Counts paths from a point to reach Origin

Difficulty Level : Easy
Last Updated : 26 Jan, 2021

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.

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

<?php
// PHP program to count total number
// of paths from a point to origin
 
// Recursive function to 
// count number of paths
function 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));
}
 
    // Driver Code
    $n = 3; 
    $m = 2;
    echo " Number of Paths "
      , countPaths($n, $m);
 
// This code is contributed by aj_36
?>

Output

 Number of Paths 10

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 po 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

<?php
// PHP program to count total number of
// paths from a point to origin
 
// DP based function to 
// count number of paths
function countPaths($n, $m)
{
     
    //$dp[$n+1][$m+1];
    // Fill entries in bottommost
    // row and leftmost columns
    for ($i = 0; $i <= $n; $i++)
        $dp[$i][0] = 1;
         
    for ($i = 0; $i <= $m; $i++)
        $dp[0][$i] = 1;
 
    // Fill DP in bottom up manner
    for ($i = 1; $i <= $n; $i++)
    for ($j = 1; $j <= $m; $j++)
        $dp[$i][$j] = $dp[$i - 1][$j] + 
                      $dp[$i][$j - 1];
 
    return $dp[$n][$m];
}
 
    // Driver Code
    $n = 3;
    $m = 2;
    echo " Number of Paths " , countPaths($n, $m);
 
// This code is contributed by m_kit
?>

Output

 Number of Paths 10

Another Approach:

Using Pascal’s Triangle Approach, we also solve the problem by calculating the value of ^n+mC_n. 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:

C++

// C++ Program for above approach
#include <iostream>
#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

Output

Number of Paths: 10

This article is contributed by Nikhil Rawat.
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