# Count the number of ways to traverse a Matrix

Given a two-dimensional matrix, in how way can someone traverse it from top-left to bottom-right?
Condition- At any particular cell the possible moves are either down or right, no other steps possible.
Stop when the end is reached.

Examples:

```Input : 3 3
Output : 6

Input : 5 5
Output : 70```

If we look closely, we will find that the number of ways a cell can be reached is = Number of ways it can reach the cell above it + number of ways it can reach the cell which is left of it.

So, start filling the 2D array according to it and return the last cell after completely filling the array.

Implementation:

## C++

 `// C++ program using recursive solution to count` `// number of ways to reach mat[m-1][n-1] from` `// mat[0][0] in a matrix mat[][]` `#include ` `using` `namespace` `std;`   `// Returns The number of way from top-left ` `// to mat[m-1][n-1]` `int` `countPaths(``int` `m, ``int` `n)` `{` `    ``// Return 1 if it is the first row or ` `    ``// first column` `    ``if` `(m == 1 || n == 1)` `        ``return` `1;` ` `  `    ``// Recursively find the no of way to ` `    ``// reach the last cell.` `    ``return` `countPaths(m - 1, n) + ` `           ``countPaths(m, n - 1);` `}`   `// Driver code` `int` `main()` `{` `    ``int` `n = 5;` `    ``int` `m = 5; ` `    ``cout << countPaths(n, m);` `    ``return` `0;` `}`

## Java

 `// Java program using recursive ` `// solution to count number of ` `// ways to reach mat[m-1][n-1] from` `// mat[0][0] in a matrix mat[][]` `import` `java.lang.*;` `import` `java.util.*;`   `class` `GFG` `{` `// Returns The number of way ` `// from top-left to mat[m-1][n-1]` `public` `int` `countPaths(``int` `m, ``int` `n)` `{` `    ``// Return 1 if it is the first` `    ``// row or first column` `    ``if` `(m == ``1` `|| n == ``1``)` `        ``return` `1``;`   `    ``// Recursively find the no of ` `    ``// way to reach the last cell.` `    ``return` `countPaths(m - ``1``, n) +` `           ``countPaths(m, n - ``1``);` `}`   `// Driver Code` `public` `static` `void` `main(String args[])` `{` `    ``GFG g = ``new` `GFG();`   `    ``int` `n = ``5``, m = ``5``;` `    ``System.out.println(g.countPaths(n, m));` `}` `}`   `// This code is contributed ` `// by Akanksha Rai(Abby_akku) `

## Python3

 `# Python3 program using recursive solution to count ` `# number of ways to reach mat[m-1][n-1] from ` `# mat[0][0] in a matrix mat[][]`   `# Returns The number of way from top-left ` `# to mat[m-1][n-1] ` `def` `countPaths(m, n) :`   `    ``# Return 1 if it is the first row or ` `    ``# first column ` `    ``if` `m ``=``=` `1` `or` `n ``=``=` `1` `:` `        ``return` `1`   `    ``# Recursively find the no of way to ` `    ``# reach the last cell. ` `    ``return` `(countPaths(m ``-` `1``, n) ``+` `            ``countPaths(m, n ``-` `1``))`     `# Driver code     ` `if` `__name__ ``=``=` `"__main__"` `:`   `    ``n ``=` `5` `    ``m ``=` `5` `    ``print``(countPaths(n, m))`   `# This code is contributed by ANKITRAI1`

## C#

 `// C# program using recursive ` `// solution to count number of ` `// ways to reach mat[m-1][n-1] from` `// mat[0][0] in a matrix mat[][]` `using` `System;`   `class` `GFG` `{` `// Returns The number of way ` `// from top-left to mat[m-1][n-1]` `public` `int` `countPaths(``int` `m, ``int` `n)` `{` `    ``// Return 1 if it is the first` `    ``// row or first column` `    ``if` `(m == 1 || n == 1)` `        ``return` `1;`   `    ``// Recursively find the no of ` `    ``// way to reach the last cell.` `    ``return` `countPaths(m - 1, n) +` `        ``countPaths(m, n - 1);` `}`   `// Driver Code` `public` `static` `void` `Main()` `{` `    ``GFG g = ``new` `GFG();`   `    ``int` `n = 5, m = 5;` `    ``Console.WriteLine(g.countPaths(n, m));` `    ``Console.Read();` `}` `}`   `// This code is contributed ` `// by SoumikMondal `

## PHP

 ``

## Javascript

 ``

Output

`70`

The above solution has exponential time complexity. It can be optimized using Dynamic Programming as there are overlapping subproblems (highlighted below in partial recursion tree for m=3, n=3)

Space Complexity: 2(m+n)

