Count ways to reach the nth stair using step 1, 2 or 3

A child is running up a staircase with n steps and can hop either 1 step, 2 steps, or 3 steps at a time. Implement a method to count how many possible ways the child can run up the stairs.

There are two methods to solve this problem
1. Recursive Method
2. Dynamic Programming

Examples :

Input : 4
Output : 7

Input : 3
Output : 4

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

1. Recursive Method
How Code is Working :
Suppose you have n stairs then you can hop either 1 step, 2 step, 3 step.
1. If you hop 1 step then remaining stairs = n-1
2. If you hop 2 step then remaining stairs = n-2
3. If you hop 3 step then remaining stairs = n-3

If you hop 1 step then again you can hop 1 step, 2 step, 3 step until n equals 0.
Repeat this process and count total number of ways to reach at nth stair using step 1, 2, 3.

C

// Program to find n-th stair using step size 
// 1 or 2 or 3. 
#include <stdio.h> 
  
// Returns count of ways to reach n-th stair 
// using 1 or 2 or 3 steps. 
int findStep(int n) 
{ 
    if (n == 1 || n == 0)  
        return 1; 
    else if (n == 2)  
        return 2; 
      
    else 
        return findStep(n - 3) +  
               findStep(n - 2) + 
               findStep(n - 1);     
} 
  
// Driver code 
int main() 
{ 
    int n = 4; 
    printf("%d\n", findStep(n)); 
    return 0; 
} 

Java

// Program to find n-th stair 
// using step size 1 or 2 or 3. 
import java.util.*; 
import java.lang.*; 
  
public class GfG{ 
      
    // Returns count of ways to reach 
    // n-th stair using 1 or 2 or 3 steps. 
    public static int findStep(int n) 
    { 
        if (n == 1 || n == 0)  
            return 1; 
        else if (n == 2)  
            return 2; 
       
        else
            return findStep(n - 3) +  
                   findStep(n - 2) + 
                   findStep(n - 1);     
    } 
      
    // Driver function 
    public static void main(String argc[]){ 
        int n = 4; 
        System.out.println(findStep(n)); 
    } 
} 
  
/* This code is contributed by Sagar Shukla */

Python

# Python program to find n-th stair   
# using step size 1 or 2 or 3. 
  
# Returns count of ways to reach n-th  
# stair using 1 or 2 or 3 steps. 
def findStep( n) : 
    if (n == 1 or n == 0) : 
        return 1
    elif (n == 2) : 
        return 2
      
    else : 
        return findStep(n - 3) + findStep(n - 2) + findStep(n - 1)  
  
  
# Driver code 
n = 4
print(findStep(n)) 
  
#This code is contributed by Nikita Tiwari. 

C#

// Program to find n-th stair 
// using step size 1 or 2 or 3. 
using System; 
  
public class GfG{ 
      
    // Returns count of ways to reach 
    // n-th stair using 1 or 2 or 3 steps. 
    public static int findStep(int n) 
    { 
        if (n == 1 || n == 0)  
            return 1; 
        else if (n == 2)  
            return 2; 
      
        else
            return findStep(n - 3) +  
                   findStep(n - 2) + 
                   findStep(n - 1);  
    } 
      
    // Driver function 
    public static void Main(){ 
        int n = 4; 
        Console.WriteLine(findStep(n)); 
    } 
} 
  
/* This code is contributed by vt_m */

PHP

<?php 
// PHP Program to find n-th stair  
// using step size 1 or 2 or 3. 
  
// Returns count of ways to  
// reach n-th stair using   
// 1 or 2 or 3 steps. 
function findStep($n) 
{ 
    if ($n == 1 || $n == 0)  
        return 1; 
    else if ($n == 2)  
        return 2; 
      
    else
        return findStep($n - 3) +  
               findStep($n - 2) + 
                findStep($n - 1);  
} 
  
// Driver code 
$n = 4; 
echo findStep($n); 
  
// This code is contributed by m_kit 
?> 

Output :

How Code is Working….

The Time Complexity of the program is Optimized using Dynamic Programming.

2. Using Dynamic Programming

C

// A C program to count number of ways to reach n't stair when 
#include <stdio.h> 
  
// A recursive function used by countWays 
int countWays(int n) 
{ 
    int res[n + 1]; 
    res[0] = 1; 
    res[1] = 1; 
    res[2] = 2; 
    for (int i = 3; i <= n; i++)  
        res[i] = res[i-1] + res[i-2]  
                          + res[i-3]; 
      
    return res[n]; 
} 
  
// Driver program to test above functions 
int main() 
{ 
    int n = 4; 
    printf("%d", countWays(n)); 
    return 0; 
} 

Java

// Program to find n-th stair 
// using step size 1 or 2 or 3. 
import java.util.*; 
import java.lang.*; 
  
public class GfG { 
  
    // A recursive function used by countWays 
    public static int countWays(int n) 
    { 
        int[] res = new int[n + 1]; 
        res[0] = 1; 
        res[1] = 1; 
        res[2] = 2; 
  
        for (int i = 3; i <= n; i++) 
            res[i] = res[i - 1] + res[i - 2] 
                                + res[i - 3]; 
  
        return res[n]; 
    } 
  
    // Driver function 
    public static void main(String argc[]) 
    { 
        int n = 4; 
        System.out.println(countWays(n)); 
    } 
} 
  
/* This code is contributed by Sagar Shukla */

Python

# Python program to find n-th stair   
# using step size 1 or 2 or 3. 
  
# A recursive function used by countWays 
def countWays(n) : 
    res = [0] * (n + 1) 
    res[0] = 1
    res[1] = 1
    res[2] = 2
      
    for i in range(3, n + 1) : 
        res[i] = res[i - 1] + res[i - 2] + res[i - 3] 
      
    return res[n] 
  
# Driver code 
n = 4
print(countWays(n)) 
  
  
# This code is contributed by Nikita Tiwari. 

C#

// Program to find n-th stair 
// using step size 1 or 2 or 3. 
using System; 
  
public class GfG { 
  
    // A recursive function used by countWays 
    public static int countWays(int n) 
    { 
        int[] res = new int[n + 1]; 
        res[0] = 1; 
        res[1] = 1; 
        res[2] = 2; 
  
        for (int i = 3; i <= n; i++) 
            res[i] = res[i - 1] + res[i - 2] 
                                + res[i - 3]; 
  
        return res[n]; 
    } 
  
    // Driver function 
    public static void Main() 
    { 
        int n = 4; 
        Console.WriteLine(countWays(n)); 
    } 
} 
  
/* This code is contributed by vt_m */

PHP

<?php 
// A PHP program to count  
// number of ways to reach  
// n'th stair when 
  
// A recursive function 
// used by countWays 
function countWays($n) 
{ 
    $res[0] = 1; 
    $res[1] = 1; 
    $res[2] = 2; 
    for ($i = 3; $i <= $n; $i++)  
        $res[$i] = $res[$i - 1] +  
                   $res[$i - 2] +  
                   $res[$i - 3]; 
      
    return $res[$n]; 
} 
  
// Driver Code 
$n = 4; 
echo countWays($n); 
  
// This code is contributed by ajit 
?> 

Output :

Output Explanation :
1 -> 1 -> 1 -> 1
1 -> 1 -> 2
1 -> 2 -> 1
1 -> 3
2 -> 1 -> 1
2 -> 2
3 -> 1

So Total ways: 7

Other Related Articles
http://www.geeksforgeeks.org/count-ways-reach-nth-stair/

This article is contributed by Prakhar Agrawal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

My Personal Notes