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**

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

#include <iostream> 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

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

#include <iostream> 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][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 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(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

Output:

6

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

Note the count can also be calculated using the formula (m-1 + n-1)!/(m-1)!(n-1)!. See this for more details.

This article is contributed by **Hariprasad NG**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above