# Count of possible paths from top left to bottom right of a M x N matrix by moving right, down or diagonally

• Last Updated : 07 Apr, 2022

Given 2 integers M and N, the task is to find the count of all the possible paths from top left to the bottom right of an M x N matrix with the constraints that from each cell you can either move only to right or down or diagonally

Examples:

Input:  M = 3, N = 3
Output: 13
Explanation: There are 13 paths as follows: VVHH, VHVH, HVVH, DVH, VDH, VHHV, HVHV, DHV, HHVV, HDV, VHD, HVD, and DD where V represents vertical, H represents horizontal, and D represents diagonal paths.

Input:  M = 2, N = 2
Output: 3

Approach: The idea is to use recursion to find the total number of paths. This approach is very similar to the one discussed in this article. Follow the steps below to solve the problem:

• If M or N equals 1, then return 1.
• Else create a recursive function numberOfPaths() call the same function for values {M-1, N}, {M, N-1} and {M-1, N-1} representing vertical, horizontal and diagonal movement respectively.

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)``{` `    ``// If either given row number or``    ``// given column number is first``    ``if` `(M == 1 || N == 1)``        ``return` `1;` `    ``// Horizontal Paths +``    ``// Vertical Paths +``    ``// Diagonal Paths``    ``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

 `// Java  program for the above approach``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` `M, ``int` `N)``{` `    ``// If either given row number or``    ``// given column number is first``    ``if` `(M == ``1` `|| N == ``1``)``        ``return` `1``;` `    ``// Horizontal Paths +``    ``// Vertical Paths +``    ``// Diagonal Paths``    ``return` `numberOfPaths(M - ``1``, N)``          ``+ numberOfPaths(M, N - ``1``)``           ``+ numberOfPaths(M - ``1``, N - ``1``);``}` `// Driver Code``public` `static` `void` `main(String args[])``{``    ``System.out.print(numberOfPaths(``3``, ``3``));` `}``}` `// This code is contributed by Samim Hossain Mondal.`

## Python

 `# Python  program for the above approach` `# 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):` `    ``# If either given row number or``    ``# given column number is first``    ``if` `(M ``=``=` `1` `or` `N ``=``=` `1``):``        ``return` `1` `    ``# Horizontal Paths +``    ``# Vertical Paths +``    ``# Diagonal Paths``    ``return` `numberOfPaths(M ``-` `1``, N) \``        ``+` `numberOfPaths(M, N ``-` `1``) \``        ``+` `numberOfPaths(M ``-` `1``, N ``-` `1``)` `# Driver Code``print``(numberOfPaths(``3``, ``3``))` `# This code is contributed by Samim Hossain Mondal.`

## C#

 `// C#  program for the above approach``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 or``    ``// given column number is first``    ``if` `(M == 1 || N == 1)``        ``return` `1;` `    ``// Horizontal Paths +``    ``// Vertical Paths +``    ``// Diagonal Paths``    ``return` `numberOfPaths(M - 1, N)``          ``+ numberOfPaths(M, N - 1)``           ``+ numberOfPaths(M - 1, N - 1);``}` `// Driver Code``public` `static` `void` `Main(String []args)``{``    ``Console.Write(numberOfPaths(3, 3));` `}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output
`13`

Time Complexity: O(3M*N)
Auxiliary Space: O(1)

Efficient Approach: Dp (using Memoization)

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `static` `int` `dp;` `// 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 or``    ``// given column number is first``    ``if` `(M == 1 || N == 1)``        ``return` `1;``    ` `    ``// If a value already present``    ``// in t[][], return it``    ``if``(dp[M][N] != -1) {``        ``return` `dp[M][N];``    ``}``    ` `    ``// Horizontal Paths +``    ``// Vertical Paths +``    ``// Diagonal Paths``    ``return` `dp[M][N] = numberOfPaths(M - 1, N)``                    ``+ numberOfPaths(M, N - 1)``                    ``+ numberOfPaths(M - 1, N - 1);``}` `// Driver Code``int` `main()``{``    ``memset``(dp,-1,``sizeof``(dp));``    ``cout << numberOfPaths(3, 3);``    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;``public` `class` `GFG {``  ``static` `int``[][] dp = ``new` `int``[``1001``][``1001``];` `  ``// 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 or``    ``// given column number is first``    ``if` `(M == ``1` `|| N == ``1``)``      ``return` `1``;` `    ``// If a value already present``    ``// in t[][], return it``    ``if` `(dp[M][N] != -``1``) {``      ``return` `dp[M][N];``    ``}` `    ``// Horizontal Paths +``    ``// Vertical Paths +``    ``// Diagonal Paths``    ``dp[M][N] = numberOfPaths(M - ``1``, N)``      ``+ numberOfPaths(M, N - ``1``)``      ``+ numberOfPaths(M - ``1``, N - ``1``);``    ``return` `dp[M][N];``  ``}` `  ``// Driver Code``  ``public` `static` `void` `main(String args[])``  ``{``    ``for` `(``int` `i = ``0``; i < ``1001``; i++)``      ``for` `(``int` `j = ``0``; j < ``1001``; j++)``        ``dp[i][j] = -``1``;``    ``System.out.println(numberOfPaths(``3``, ``3``));``  ``}``}` `// This code is contributed by Samim Hossain Mondal.`

## Python

 `# Python program for the above approach` `# Taking the matrix as globally``dp ``=` `[[``-``1` `for` `i ``in` `range``(``1001``)] ``for` `j ``in` `range``(``1001``)]` `# 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):` `    ``# If either given row number or``    ``# given column number is first``    ``if` `(M ``=``=` `1` `or` `N ``=``=` `1``):``        ``return` `1` `    ``# If a value already present``    ``# in t[][], return it``    ``if``(dp[M][N] !``=` `-``1``):``        ``return` `dp[M][N]` `    ``# Horizontal Paths +``    ``# Vertical Paths +``    ``# Diagonal Paths``    ``dp[M][N] ``=` `numberOfPaths(M ``-` `1``, N) ``+` `numberOfPaths(M,``                     ``N ``-` `1``) ``+` `numberOfPaths(M ``-` `1``, N ``-` `1``)``    ``return` `dp[M][N]` `# Driver Code``print``(numberOfPaths(``3``, ``3``))` `# This code is contributed by Samim Hossain Mondal.`

## C#

 `// C# program for the above approach``using` `System;``class` `GFG {``    ``static` `int``[, ] dp = ``new` `int``[1001, 1001];` `    ``// 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 or``        ``// given column number is first``        ``if` `(M == 1 || N == 1)``            ``return` `1;` `        ``// If a value already present``        ``// in t[][], return it``        ``if` `(dp[M, N] != -1) {``            ``return` `dp[M, N];``        ``}` `        ``// Horizontal Paths +``        ``// Vertical Paths +``        ``// Diagonal Paths``        ``dp[M, N] = numberOfPaths(M - 1, N)``                   ``+ numberOfPaths(M, N - 1)``                   ``+ numberOfPaths(M - 1, N - 1);``        ``return` `dp[M, N];``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``for` `(``int` `i = 0; i < 1001; i++)``            ``for` `(``int` `j = 0; j < 1001; j++)``                ``dp[i, j] = -1;``        ``Console.Write(numberOfPaths(3, 3));``    ``}``}` `// This code is contributed by ukasp.`

## Javascript

 ``
Output
`13`

Time Complexity: O(M*N)

Auxiliary Space: O(M*N)

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