```            CP(3, 3)
/        \
CP(2, 3)     CP(3, 2)
/    \        /     \
CP(1,3) CP(2,2) CP(2,2) CP(3,1)```

Implementation:

## C++

 `// A simple recursive solution to count` `// number of ways to reach mat[m-1][n-1] from` `// mat[0][0] in a matrix mat[][]` `#include ` `using` `namespace` `std;`   `// Returns The number of way from top-left ` `// to mat[m-1][n-1]` `int` `countPaths(``int` `m, ``int` `n)` `{` `    ``int` `dp[m+1][n+1];` `    `  `    ``for` `(``int` `i=1; i<=m; i++)` `    ``{` `        ``for` `(``int` `j=1; j<=n; j++)` `        ``{` `           ``if` `(i==1 || j == 1)` `              ``dp[i][j] = 1;` `            ``else` `              ``dp[i][j] = dp[i-1][j] + dp[i][j-1];              ` `        ``}` `    ``}` `  `  `    ``return` `dp[m][n];` `}`   `// Driver code` `int` `main()` `{` `    ``int` `n = 5;` `    ``int` `m = 5; ` `    ``cout << countPaths(n, m);` `    ``return` `0;` `}`

## Java

 `// A simple recursive solution to count` `// number of ways to reach mat[m-1][n-1] from` `// mat[0][0] in a matrix mat[][]`   `class` `GFG` `{` `    ``// Returns The number of way from top-left ` `    ``// to mat[m-1][n-1]` `    ``static` `int` `countPaths(``int` `m, ``int` `n)` `    ``{` `        ``int` `[][]dp=``new` `int``[m+``1``][n+``1``];` `        `  `        ``for` `(``int` `i=``1``; i<=m; i++)` `        ``{` `            ``for` `(``int` `j=``1``; j<=n; j++)` `            ``{` `            ``if` `(i==``1` `|| j == ``1``)` `                ``dp[i][j] = ``1``;` `                ``else` `                ``dp[i][j] = dp[i-``1``][j] + dp[i][j-``1``];             ` `            ``}` `        ``}` `    `  `        ``return` `dp[m][n];` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `main(String []args)` `    ``{` `        ``int` `n = ``5``;` `        ``int` `m = ``5``; ` `        ``System.out.println(countPaths(n, m));` `        `  `    ``}` `}`   `// This code is contributed ` `// by ihritik (Hritik Raj)`

## Python3

 `# A simple recursive solution to ` `# count number of ways to reach ` `# mat[m-1][n-1] from mat[0][0] ` `# in a matrix mat[][]`   `# Returns The number of way ` `# from top-left to mat[m-1][n-1]` `def` `countPaths(m, n):`   `    ``dp ``=` `[[``0` `for` `i ``in` `range``(m ``+` `1``)] ` `             ``for` `j ``in` `range``(n ``+` `1``)]` `    `  `    ``for` `i ``in` `range``(``1``, m ``+` `1``):` `        ``for` `j ``in` `range``(``1``, n ``+` `1``):` `            ``if` `(i ``=``=` `1` `or` `j ``=``=` `1``):` `                ``dp[i][j] ``=` `1` `            ``else``:` `                ``dp[i][j] ``=` `(dp[i ``-` `1``][j] ``+` `                            ``dp[i][j ``-` `1``])             ` `    `  `    ``return` `dp[m][n]`   `# Driver code` `if` `__name__ ``=``=``"__main__"``:` `    `  `    ``n ``=` `5` `    ``m ``=` `5` `    ``print``(countPaths(n, m))`   `# This code is contributed ` `# by ChitraNayal`

## C#

 `// A simple recursive solution to count` `// number of ways to reach mat[m-1][n-1] from` `// mat[0][0] in a matrix mat[][]`   `using` `System;` `class` `GFG` `{` `    ``// Returns The number of way from top-left ` `    ``// to mat[m-1][n-1]` `    ``static` `int` `countPaths(``int` `m, ``int` `n)` `    ``{` `        ``int` `[,]dp=``new` `int``[m+1,n+1];` `        `  `        ``for` `(``int` `i=1; i<=m; i++)` `        ``{` `            ``for` `(``int` `j=1; j<=n; j++)` `            ``{` `            ``if` `(i==1 || j == 1)` `                ``dp[i,j] = 1;` `                ``else` `                ``dp[i,j] = dp[i-1,j] + dp[i,j-1];             ` `            ``}` `        ``}` `    `  `        ``return` `dp[m,n];` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `n = 5;` `        ``int` `m = 5; ` `        ``Console.WriteLine(countPaths(n, m));` `        `  `    ``}` `}`   `// This code is contributed ` `// by ihritik (Hritik Raj)`

## PHP

 ``

## Javascript

 ``

Output

`70`

Time Complexity: O(m * n)

Auxiliary Space: O(m * n)

Another Method(Efficient):

There is one more efficient way of reaching the solution in O(m) or O(n) whichever is greater.

