible paths // from top left to bottom right   #include  using namespace std;   // Returns count of possible paths to reach cell at row // number m and column number n from the topmost leftmost // cell (cell at 1, 1) int numberOfPaths(int m, int n) {     // If either given row number is first or given column     // number is first     if (m == 1 || n == 1)         return 1;       // If diagonal movements are allowed then the last     // addition is required.     return numberOfPaths(m - 1, n)            + numberOfPaths(m, n - 1);     // + numberOfPaths(m-1, n-1); }   // Driver code int main() {     cout << numberOfPaths(3, 3);     return 0; }

## Java

 // A Java program to count all possible paths // from top left to bottom right   class GFG {       // Returns count of possible paths to reach     // cell at row number m and column number n     // from the topmost leftmost cell (cell at 1, 1)     static int numberOfPaths(int m, int n)     {         // If either given row number is first or         // given column number is first         if (m == 1 || n == 1)             return 1;           // If diagonal movements are allowed then         // the last addition is required.         return numberOfPaths(m - 1, n)             + numberOfPaths(m, n - 1);         // + numberOfPaths(m-1, n-1);     }         // Driver code     public static void main(String args[])     {         System.out.println(numberOfPaths(3, 3));     } }   // This code is contributed by Sumit Ghosh

## Python3

 # Python program to count all possible paths # from top left to bottom right   # function to return count of possible paths # to reach cell at row number m and column # number n from the topmost leftmost # cell (cell at 1, 1)     def numberOfPaths(m, n):     # If either given row number is first     # or given column number is first     if(m == 1 or n == 1):         return 1   # If diagonal movements are allowed # then the last addition # is required.     return numberOfPaths(m-1, n) + numberOfPaths(m, n-1)     # Driver program to test above function if __name__ == '__main__':   m = 3   n = 3   print(numberOfPaths(m, n))   # This code is contributed by Aditi Sharma

## C#

 // A C# program to count all possible paths // from top left to bottom right   using System;   public class GFG {     // Returns count of possible paths to reach     // cell at row number m and column number n     // from the topmost leftmost cell (cell at 1, 1)     static int numberOfPaths(int m, int n)     {         // If either given row number is first or         // given column number is first         if (m == 1 || n == 1)             return 1;           // If diagonal movements are allowed then         // the last addition is required.         return numberOfPaths(m - 1, n)             + numberOfPaths(m, n - 1);         // + numberOfPaths(m-1, n-1);     }         // Driver code     static public void Main()     {         Console.WriteLine(numberOfPaths(3, 3));     } }   // This code is contributed by ajit

## PHP

 

## Javascript

 

Output

6

Time Complexity: O(2N)
Auxiliary Space: O(N + M)

## Count all possible paths from top left to the bottom right of a M X N matrix using Memoization:

To solve the problem follow the below idea:

As the above recursive solution has overlapping subproblems so we can declare a 2-D array to save the values for different states of the recursive function and later on use the values of this dp array to get the answer for already solved subproblems

Follow the below steps to solve the problem:

• Declare a 2-D array of size N X M
• Create a recursive function with parameters as row and column index and 2-D array
• Call this recursive function for N-1 and M-1
• In the recursive function
• If N == 1 or M == 1 then return 1
• If the value of this recursive function is not stored in the 2-D array then call the recursive function for (N-1, M, dp) and (N, M-1, dp) and assign the sum of answers of these functions in the 2-D array and return this value
• else return the value of this function stored in the 2-D array

Below is the implementation of the above approach:

## C++

 // A C++ program to count all possible paths from // top left to top bottom right using // Recursive Dynamic Programming #include  using namespace std;   // Returns count of possible paths to reach // cell at row number m and column number n from // the topmost leftmost cell (cell at 1, 1) int numberOfPaths(int n, int m, int DP) {       if (n == 1 || m == 1)         return DP[n][m] = 1;       // Add the element in the DP table     // If it was not computed before     if (DP[n][m] == 0)         DP[n][m] = numberOfPaths(n - 1, m, DP)                    + numberOfPaths(n, m - 1, DP);       return DP[n][m]; }   // Driver code int main() {     // Create an empty 2D table     int DP = { 0 };     memset(DP, 0, sizeof(DP));       cout << numberOfPaths(3, 3, DP);       return 0; } // This code is contributed // by Gatea David

