# Print all possible paths from the first row to the last row in a 2D array

Given a 2D array of characters with M rows and N columns. The task is to print all the possible paths from top (first row) to bottom (last row).

Examples:

Input: arr[][] = {
{‘a’, ‘b’, ‘c’},
{‘d’, ‘e’, ‘f’},
{‘g’, ‘h’, ‘i’}}
Output:
bdg bdh bdi beg beh bei bfg bfh bfi
cdg cdh cdi ceg ceh cei cfg cfh cfi

Input: arr[][] = {
{‘a’, ‘b’},
{‘d’, ‘e’}, }
Output:
bd be

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

1. This problem can be solved using Depth First Traversal of an array with a slight modification.
2. The modification here is to only iterate through the columns in the array until the target row is reached.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach  ` `#include ` `using` `namespace` `std; ` ` `  `void` `dfs(``char` `inputchar[], string res,  ` `                ``int` `i, ``int` `j, ``int` `R, ``int` `C)  ` `{ ` `     `  `    ``// If the current row index equals to R, it  ` `    ``// indicates we have reached the bottom of  ` `    ``// the array and hence we print the result  ` `    ``if` `(i == R)  ` `    ``{ ` `        ``cout << res << ``" "``; ` `        ``return``;  ` `    ``} ` ` `  `    ``res = res + (inputchar[i][j]);  ` ` `  `    ``// Iterate over each of the columns  ` `    ``// in the array  ` `    ``for` `(``int` `k = 0; k < C ; k++) ` `    ``{ ` `        ``dfs(inputchar, res, i + 1, k, R, C); ` `        ``if` `(i + 1 == R)  ` `            ``break``;  ` `    ``} ` `} ` ` `  `// Function to print all the possible paths  ` `void` `printPaths(``char` `inputchar[], ``int` `R, ``int` `C) ` `{ ` `    ``for` `(``int` `i = 0; i < C; i++) ` `    ``{ ` `        ``dfs(inputchar, ``""``, 0, i, R, C); ` `        ``cout<

