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 C++ implementation of above steps.
C++
// CPP 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
# Python 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++
// CPP 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
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
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