## Java

 // Java program to count all possible paths from // top left to top bottom right using // Recursive Dynamic Programming import java.util.*; public class GFG {       // Returns count of possible paths to reach     // cell at row number m and column number n from     // the topmost leftmost cell (cell at 1, 1)     static int numberOfPaths(int n, int m, int DP[][])     {           if (n == 1 || m == 1)             return DP[n][m] = 1;           // Add the element in the DP table         // If it was not computed before         if (DP[n][m] == 0)             DP[n][m] = numberOfPaths(n - 1, m, DP)                        + numberOfPaths(n, m - 1, DP);           return DP[n][m];     }       // Driver code     public static void main(String args[])     {         // Create an empty 2D table         int DP[][] = new int;         for (int i = 0; i < 4; i++) {             for (int j = 0; j < 4; j++) {                 DP[i][j] = 0;             }         }           System.out.println(numberOfPaths(3, 3, DP));     } }   // This code is contributed // by Samim Hossain Mondal.

## Python3

 # Python program to count all possible paths from # top left to top bottom right using # Recursive Dynamic Programming   # Returns count of possible paths to reach # cell at row number m and column number n from # the topmost leftmost cell (cell at 1, 1)     def numberOfPaths(n, m, DP):       if (n == 1 or m == 1):         DP[n][m] = 1         return 1       # Add the element in the DP table     # If it was not computed before     if (DP[n][m] == 0):         DP[n][m] = numberOfPaths(n - 1, m, DP) + numberOfPaths(n, m - 1, DP)       return DP[n][m]     # Driver code if __name__ == '__main__':       # Create an empty 2D table     DP = [[0 for i in range(4)] for j in range(4)]       print(numberOfPaths(3, 3, DP))   # This code is contributed by gauravrajput1

## C#

 // C# program to count all possible paths from // top left to top bottom right using // Recursive Dynamic Programming using System; class GFG {       // Returns count of possible paths to reach     // cell at row number m and column number n from     // the topmost leftmost cell (cell at 1, 1)     static int numberOfPaths(int n, int m, int[, ] DP)     {           if (n == 1 || m == 1)             return DP[n, m] = 1;           // Add the element in the DP table         // If it was not computed before         if (DP[n, m] == 0)             DP[n, m] = numberOfPaths(n - 1, m, DP)                        + numberOfPaths(n, m - 1, DP);           return DP[n, m];     }       // Driver code     public static void Main()     {         // Create an empty 2D table         int[, ] DP = new int[4, 4];         for (int i = 0; i < 4; i++) {             for (int j = 0; j < 4; j++) {                 DP[i, j] = 0;             }         }           Console.WriteLine(numberOfPaths(3, 3, DP));     } }   // This code is contributed // by Samim Hossain Mondal.

## Javascript

 

Output

6

Time Complexity: O(N * M)
Auxiliary Space: (N * M)

## Count all possible paths from the top left to the bottom right of a M X N matrix using DP:

To solve the problem follow the below idea:

So this problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, recomputations of the same subproblems can be avoided by constructing a temporary array count[][] in a bottom-up manner using the above recursive formula

Follow the below steps to solve the problem:

• Declare a 2-D array count of size M * N
• Set value of count[i] equal to 1 for 0 <= i < M as the answer of subproblem with a single column is equal to 1
• Set value of count[j] equal to 1 for 0 <= j < N as the answer of subproblem with a single row is equal to 1
• Create a nested for loop for 0 <= i < M and 0 <= j < N and assign count[i][j] equal to count[i-1][j] + count[i][j-1]
• Print value of count[M-1][N-1]

Below is the implementation of the above approach:

