Counts paths from a point to reach Origin (original) (raw)

Last Updated : 23 Jul, 2025

You are standing on a point (n, m) and you want to go to origin (0, 0) by taking steps either left or down i.e. from each point you are allowed to move either in (n-1, m) or (n, m-1). Find the number of paths from point to origin.

Examples:

Input : 3 6 Output : Number of Paths 84

Input : 3 0 Output : Number of Paths 1

As we are restricted to move down and left only we would run a recursive loop for each of the combinations of the
steps that can be taken.

// Recursive function to count number of paths countPaths(n, m) { // If we reach bottom or top left, we are // have only one way to reach (0, 0) if (n==0 || m==0) return 1;

// Else count sum of both ways
return (countPaths(n-1, m) + countPaths(n, m-1));

}

Below is the implementation of the above steps.

Try It Yourself

C++ `

// C++ program to count total number of // paths from a point to origin #include<bits/stdc++.h> using namespace std;

// Recursive function to count number of paths int countPaths(int n, int m) { // If we reach bottom or top left, we are // have only one way to reach (0, 0) if (n==0 || m==0) return 1;

// Else count sum of both ways
return (countPaths(n-1, m) + countPaths(n, m-1));

}

// Driver Code int main() { int n = 3, m = 2; cout << " Number of Paths " << countPaths(n, m); return 0; }

Java

// Java program to count total number of // paths from a point to origin import java.io.*;

class GFG {

// Recursive function to count number of paths
static int countPaths(int n, int m)
{
    // If we reach bottom or top left, we are
    // have only one way to reach (0, 0)
    if (n == 0 || m == 0)
        return 1;

    // Else count sum of both ways
    return (countPaths(n - 1, m) + countPaths(n, m - 1));
}

// Driver Code
public static void main (String[] args)
{
    int n = 3, m = 2;
    System.out.println (" Number of Paths "
                        + countPaths(n, m));
    
}

}

// This code is contributed by vt_m

Python3

Python3 program to count

total number of

paths from a point to origin

Recursive function to

count number of paths

def countPaths(n,m):

# If we reach bottom
# or top left, we are
# have only one way to reach (0, 0)
if (n==0 or m==0):
    return 1

# Else count sum of both ways
return (countPaths(n-1, m) + countPaths(n, m-1))

Driver Code

n = 3 m = 2 print(" Number of Paths ", countPaths(n, m))

This code is contributed

by Azkia Anam.

C#

// C# program to count total number of // paths from a point to origin using System;

public class GFG {

// Recursive function to count number
// of paths
static int countPaths(int n, int m)
{
    
    // If we reach bottom or top left,
    // we are have only one way to
    // reach (0, 0)
    if (n == 0 || m == 0)
        return 1;

    // Else count sum of both ways
    return (countPaths(n - 1, m) 
             + countPaths(n, m - 1));
}

// Driver Code
public static void Main ()
{
    int n = 3, m = 2;
    
    Console.WriteLine (" Number of"
     + " Paths " + countPaths(n, m));
    
}

}

// This code is contributed by Sam007.

PHP

JavaScript

`

Output

Number of Paths 10

Time Complexity: O(min(m,n))

Auxiliary Space: O(min(m,n))

We can use Dynamic Programming as there are overlapping subproblems. We can draw recursion tree to see overlapping problems. For example, in case of countPaths(4, 4), we compute countPaths(3, 3) multiple times.

C++ `

// C++ program to count total number of // paths from a point to origin #include<bits/stdc++.h> using namespace std;

// DP based function to count number of paths int countPaths(int n, int m) { int dp[n+1][m+1];

// Fill entries in bottommost row and leftmost
// columns
for (int i=0; i<=n; i++)
  dp[i][0] = 1;
for (int i=0; i<=m; i++)
  dp[0][i] = 1;

// Fill DP in bottom up manner
for (int i=1; i<=n; i++)
   for (int j=1; j<=m; j++)
      dp[i][j] = dp[i-1][j] + dp[i][j-1];

return dp[n][m];

}

// Driver Code int main() { int n = 3, m = 2; cout << " Number of Paths " << countPaths(n, m); return 0; }

Java

// Java program to count total number of // paths from a point to origin import java.io.*;

class GFG {

// DP based function to count number of paths
static int countPaths(int n, int m)
{
    int dp[][] = new int[n + 1][m + 1];

    // Fill entries in bottommost row and leftmost
    // columns
    for (int i = 0; i <= n; i++)
        dp[i][0] = 1;
    for (int i = 0; i <= m; i++)
        dp[0][i] = 1;

