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

Difficulty Level : Easy
Last Updated : 09 Nov, 2020

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.

Examples:

Input : 4
Output : 7
Explantion:
Below are the four ways
 1 step + 1 step + 1 step + 1 step
 1 step + 2 step + 1 step
 2 step + 1 step + 1 step 
 1 step + 1 step + 2 step
 2 step + 2 step
 3 step + 1 step
 1 step + 3 step

Input : 3
Output : 4
Explantion:
Below are the four ways
 1 step + 1 step + 1 step
 1 step + 2 step
 2 step + 1 step
 3 step

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

There are two methods to solve this problem:

Recursive Method
Dynamic Programming

Method 1: Recursive.
There are n stairs, and a person is allowed to jump next stair, skip one stair or skip two stairs. So there are n stairs. So if a person is standing at i-th stair, the person can move to i+1, i+2, i+3-th stair. A recursive function can be formed where at current index i the function is recursively called for i+1, i+2 and i+3 th stair.
There is another way of forming the recursive function. To reach a stair i, a person has to jump either from i-1, i-2 or i-3 th stair or i is the starting stair.

Algorithm:

Create a recursive function (count(int n)) which takes only one parameter.
Check the base cases. If the value of n is less than 0 then return 0, and if the value of n is equal to zero then return 1 as it is the starting stair.
Call the function recursively with values n-1, n-2 and n-3 and sum up the values that are returned, i.e. sum = count(n-1) + count(n-2) + count(n-3)
Return the value of the sum.

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 :

Working:

Complexity Analysis:

Time Complexity: O(3ⁿ).
The time complexity of the above solution is exponential, a close upper bound will be O(3ⁿ). From each state, 3 recursive function are called. So the upperbound for n states is O(3ⁿ).
Space Complexity:O(1).
As no extra space is required.

Note: The Time Complexity of the program can be optimized using Dynamic Programming.

Method 2: Dynamic Programming.

The idea is similar, but it can be observed that there are n states but the recursive function is called 3 ^ n times. That means that some states are called repeatedly. So the idea is to store the value of states. This can be done in two ways.

Top-Down Approach: The first way is to keep the recursive structure intact and just store the value in a HashMap and whenever the function is called again return the value store without computing ().
Bottom-Up Approach: The second way is to take an extra space of size n and start computing values of states from 1, 2 .. to n, i.e. compute values of i, i+1, i+2 and then use them to calculate the value of i+3.

Algorithm:

Create an array of size n + 1 and initialize the first 3 variables with 1, 1, 2. The base cases.
Run a loop from 3 to n.
For each index i, computer value of ith position as dp[i] = dp[i-1] + dp[i-2] + dp[i-3].
Print the value of dp[n], as the Count of the number of ways to reach n th step.

C++

// A C++ program to count number of ways 
// to reach n't stair when 
#include <iostream> 
using namespace std; 
  
// 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; 
    cout << countWays(n); 
    return 0; 
} 
//This code is contributed by shubhamsingh10 

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 + 2) 
    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 + 2]; 
        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 :

Working:

1 -> 1 -> 1 -> 1
1 -> 1 -> 2
1 -> 2 -> 1
1 -> 3
2 -> 1 -> 1
2 -> 2
3 -> 1

So Total ways: 7

Complexity Analysis:

Time Complexity: O(n).
Only one traversal of the array is needed. So Time Complexity is O(n).
Space Complexity: O(n).
To store the values in a DP, n extra space is needed.

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

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