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++

// C++ Program to find n-th stair using step size 
// 1 or 2 or 3. 
#include <iostream> 
using namespace std; 
  
class GFG 
{ 
      
// Returns count of ways to reach n-th stair 
// using 1 or 2 or 3 steps. 
public :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() 
{ 
    GFG g ; 
    int n = 4; 
    cout << g.findStep(n); 
    return 0; 
} 
  
// This code is contributed by SoM15242 

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/

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