Count ways to reach the n'th stair

There are n stairs, a person standing at the bottom wants to reach the top. The person can climb either 1 stair or 2 stairs at a time. Count the number of ways, the person can reach the top.

Consider the example shown in the diagram. The value of n is 3. There are 3 ways to reach the top. The diagram is taken from Easier Fibonacci puzzles

Examples:

Input: n = 1
Output: 1
There is only one way to climb 1 stair

Input: n = 2
Output: 2
There are two ways: (1, 1) and (2)

Input: n = 4
Output: 5
(1, 1, 1, 1), (1, 1, 2), (2, 1, 1), (1, 2, 1), (2, 2)

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Method 1: The first method uses the technique of recursion to solve this problem.

Approach: We can easily find the recursive nature in the above problem. The person can reach n^th stair from either (n-1)^th stair or from (n-2)^th stair. Hence, for each stair n, we try to find out the number of ways to reach n-1^th stair and n-2^th stair and add them to give the answer for the n^th stair. Therefore the expression for such an approach comes out to be :

ways(n) = ways(n-1) + ways(n-2)

The above expression is actually the expression for Fibonacci numbers, but there is one thing to notice, the value of ways(n) is equal to fibonacci(n+1).

ways(1) = fib(2) = 1
ways(2) = fib(3) = 2
ways(3) = fib(4) = 3

For a better understanding, let’s refer to the recursion tree below -:

Input: N = 4

                  fib(5)
           '3'  /        \   '2'
               /          \
           fib(4)         fib(3)
     '2'  /      \ '1'   /      \  
         /        \     /        \ 
     fib(3)     fib(2)fib(2)      fib(1) 
     /    \ '1' /   \ '0'
'1' /   '1'\   /     \ 
   /        \ fib(1) fib(0) 
fib(2)     fib(1)

So we can use the function for Fibonacci numbers to find the value of ways(n). Following is C++ implementation of the above idea.

C++

// C++ program to count number of 
// ways to reach Nth stair 
#include <bits/stdc++.h> 
using namespace std; 
  
// A simple recursive program to 
// find N'th fibonacci number 
int fib(int n) 
{ 
    if (n <= 1) 
        return n; 
    return fib(n - 1) + fib(n - 2); 
} 
  
// Returns number of ways to 
// reach s'th stair 
int countWays(int s) 
{ 
    return fib(s + 1); 
} 
  
// Driver C 
int main() 
{ 
    int s = 4; 
  
    cout << "Number of ways = " << countWays(s); 
  
    return 0; 
} 
  
// This code is contributed by shubhamsingh10 

C

// C Program to count number of 
// ways to reach Nth stair 
#include <stdio.h> 
  
// A simple recursive program to 
// find n'th fibonacci number 
int fib(int n) 
{ 
    if (n <= 1) 
        return n; 
    return fib(n - 1) + fib(n - 2); 
} 
  
// Returns number of ways to reach s'th stair 
int countWays(int s) 
{ 
    return fib(s + 1); 
} 
  
// Driver program to test above functions 
int main() 
{ 
    int s = 4; 
    printf("Number of ways = %d", countWays(s)); 
    getchar(); 
    return 0; 
}

Java

class stairs { 
    // A simple recursive program to find 
    // n'th fibonacci number 
    static int fib(int n) 
    { 
        if (n <= 1) 
            return n; 
        return fib(n - 1) + fib(n - 2); 
    } 
  
    // Returns number of ways to reach s'th stair 
    static int countWays(int s) 
    { 
        return fib(s + 1); 
    } 
  
    /* Driver program to test above function */
    public static void main(String args[]) 
    { 
        int s = 4; 
        System.out.println("Number of ways = " + countWays(s)); 
    } 
} /* This code is contributed by Rajat Mishra */

Python

# Python program to count 
# ways to reach nth stair 
  
# Recursive function to find  
# Nth fibonacci number 
def fib(n): 
    if n <= 1: 
        return n 
    return fib(n-1) + fib(n-2) 
  
# Returns no. of ways to  
# reach sth stair 
def countWays(s): 
    return fib(s + 1) 
  
# Driver program 
s = 4
print "Number of ways = ", 
print countWays(s) 
  
# Contributed by Harshit Agrawal 

C#

// C# program to count the 
// number of ways to reach 
// n'th stair 
using System; 
  
class GFG { 
    // A simple recursive 
    // program to find n'th 
    // fibonacci number 
    static int fib(int n) 
    { 
        if (n <= 1) 
            return n; 
        return fib(n - 1) + fib(n - 2); 
    } 
  
    // Returns number of ways 
    // to reach s'th stair 
    static int countWays(int s) 
    { 
        return fib(s + 1); 
    } 
  
    // Driver Code 
    static public void Main() 
    { 
        int s = 4; 
        Console.WriteLine("Number of ways = " + countWays(s)); 
    } 
} 
  
// This code is contributed 
// by akt_mit 

PHP

<?php 
// A PHP program to count  
// number of ways to reach  
// n'th stair when a person 
// can climb 1, 2, ..m stairs  
// at a time.  
  