## Java

 `// Java implementation of the approach ` `public` `class` `GFG { ` ` `  `    ``// Function to print all the possible paths ` `    ``private` `static` `void` `printPaths(``char``[][] input, ` `                                   ``int` `R, ``int` `C) ` `    ``{ ` `        ``for` `(``int` `i = ``0``; i < C; i++) { ` `            ``dfs(input, ``""``, ``0``, i, R, C); ` `            ``System.out.println(); ` `        ``} ` `    ``} ` ` `  `    ``/** ` `    ``* Depth first traversal of the array ` `    ``* ` `    ``* @param input array of characters ` `    ``* @param res to be printed in console ` `    ``* @param i     current row index ` `    ``* @param j     current column index ` `    ``* @param R     number of rows in the input array ` `    ``* @param C     number of rows in the output array ` `    ``*/` `    ``private` `static` `void` `dfs(``char``[][] input, String res, ` `                            ``int` `i, ``int` `j, ``int` `R, ``int` `C) ` `    ``{ ` `        ``// If the current row index equals to R, it ` `        ``// indicates we have reached the bottom of ` `        ``// the array and hence we print the result ` `        ``if` `(i == R) { ` `            ``System.out.print(res + ``" "``); ` `            ``return``; ` `        ``} ` ` `  `        ``res = res + input[i][j]; ` ` `  `        ``// Iterate over each of the columns ` `        ``// in the array ` `        ``for` `(``int` `k = ``0``; k < C; k++) { ` `            ``dfs(input, res, i + ``1``, k, R, C); ` `            ``if` `(i + ``1` `== R) { ` `                ``break``; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``char``[][] input = { ` `            ``{ ``'a'``, ``'b'` `}, ` `            ``{ ``'d'``, ``'e'` `} ` `        ``}; ` `        ``int` `R = input.length; ` `        ``int` `C = input[``0``].length; ` `        ``printPaths(input, R, C); ` `    ``} ` `} `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to print all the possible paths  ` `def` `printPaths(inputchar, R, C) : ` `    ``for` `i ``in` `range``(C) : ` `        ``dfs(inputchar, "", ``0``, i, R, C); ` `        ``print``() ` `         `  `    ``""" ` `    ``* Depth first traversal of the array  ` `    ``*  ` `    ``* @param input array of characters  ` `    ``* @param res to be printed in console  ` `    ``* @param i current row index  ` `    ``* @param j current column index  ` `    ``* @param R number of rows in the input array  ` `    ``* @param C number of rows in the output array  ` `    ``* """` `def` `dfs(inputchar, res, i, j, R, C) : ` `     `  `    ``# If the current row index equals to R, it  ` `    ``# indicates we have reached the bottom of  ` `    ``# the array and hence we print the result  ` `    ``if` `(i ``=``=` `R) : ` `        ``print``(res, end ``=` `" "``); ` `        ``return``;  ` ` `  `    ``res ``=` `res ``+` `inputchar[i][j];  ` ` `  `    ``# Iterate over each of the columns  ` `    ``# in the array  ` `    ``for` `k ``in` `range``(C) : ` `        ``dfs(inputchar, res, i ``+` `1``, k, R, C); ` `        ``if` `(i ``+` `1` `=``=` `R) : ` `            ``break``;  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `: ` `     `  `    ``inputchar ``=` `[ ` `            ``[ ``'a'``, ``'b'` `],  ` `            ``[ ``'d'``, ``'e'` `]  ` `            ``]; ` `             `  `    ``R ``=` `len``(inputchar); ` `    ``C ``=` `len``(inputchar[``0``]); ` `     `  `    ``printPaths(inputchar, R, C);  ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` ` `  `    ``// Function to print all the possible paths  ` `    ``private` `static` `void` `printPaths(``char``[,] input,  ` `                                ``int` `R, ``int` `C)  ` `    ``{  ` `        ``for` `(``int` `i = 0; i < C; i++) ` `        ``{  ` `            ``dfs(input, ``""``, 0, i, R, C);  ` `            ``Console.WriteLine();  ` `        ``}  ` `    ``}  ` ` `  `    ``/**  ` `    ``* Depth first traversal of the array  ` `    ``*  ` `    ``* @param input array of characters  ` `    ``* @param res to be printed in console  ` `    ``* @param i current row index  ` `    ``* @param j current column index  ` `    ``* @param R number of rows in the input array  ` `    ``* @param C number of rows in the output array  ` `    ``*/` `    ``private` `static` `void` `dfs(``char``[,] input, ``string` `res,  ` `                            ``int` `i, ``int` `j, ``int` `R, ``int` `C)  ` `    ``{  ` `        ``// If the current row index equals to R, it  ` `        ``// indicates we have reached the bottom of  ` `        ``// the array and hence we print the result  ` `        ``if` `(i == R)  ` `        ``{  ` `            ``Console.Write(res + ``" "``);  ` `            ``return``;  ` `        ``}  ` ` `  `        ``res = res + input[i,j];  ` ` `  `        ``// Iterate over each of the columns  ` `        ``// in the array  ` `        ``for` `(``int` `k = 0; k < C; k++)  ` `        ``{  ` `            ``dfs(input, res, i + 1, k, R, C);  ` `            ``if` `(i + 1 == R)  ` `            ``{  ` `                ``break``;  ` `            ``}  ` `        ``}  ` `    ``}  ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` `        ``char``[,] input = {  ` `            ``{ ``'a'``, ``'b'` `},  ` `            ``{ ``'d'``, ``'e'` `}  ` `        ``};  ` `        ``int` `R = input.GetLength(0);  ` `        ``int` `C = input.GetLength(1);  ` `        ``printPaths(input, R, C);  ` `    ``}  ` `}  ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```ad ae
bd be
```

Time Complexity: O(R * C)
Space Complexity: O(1)

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