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

• Difficulty Level : Easy
• Last Updated : 14 Sep, 2022

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

Input :  m = 2, n = 2;
Output : 2
There are two paths
(0, 0) -> (0, 1) -> (1, 1)
(0, 0) -> (1, 0) -> (1, 1)

Input :  m = 2, n = 3;
Output : 3
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)

We have discussed a solution to print all possible paths, counting all paths is easier. Let NumberOfPaths(m, n) be the count of paths to reach row number m and column number n in the matrix, NumberOfPaths(m, n) can be recursively written as following.

Implementation:

## 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);}  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);    }      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 m = 3n = 3print(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);    }      static public void Main()    {        Console.WriteLine(numberOfPaths(3, 3));    }}  // This code is contributed by ajit

## PHP

 

## Javascript

 

Output

6

The time complexity of above recursive solution is exponential – O(2^n). There are many overlapping subproblems. We can draw a recursion tree for numberOfPaths(3, 3) and see many overlapping subproblems. The recursion tree would be similar to Recursion tree for Longest Common Subsequence problem

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

Implementation:

## 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 program to test above functionsint main(){    cout << numberOfPaths(3, 3);    return 0;}

## Java

 // A Java program to count all possible paths// from top left to bottom rightclass 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 column 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 program to test above function    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  # 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 column 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 program to test above function m = 3n = 3print( numberOfPaths(m, n))  # This code is contributed by Aditi Sharma

## C#

 // A C#  program to count all possible paths// from top left to bottom rightusing 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 column 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 program to test above function    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 MxN.

Recursive Dynamic Programming solution. The following implementation is based on the Top-Down approach.

## C++

 // A C++ program to count all possible paths from// top left to top bottom right using // Recursive Dynamic Programming#include #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 is 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 codeint 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 Programmingimport 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 is 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 is 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 codeif __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 Programmingusing 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 is 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 of the above 2 dynamic programming solutions is O(mn).
The space complexity of the above 2 solutions is O(mn) required for the 2D array.
Space Optimization of DP solution.

Above solution is more intuitive but we can also reduce the space by O(n); where n is column size.
Implementation:

## C++

 #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 codeint main(){    cout << numberOfPaths(3, 3);}  // This code is contributed by mohit kumar 29

## Java

 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 program to test above function    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 Codeprint(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)
Auxiliary Space: 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)!.

Another Approach: (Using combinatorics) In this approach We have to calculate m+n-2Cn-1 here which will be (m+n-2)! / (n-1)! (m-1)!

Now, let us see how this formula is giving the correct answer (Reference 1) (Reference 2) read reference 1 and reference 2 for a better understanding

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 moves as a string of ‘R’ and ‘D’ character 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 automatic choosing of (m – 1) ‘D’ character positions

Calculate nCr
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) != Another way to think about this problem: 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)

## 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 codeint 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 combinatoricsclass 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 combinatoricsdef 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 codeprint(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 combinatoricsusing 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+n)

Auxiliary Space: O(1)