## C++

 // A C++ program to count all possible paths // from top left to bottom right #include  using namespace std;   // Returns count of possible paths to reach cell at // row number m and column  number n from the topmost // leftmost cell (cell at 1, 1) int numberOfPaths(int m, int n) {     // Create a 2D table to store results of subproblems     int count[m][n];       // Count of paths to reach any cell in first column is 1     for (int i = 0; i < m; i++)         count[i] = 1;       // Count of paths to reach any cell in first row is 1     for (int j = 0; j < n; j++)         count[j] = 1;       // Calculate count of paths for other cells in     // bottom-up manner using the recursive solution     for (int i = 1; i < m; i++) {         for (int j = 1; j < n; j++)               // By uncommenting the last part the code             // calculates the total possible paths if the             // diagonal Movements are allowed             count[i][j]                 = count[i - 1][j]                   + count[i][j - 1]; //+ count[i-1][j-1];     }     return count[m - 1][n - 1]; }   // Driver code int main() {     cout << numberOfPaths(3, 3);     return 0; }

## Java

 // A Java program to count all possible paths // from top left to bottom right   import java.io.*;   class GFG {     // Returns count of possible paths to reach     // cell at row number m and column number n from     // the topmost leftmost cell (cell at 1, 1)     static int numberOfPaths(int m, int n)     {         // Create a 2D table to store results         // of subproblems         int count[][] = new int[m][n];           // Count of paths to reach any cell in         // first column is 1         for (int i = 0; i < m; i++)             count[i] = 1;           // Count of paths to reach any cell in         // first row is 1         for (int j = 0; j < n; j++)             count[j] = 1;           // Calculate count of paths for other         // cells in bottom-up manner using         // the recursive solution         for (int i = 1; i < m; i++) {             for (int j = 1; j < n; j++)                   // By uncommenting the last part the                 // code calculates the total possible paths                 // if the diagonal Movements are allowed                 count[i][j]                     = count[i - 1][j]                       + count[i]                              [j - 1]; //+ count[i-1][j-1];         }         return count[m - 1][n - 1];     }       // Driver code     public static void main(String args[])     {         System.out.println(numberOfPaths(3, 3));     } }   // This code is contributed by Sumit Ghosh

## Python3

 # Python3 program to count all possible paths # from top left to bottom right   # Returns count of possible paths to reach cell # at row number m and column number n from the # topmost leftmost cell (cell at 1, 1)     def numberOfPaths(m, n):     # Create a 2D table to store     # results of subproblems     # one-liner logic to take input for rows and columns     # mat = [[int(input()) for x in range (C)] for y in range(R)]       count = [[0 for x in range(n)] for y in range(m)]       # Count of paths to reach any     # cell in first column is 1     for i in range(m):         count[i] = 1       # Count of paths to reach any     # cell in first row is 1     for j in range(n):         count[j] = 1       # Calculate count of paths for other     # cells in bottom-up     # manner using the recursive solution     for i in range(1, m):         for j in range(1, n):             count[i][j] = count[i-1][j] + count[i][j-1]     return count[m-1][n-1]     # Driver code if __name__ == '__main__':   m = 3   n = 3   print(numberOfPaths(m, n))   # This code is contributed by Aditi Sharma

## C#

 // A C#  program to count all possible paths // from top left to bottom right using System;   public class GFG {       // Returns count of possible paths to reach     // cell at row number m and column number n from     // the topmost leftmost cell (cell at 1, 1)     static int numberOfPaths(int m, int n)     {         // Create a 2D table to store results         // of subproblems         int[, ] count = new int[m, n];           // Count of paths to reach any cell in         // first column is 1         for (int i = 0; i < m; i++)             count[i, 0] = 1;           // Count of paths to reach any cell in         // first row is 1         for (int j = 0; j < n; j++)             count[0, j] = 1;           // Calculate count of paths for other         // cells in bottom-up manner using         // the recursive solution         for (int i = 1; i < m; i++) {             for (int j = 1; j < n; j++)                   // By uncommenting the last part the                 // code calculates the total possible paths                 // if the diagonal Movements are allowed                 count[i, j]                     = count[i - 1, j]                       + count[i,                               j - 1]; //+ count[i-1][j-1];         }         return count[m - 1, n - 1];     }       // Driver code     static public void Main()     {         Console.WriteLine(numberOfPaths(3, 3));     } }   // This code is contributed by akt_mit

