Given a 2D matrix of size n*m, a source ‘s’ and a destination ‘d’, print the count of all unique paths from given ‘s’ to ‘d’. From each cell, you can either move only to the right or down.
Examples:
Input: arr[][] = { {1, 2, 3}, {4, 5, 6} }, s = {0, 0}, d = {1, 2}
Output: 3
Explanation: All possible paths from source to destination are:
- 1 -> 4 -> 5 -> 6
- 1 -> 2 -> 5 -> 6
- 1 -> 2 -> 3 -> 6
Input: arr[][] = { {1, 2}, {3, 4} }, s = {0, 1}, d = {1, 1}
Output: 1
Approach: Use recursion to move right first & then down from each cell in the path of the matrix, starting from the source. If the destination is reached, increment the count of possible paths.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the// count of all possible pathsint countPaths(int i, int j, int count, int p, int q){ // Destination is reached if (i == p || j == q) { count++; return count; } // Move right count = countPaths(i, j + 1, count, p, q); // Move down count = countPaths(i + 1, j, count, p, q); return count;}// Driver program to test above functionsint main(){ vector<vector<int> > mat = { { 1, 2, 3 }, { 4, 5, 6 } }; vector<int> s = { 0, 0 }; vector<int> d = { 1, 2 }; cout << countPaths(s[0], s[1], 0, d[0], d[1]); return 0;} |
Java
// Java program for the above approachclass GFG { // Function to find the // count of all possible paths static int countPaths(int i, int j, int count, int p, int q) { // Destination is reached if (i == p || j == q) { count++; return count; } // Move right count = countPaths(i, j + 1, count, p, q); // Move down count = countPaths(i + 1, j, count, p, q); return count; } // Driver program to test above functions public static void main(String args[]) { int[] s = { 0, 0 }; int[] d = { 1, 2 }; System.out.println(countPaths(s[0], s[1], 0, d[0], d[1])); }}// This code is contributed by gfgking. |
Python3
# python program for the above approach# Function to find the# count of all possible pathsdef countPaths(i, j, count, p, q): # Destination is reached if (i == p or j == q): count += 1 return count # Move right count = countPaths(i, j + 1, count, p, q) # Move down count = countPaths(i + 1, j, count, p, q) return count# Driver program to test above functionsif __name__ == "__main__": mat = [[1, 2, 3], [4, 5, 6]] s = [0, 0] d = [1, 2] print(countPaths(s[0], s[1], 0, d[0], d[1])) # This code is contributed by rakeshsahni |
C#
// C# program for the above approachusing System;class GFG { // Function to find the // count of all possible paths static int countPaths(int i, int j, int count, int p, int q) { // Destination is reached if (i == p || j == q) { count++; return count; } // Move right count = countPaths(i, j + 1, count, p, q); // Move down count = countPaths(i + 1, j, count, p, q); return count; } // Driver program to test above functions public static void Main() { int[] s = { 0, 0 }; int[] d = { 1, 2 }; Console.Write(countPaths(s[0], s[1], 0, d[0], d[1])); }}// This code is contributed by gfgking. |
Javascript
<script> // JavaScript code for the above approach // Function to find the // count of all possible paths function countPaths(i, j, count, p, q) { // Destination is reached if (i == p || j == q) { count++; return count; } // Move right count = countPaths(i, j + 1, count, p, q); // Move down count = countPaths(i + 1, j, count, p, q); return count; } // Driver program to test above functions let mat = [[1, 2, 3], [4, 5, 6]]; let s = [0, 0]; let d = [1, 2]; document.write(countPaths(s[0], s[1], 0, d[0], d[1])); // This code is contributed by Potta Lokesh </script> |
Output
3
Time Complexity: O(n+m)
Auxiliary Space: O(1)
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