// A simple recursive 
// function to find n'th  
// fibonacci number 
function fib($n) 
{ 
if ($n <= 1) 
    return $n; 
return fib($n - 1) +  
       fib($n - 2); 
} 
  
// Returns number of  
// ways to reach s'th stair 
function countWays($s) 
{ 
    return fib($s + 1); 
} 
  
// Driver Code 
$s = 4; 
echo "Number of ways = ",  
           countWays($s); 
  
// This code is contributed 
// by m_kit 
?> 

Output:

Number of ways = 5

Complexity Analysis:

Time Complexity: O(2^n)
The time complexity of the above implementation is exponential (golden ratio raised to power n) due to redundant calculations.It can be optimized to work in O(Logn) time using the previously discussed Fibonacci function optimizations.
Auxiliary Space: O(1)

Generalization of the Problem
How to count the number of ways if the person can climb up to m stairs for a given value m. For example, if m is 4, the person can climb 1 stair or 2 stairs or 3 stairs or 4 stairs at a time.

Approach: For the generalization of above approach the following recursive relation can be used.

ways(n, m) = ways(n-1, m) + ways(n-2, m) + ... ways(n-m, m)

In this approach to reach n^th stair, try climbing all possible number of stairs lesser than equal to n from present stair.

Following is the implementation of above recurrence.

C++

// C++ program to count number of ways 
// to reach nth stair when a person 
// can climb either 1 or 2 stairs at a time 
#include <bits/stdc++.h> 
using namespace std; 
  
// A recursive function used by countWays 
int countWaysUtil(int n, int m) 
{ 
    if (n <= 1) 
    { 
        return n; 
    } 
      
    int res = 0; 
    for(int i = 1; i <= m && i <= n; i++) 
    { 
       res += countWaysUtil(n - i, m); 
    } 
    return res; 
} 
  
// Returns number of ways to reach s'th stair 
int countWays(int s, int m) 
{ 
    return countWaysUtil(s + 1, m); 
} 
  
// Driver code 
int main() 
{ 
    int s = 4, m = 2; 
    cout << "Nuber of ways = " << countWays(s, m); 
  
    return 0; 
} 
  
// This code is contribute by shubhamsingh10 

C

// C program to count number of ways 
// to reach nth stair when a person 
// can climb either 1 or 2 stairs at a time 
#include <stdio.h> 
  
// A recursive function used by countWays 
int countWaysUtil(int n, int m) 
{ 
    if (n <= 1) 
        return n; 
    int res = 0; 
    for (int i = 1; i <= m && i <= n; i++) 
        res += countWaysUtil(n - i, m); 
    return res; 
} 
  
// Returns number of ways to reach s'th stair 
int countWays(int s, int m) 
{ 
    return countWaysUtil(s + 1, m); 
} 
  
// Driver program to test above functions- 
int main() 
{ 
    int s = 4, m = 2; 
    printf("Nuber of ways = %d", countWays(s, m)); 
    return 0; 
} 

Java

class stairs { 
    // A recursive function used by countWays 
    static int countWaysUtil(int n, int m) 
    { 
        if (n <= 1) 
            return n; 
        int res = 0; 
        for (int i = 1; i <= m && i <= n; i++) 
            res += countWaysUtil(n - i, m); 
        return res; 
    } 
  
    // Returns number of ways to reach s'th stair 
    static int countWays(int s, int m) 
    { 
        return countWaysUtil(s + 1, m); 
    } 
  
    /* Driver program to test above function */
    public static void main(String args[]) 
    { 
        int s = 4, m = 2; 
        System.out.println("Number of ways = "
                           + countWays(s, m)); 
    } 
} /* This code is contributed by Rajat Mishra */

Python

# A program to count the number of ways 
# to reach n'th stair 
  
# Recursive function used by countWays 
def countWaysUtil(n, m): 
    if n <= 1: 
        return n 
    res = 0
    i = 1
    while i<= m and i<= n: 
        res = res + countWaysUtil(n-i, m) 
        i = i + 1
    return res 
      
