Given two integers N and M, which denotes the position of in the coordinate system, the task is to count path from this point to reach origin by taking steps either left or down.
Examples:
Input: 3 6
Output: Number of Paths 84
Input: 3 3
Output: Number of Paths 20
We have already discussed this problem in this article –
Counts paths from a point to reach Origin
Approach: The idea is to use the Dynamic programming paradigm to solve this problem. In this problem, there is a restriction to move only left or down in a path, due to which for any point number of paths is equal to the sum of the number of paths for the left point and the number of paths for the right point and in this way we compute the path for every point in a bottom-up manner from (0, 0) to (N, M). The key observation in this problem to reduce the space is for computing the solution for a point say (i, j) we only need the value of (i-1, j) and (i, j-1), that is we don’t need the portion of (i-1, j-1) computed values once they are used.
Below is the implementation of the above approach:
C++
// C++ implementation to count the // paths from a point to origin #include <iostream> #include<cstring> using namespace std; // Function to count the paths from // a point to the origin long Count_Paths( int x, int y) { // Base Case when the point // is already at origin if (x == 0 && y == 0) return 0; // Base Case when the point // is on the x or y-axis if (x == 0 || y == 0) return 1; long dp[max(x,y)+1], p=max(x,y),q=min(x,y); // Loop to fill all the // position as 1 in the array for ( int i=0;i<=p;i++) dp[i]=1; // Loop to count the number of // paths from a point to origin // in bottom-up manner for ( int i=1;i<=q;i++) for ( int j=1;j<=p;j++) dp[j]+=dp[j-1]; return dp[p]; } // Driver Code int main() { int x = 3,y = 3; // Function Call cout<< "Number of Paths " << Count_Paths(x,y); return 0; } |
Java
// Java implementation to count the // paths from a point to origin class GFG { // Function to count the paths from // a point to the origin static long Count_Paths( int x, int y) { // Base Case when the point // is already at origin if (x == 0 && y == 0 ) return 0 ; // Base Case when the point // is on the x or y-axis if (x == 0 || y == 0 ) return 1 ; int [] dp = new int [(Math.max(x, y) + 1 )]; int p = Math.max(x, y), q = Math.min(x, y); // Loop to fill all the // position as 1 in the array for ( int i = 0 ; i <= p; i++) dp[i] = 1 ; // Loop to count the number of // paths from a point to origin // in bottom-up manner for ( int i = 1 ; i <= q; i++) for ( int j = 1 ; j <= p; j++) dp[j] += dp[j - 1 ]; return dp[p]; } // Driver Code public static void main(String[] args) { int x = 3 , y = 3 ; // Function Call System.out.print( "Number of Paths " + Count_Paths(x, y)); } } // This code contributed by Rajput-Ji |
Python3
# Python3 implementation to count the # paths from a point to origin # Function to count the paths from # a point to the origin def Count_Paths(x, y): # Base Case when the point # is already at origin if (x = = 0 and y = = 0 ): return 0 # Base Case when the point # is on the x or y-axis if (x = = 0 and y = = 0 ): return 1 dp = [ 0 ] * ( max (x, y) + 1 ) p = max (x, y) q = min (x, y) # Loop to fill all the # position as 1 in the array for i in range (p + 1 ): dp[i] = 1 # Loop to count the number of # paths from a point to origin # in bottom-up manner for i in range ( 1 , q + 1 ): for j in range ( 1 , p + 1 ): dp[j] + = dp[j - 1 ] return dp[p] # Driver Code x = 3 y = 3 # Function call print ( "Number of Paths " , Count_Paths(x, y)) # This code is contributed by sanjoy_62 |
C#
// C# implementation to count the // paths from a point to origin using System; class GFG { // Function to count the paths from // a point to the origin static long Count_Paths( int x, int y) { // Base Case when the point // is already at origin if (x == 0 && y == 0) return 0; // Base Case when the point // is on the x or y-axis if (x == 0 || y == 0) return 1; int [] dp = new int [(Math.Max(x, y) + 1)]; int p = Math.Max(x, y), q = Math.Min(x, y); // Loop to fill all the // position as 1 in the array for ( int i = 0; i <= p; i++) dp[i] = 1; // Loop to count the number of // paths from a point to origin // in bottom-up manner for ( int i = 1; i <= q; i++) for ( int j = 1; j <= p; j++) dp[j] += dp[j - 1]; return dp[p]; } // Driver Code public static void Main(String[] args) { int x = 3, y = 3; // Function Call Console.Write( "Number of Paths " + Count_Paths(x, y)); } } // This code is contributed by sapnasingh4991 |
Javascript
// Function to count the paths from // a point to the origin function count_paths(x, y){ // Base case when the point // is already at origin if (x == 0 && y == 0){ return 0; } // Base case when the point // is already at origin if (x == 0 || y == 0){ return 1; } let dp = []; let p = Math.max(x, y); let q = Math.min(x, y); // Loop to fill all the position // as 1 in the array for (let i = 0; i <= Math.max(x, y); i++){ if (i <= p){ dp[i] = 1; } else { dp[i] = 0; } } // Loop to count the number of // paths from a point to origin for (let i = 1; i <=q; i++){ for (let j = 1; j <= q; j++){ dp[j] += dp[j-1]; } } return dp[p]; } // Driver code let x = 3; let y = 3; // Function call console.log( "Number of Paths " ,count_paths(x, y)); // This code is contributed BY Aditya Sharma |
Number of Paths 20
Performance Analysis:
- Time Complexity: In the above-given approach, there are two loops of which takes O(N*M) time in the worst case. Therefore, the time complexity for this approach will be O(N*M).
- Auxiliary Space: In the above-given approach, there is a single array of the size maximum value of N and M. Therefore the space complexity for the above approach will be O(max(N, M))
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