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

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)
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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.

## 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 `

## Python

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

## PHP

 ` `

Output:

`6`

The time complexity of above recursive solution is exponential. 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.

## 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 calculatest he 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 functions ` `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) ` `    ``{ ` `        ``// 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 calculatest he 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 `

## Python

 `# 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 ` `    ``count ``=` `[[``0` `for` `x ``in` `range``(m)] ``for` `y ``in` `range``(n)] ` `   `  `    ``# Count of paths to reach any  ` `    ``# cell in first column is 1 ` `    ``for` `i ``in` `range``(m): ` `        ``count[i][``0``] ``=` `1``; ` `   `  `    ``# Count of paths to reach any  ` `    ``# cell in first column is 1 ` `    ``for` `j ``in` `range``(n): ` `        ``count[``0``][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 ``=` `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 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 calculatest he 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

 ` `

Output:

`6`

Time complexity of the above dynamic programming solution is O(mn).
The space complexity of the above solution is O(mn).

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.

## 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 code ` `int` `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[``0``] = ``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 Code ` `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

 ` `

Output:

`6`

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-2 C n-1 here which will be (m+n-2)! / (n-1)! (m-1)!

## 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 ` `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) : ` ` `  `    ``# 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

 `

Output:

`6`