    // Fill DP in bottom up manner
    for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
        dp[i][j] = dp[i - 1][j] + dp[i][j - 1];

    return dp[n][m];
}

// Driver Code
public static void main (String[] args) {
    int n = 3, m = 2;
    System.out.println(" Number of Paths "
                       + countPaths(n, m));
    
}

}

// This code is contributed by vt_m

Python3

Python3 program to count total

number of paths from a point to origin

Recursive function to count

number of paths

def countPaths(n, m):

# If we reach bottom or top 
# left, we are have only one 
# way to reach (0, 0)
if (n == 0 or m == 0):
    return 1

# Else count sum of both ways
return (countPaths(n - 1, m) + 
        countPaths(n, m - 1))

Driver Code

n = 3 m = 2 print("Number of Paths", countPaths(n, m))

This code is contributed by ash264

C#

// C# program to count total number of // paths from a point to origin using System;

public class GFG {

// DP based function to count number
// of paths
static int countPaths(int n, int m)
{
    int [,]dp = new int[n + 1,m + 1];

    // Fill entries in bottommost row
    // and leftmost columns
    for (int i = 0; i <= n; i++)
        dp[i,0] = 1;
    for (int i = 0; i <= m; i++)
        dp[0,i] = 1;

    // Fill DP in bottom up manner
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            dp[i,j] = dp[i - 1,j]
                     + dp[i,j - 1];
    
    return dp[n,m];
}

// Driver Code
public static void Main ()
{
    int n = 3, m = 2;
    
    Console.WriteLine(" Number of"
    + " Paths " + countPaths(n, m));
    
}

}

// This code is contributed by Sam007.

PHP

JavaScript

`

Output

Number of Paths 10

Time Complexity: O(n*m)
Auxiliary Space: O(n*m)

Another Approach:

Using Pascal's Triangle Approach, we also solve the problem by calculating the value of n+mCn. It can be observed as a pattern when you increase the value of m keeping the value of n constant.
Below is the implementation of the above approach:

Implementation:

C++ `

// C++ Program for above approach #include #include <bits/stdc++.h> using namespace std;

// Function to find // binomial Coefficient int binomialCoeff(int n, int k) { int C[k+1]; memset(C, 0, sizeof(C)); C[0] = 1;

// Constructing Pascal's Triangle
for (int i = 1; i <= n; i++) 
{
    for (int j = min(i, k); j > 0; j--) 
        C[j] = C[j] + C[j-1]; 
} 
return C[k]; 

}

//Driver Code int main() { int n=3, m=2; cout<<"Number of Paths: "<< binomialCoeff(n+m,n)<<endl; return 0; }

//Contributed by Vismay_7

Java

// Java Program for above approach import java.io.; import java.util.;

class GFG { static int min(int a,int b) { return a<b?a:b; }

// Function for binomial 
// Coefficient
static int binomialCoeff(int n, int k) 
{ 
    int C[] = new int[k + 1]; 
    C[0] = 1;  
    
    //Constructing Pascal's Triangle
    for (int i = 1; i <= n; i++) 
    { 
        for (int j = min(i,k); j > 0; j--) 
            C[j] = C[j] + C[j-1]; 
    } 
    return C[k]; 
}

// Driver Code
public static void main (String[] args) 
{
    int n=3,m=2;
    System.out.println("Number of Paths: " + 
                       binomialCoeff(n+m,n));
}

} //Contributed by Vismay_7

Python3

Python3 program for above approach

def binomialCoeff(n,k):

C = [0]*(k+1)
C[0] = 1

# Computing Pascal's Triangle
for i in range(1, n + 1): 
    
    j = min(i ,k) 
    while (j > 0): 
        C[j] = C[j] + C[j-1] 
        j -= 1

return C[k] 

Driver Code

n=3 m=2 print("Number of Paths:",binomialCoeff(n+m,n))

Contributed by Vismay_7

C#

// C# program for above approach using System;

class GFG{

// Function to find // binomial Coefficient static int binomialCoeff(int n, int k) { int[] C = new int[k + 1]; C[0] = 1;

// Constructing Pascal's Triangle
for(int i = 1; i <= n; i++) 
{
    for(int j = Math.Min(i, k); j > 0; j--) 
        C[j] = C[j] + C[j - 1]; 
} 
return C[k]; 

}

// Driver code static void Main() { int n = 3, m = 2;

Console.WriteLine("Number of Paths: " + 
                  binomialCoeff(n + m, n));

} }

// This code is contributed by divyesh072019

JavaScript

`

Output

Number of Paths: 10

Time Complexity : O((m+n)*n)
Auxiliary Space : O(n)