## PHP

 

## Javascript

 

Output

6

Time Complexity: O(M * N) – Due to nested for loops.
Auxiliary Space: O(M * N) – We have used a 2D array of size M x N

### Space optimization of the above approach:

To solve the problem follow the below idea:

We can space optimize the above dp approach as for calculating the values of the current row we require only previous row

Follow the below steps to solve the problem:

• Declare an array dp of size N
• Set dp = 1
• Create a nested for loop for 0 <= i < M and 0 <= j < N and add dp[j-1] to dp[j]
• Print value of dp[n – 1]

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach   #include  using namespace std;   // Returns count of possible paths to reach // cell at row number m and column number n from // the topmost leftmost cell (cell at 1, 1) int numberOfPaths(int m, int n) {     // Create a 1D array to store results of subproblems     int dp[n] = { 1 };     dp = 1;       for (int i = 0; i < m; i++) {         for (int j = 1; j < n; j++) {             dp[j] += dp[j - 1];         }     }       return dp[n - 1]; }   // Driver code int main() { cout << numberOfPaths(3, 3); }   // This code is contributed by mohit kumar 29

## Java

 // Java program for the above approach import java.io.*;   class GFG {     // Returns count of possible paths to reach     // cell at row number m and column number n from     // the topmost leftmost cell (cell at 1, 1)     static int numberOfPaths(int m, int n)     {         // Create a 1D array to store results of subproblems         int[] dp = new int[n];         dp = 1;           for (int i = 0; i < m; i++) {             for (int j = 1; j < n; j++) {                 dp[j] += dp[j - 1];             }         }           return dp[n - 1];     }       // Driver code     public static void main(String args[])     {         System.out.println(numberOfPaths(3, 3));     } }

## Python3

 # Returns count of possible paths # to reach cell at row number m and # column number n from the topmost # leftmost cell (cell at 1, 1)     def numberOfPaths(p, q):       # Create a 1D array to store     # results of subproblems     dp = [1 for i in range(q)]     for i in range(p - 1):         for j in range(1, q):             dp[j] += dp[j - 1]     return dp[q - 1]     # Driver Code if __name__ == '__main__':   print(numberOfPaths(3, 3))   # This code is contributed # by Ankit Yadav

## C#

 using System;   class GFG {     // Returns count of possible paths     // to reach cell at row number m     // and column number n from the     // topmost leftmost cell (cell at 1, 1)     static int numberOfPaths(int m, int n)     {         // Create a 1D array to store         // results of subproblems         int[] dp = new int[n];         dp = 1;           for (int i = 0; i < m; i++) {             for (int j = 1; j < n; j++) {                 dp[j] += dp[j - 1];             }         }           return dp[n - 1];     }       // Driver Code     public static void Main()     {         Console.Write(numberOfPaths(3, 3));     } }   // This code is contributed // by ChitraNayal

## PHP

 

## Javascript

 

Output

6

Time Complexity: O(M * N), The program uses nested loops to fill the 1D array “dp”. The outer loop runs “m” times, and the inner loop runs “n-1” times. Therefore, the time complexity of the program is O(M*N).
Auxiliary Space: O(N), The program uses a 1D array “dp” of size “n” to store the results of subproblems. Hence, the space complexity of the program is O(N).

This code is contributed by Vivek Singh

Note: the count can also be calculated using the formula (M-1 + N-1)!/(M-1)! * (N-1)!