We can permute the number of right operations and down operations.

Explanation:

In the given figure, we can see that for a matrix of 3×3, we need 2 right operations and 2 down operations.

Thus, we can permute these operations in any order, still we can reach the bottom-right.

“RRDD”,”RDRD”,”RDDR”,”DRDR”,”DDRR”,”DRRD”

Let the number of columns be m, and the number of rows is n, then no of permutations = (m+n)!/ (m!*n!)

Implementation:

## C++

 `// C++ program for above approach` `#include` `using` `namespace` `std;`   `// Find factorial` `int` `factorial(``int` `n)` `{` `    ``int` `res = 1, i;` `    `  `    ``for``(i = 2; i <= n; i++)` `        ``res *= i;` `        `  `    ``return` `res;` `}`   `// Find number of ways to reach ` `// mat[m-1][n-1] from mat[0][0]` `// in a matrix mat[][]]` ` ``int` `countWays(``int` `m, ``int` `n)` `{` `    ``m = m - 1;` `    ``n = n - 1;` `    `  `    ``return` `factorial(m + n) / ` `          ``(factorial(m) * ` `           ``factorial(n));` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `m = 5;` `    ``int` `n = 5;` `   `  `    ``// Function call` `    ``int` `result = countWays(m, n);` `    `  `    ``cout << result;` `}`   `// This code is contributed by chahattekwani71`

## Java

 `// Java Program for above approach` `import` `java.io.*;`   `class` `GFG {`   `    ``// Find factorial` `    ``static` `int` `factorial(``int` `n)` `    ``{` `        ``int` `res = ``1``, i;` `        ``for` `(i = ``2``; i <= n; i++)` `            ``res *= i;` `        ``return` `res;` `    ``}` `    ``// Find number of ways to reach ` `    ``// mat[m-1][n-1] from mat[0][0]` `    ``// in a matrix mat[][]]` `    ``static` `int` `countWays(``int` `m, ``int` `n)` `    ``{` `        ``m = m - ``1``;` `        ``n = n - ``1``;` `        ``return` `factorial(m + n)` `            ``/ (factorial(m) * factorial(n));` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `m = ``5``;` `        ``int` `n = ``5``;` `      `  `        ``// Function Call` `        ``int` `result = countWays(m, n);` `        ``System.out.println(result);` `    ``}` `}`

## Python3

 `# Python3 program for ` `# the above approach`   `# Find factorial` `def` `factorial(n):`   `    ``res ``=` `1` `    `  `    ``for` `i ``in` `range``(``2``, n ``+` `1``):` `        ``res ``*``=` `i` `        `  `    ``return` `res`   `# Find number of ways to reach ` `# mat[m-1][n-1] from mat[0][0]` `# in a matrix mat[][]]` `def` `countWays(m, n):`   `    ``m ``=` `m ``-` `1` `    ``n ``=` `n ``-` `1`    `    ``return` `(factorial(m ``+` `n) ``/``/` `           ``(factorial(m) ``*` `            ``factorial(n)))`   `# Driver code   ` `m ``=` `5` `n ``=` `5`   `# Function call` `result ``=` `countWays(m, n)`   `print``(result)`   `# This code is contributed by divyeshrabadiya07`

## C#

 `// C# Program for above approach` `using` `System;`   `class` `GFG{` `    `  `// Find factorial` `static` `int` `factorial(``int` `n)` `{` `    ``int` `res = 1, i;` `    ``for``(i = 2; i <= n; i++)` `        ``res *= i;` `        `  `    ``return` `res;` `}`   `// Find number of ways to reach ` `// mat[m-1][n-1] from mat[0][0]` `// in a matrix mat[][]]` `static` `int` `countWays(``int` `m, ``int` `n)` `{` `    ``m = m - 1;` `    ``n = n - 1;` `    `  `    ``return` `factorial(m + n) / ` `          ``(factorial(m) * factorial(n));` `}`   `// Driver code` `static` `void` `Main() ` `{` `    ``int` `m = 5;` `    ``int` `n = 5;` `   `  `    ``// Function Call` `    ``int` `result = countWays(m, n);` `    `  `    ``Console.WriteLine(result);` `}` `}`   `// This code is contributed by divyesh072019`

## Javascript

 ``

Output

`70`

Time Complexity: O(n)
Auxiliary Space: O(1)