## Count all possible paths from top left to the bottom right of a M X N matrix using combinatorics:

To solve the problem follow the below idea:

In this approach, We have to calculate m+n-2Cn-1 here which will be (m+n-2)! / (n-1)! (m-1)!
m = number of rows, n = number of columns

Total number of moves in which we have to move down to reach the last row = m – 1 (m rows, since we are starting from (1, 1) that is not included)
Total number of moves in which we have to move right to reach the last column = n – 1 (n column, since we are starting from (1, 1) that is not included)

Down moves = (m – 1)
Right moves = (n – 1)
Total moves = Down moves + Right moves = (m – 1) + (n – 1)

Now think of moves as a string of ‘R’ and ‘D’ characters where ‘R’ at any ith index will tell us to move ‘Right’ and ‘D’ will tell us to move ‘Down’. Now think of how many unique strings (moves) we can make where in total there should be (n – 1 + m – 1) characters and there should be (m – 1) ‘D’ character and (n – 1) ‘R’ character?

Choosing positions of (n – 1) ‘R’ characters results in the automatic choosing of (m – 1) ‘D’ character positions

The number of ways to choose positions for (n – 1) ‘R’ character = Total positions C n – 1 = Total positions C m – 1 = (n – 1 + m – 1) != Count the Number of ways to make an N digit Binary String (String with 0s and 1s only) with ‘X’ zeros and ‘Y’ ones (here we have replaced ‘R’ with ‘0’ or ‘1’ and ‘D’ with ‘1’ or ‘0’ respectively whichever suits you better)

Follow the below steps to solve the problem:

• Declare a variable path equal to 1
• Create a for loop from i equal to n to (m + n – 1)
• Set path equal to path * i
• Set path equal to path divided by (i – n + 1)
• Return path

Below is the implementation of the above approach:

## C++

 // A C++ program to count all possible paths from // top left to top bottom using combinatorics   #include  using namespace std;   int numberOfPaths(int m, int n) {     // We have to calculate m+n-2 C n-1 here     // which will be (m+n-2)! / (n-1)! (m-1)!     int path = 1;     for (int i = n; i < (m + n - 1); i++) {         path *= i;         path /= (i - n + 1);     }     return path; }   // Driver code int main() {     cout << numberOfPaths(3, 3);     return 0; }   // This code is suggested by Kartik Sapra

## Java

 // Java program to count all possible paths from // top left to top bottom using combinatorics   import java.io.*;   class GFG {       static int numberOfPaths(int m, int n)     {         // We have to calculate m+n-2 C n-1 here         // which will be (m+n-2)! / (n-1)! (m-1)!         int path = 1;         for (int i = n; i < (m + n - 1); i++) {             path *= i;             path /= (i - n + 1);         }         return path;     }       // Driver code     public static void main(String[] args)     {         System.out.println(numberOfPaths(3, 3));     } }   // This code is contributed by Code_Mech.

## Python3

 # Python3 program to count all possible # paths from top left to top bottom # using combinatorics     def numberOfPaths(m, n):     path = 1     # We have to calculate m + n-2 C n-1 here     # which will be (m + n-2)! / (n-1)! (m-1)! path = 1;     for i in range(n, (m + n - 1)):         path *= i         path //= (i - n + 1)       return path     # Driver code print(numberOfPaths(3, 3))   # This code is contributed # by Akanksha Rai

## C#

 // C# program to count all possible paths from // top left to top bottom using combinatorics using System;   class GFG {       static int numberOfPaths(int m, int n)     {         // We have to calculate m+n-2 C n-1 here         // which will be (m+n-2)! / (n-1)! (m-1)!         int path = 1;         for (int i = n; i < (m + n - 1); i++) {             path *= i;             path /= (i - n + 1);         }         return path;     }       // Driver code     public static void Main()     {         Console.WriteLine(numberOfPaths(3, 3));     } }   // This code is contributed by Code_Mech.

## PHP

 

## Javascript

 

Output

6`

Time Complexity: O(M), The time complexity is O(M), where n is the maximum of m and n. This is because the for loop iterates n times, and each iteration involves a constant number of arithmetic operations.
Auxiliary Space: O(1), The space complexity of the program is also O(1), which means that it uses a constant amount of memory regardless of the input size. This is because the program does not create any data structures or arrays that grow with the input size, and all calculations are done using only a few variables.