# Returns number of ways to reach s'th stair     
def countWays(s, m): 
    return countWaysUtil(s + 1, m) 
      
  
# Driver program 
s, m = 4, 2
print "Number of ways =", countWays(s, m) 
  
# Contributed by Harshit Agrawal 

C#

// C# program to Count ways to reach 
// the n’th stair 
using System; 
  
class GFG { 
  
    // A recursive function used by 
    // countWays 
    static int countWaysUtil(int n, int m) 
    { 
        if (n <= 1) 
            return n; 
        int res = 0; 
  
        for (int i = 1; i <= m && i <= n; i++) 
            res += countWaysUtil(n - i, m); 
        return res; 
    } 
  
    // Returns number of ways to reach 
    // s'th stair 
    static int countWays(int s, int m) 
    { 
        return countWaysUtil(s + 1, m); 
    } 
  
    /* Driver program to test above function */
    public static void Main() 
    { 
        int s = 4, m = 2; 
        Console.Write("Number of ways = "
                      + countWays(s, m)); 
    } 
} 
  
// This code is contributed by nitin mittal. 

PHP

<?php 
// A PHP program to count  
// number of ways to reach  
// n'th stair when a person  
// can climb either 1 or 2  
// stairs at a time  
  
// A recursive function 
// used by countWays 
function countWaysUtil($n, $m) 
{ 
    if ($n <= 1) 
        return $n; 
    $res = 0; 
    for ($i = 1; $i <= $m &&  
                 $i <= $n; $i++) 
        $res += countWaysUtil($n - $i, $m); 
    return $res; 
} 
  
// Returns number of ways 
// to reach s'th stair 
function countWays($s, $m) 
{ 
    return countWaysUtil($s + 1, $m); 
} 
  
// Driver Code 
$s = 4; $m = 2; 
echo "Nuber of ways = ", 
      countWays($s, $m); 
      
// This code is contributed 
// by akt_mit 
?> 

Output:

Number of ways = 5

Complexity Analysis:

Time Complexity: O(2^n)
The time complexity of the above implementation is exponential (golden ratio raised to power n) due to redundant calculations. It can be optimized to O(m*n) by using dynamic programming.
Auxiliary Space: O(1)

Method 2: This method uses the technique of Dynamic Programming to arrive at the solution.

Approach: We create a table res[] in bottom up manner using the following relation:

res[i] = res[i] + res[i-j] for every (i-j) >= 0

such that the i^th index of the array will contain the number of ways required to reach the i^th step considering all the possibilities of climbing (i.e. from 1 to i).

Below code implements the above approach:

C++

// C++ program to count number of ways 
// to reach n'th stair when a person 
// can climb 1, 2, ..m stairs at a time 
#include <iostream> 
using namespace std; 
  
// A recursive function used by countWays 
int countWaysUtil(int n, int m) 
{ 
    int res[n]; 
    res[0] = 1; 
    res[1] = 1; 
      
    for(int i = 2; i < n; i++)  
    { 
       res[i] = 0; 
         
       for(int j = 1; j <= m && j <= i; j++) 
          res[i] += res[i - j]; 
    } 
    return res[n - 1]; 
} 
  
// Returns number of ways to reach s'th stair 
int countWays(int s, int m) 
{ 
    return countWaysUtil(s + 1, m); 
} 
  
// Driver code 
int main() 
{ 
    int s = 4, m = 2; 
      
    cout << "Number of ways = " 
         << countWays(s, m); 
           
    return 0; 
} 
  
// This code is contributed by shubhamsingh10 

C

// A C program to count number of ways 
// to reach n'th stair when 
// a person can climb 1, 2, ..m stairs at a time 
#include <stdio.h> 
  
// A recursive function used by countWays 
int countWaysUtil(int n, int m) 
{ 
    int res[n]; 
    res[0] = 1; 
    res[1] = 1; 
    for (int i = 2; i < n; i++) { 
        res[i] = 0; 
        for (int j = 1; j <= m && j <= i; j++) 
            res[i] += res[i - j]; 
    } 
    return res[n - 1]; 
} 
  
// Returns number of ways to reach s'th stair 
int countWays(int s, int m) 
{ 
    return countWaysUtil(s + 1, m); 
} 
  
// Driver program to test above functions 
int main() 
{ 
    int s = 4, m = 2; 
    printf("Number of ways = %d", countWays(s, m)); 
    return 0; 
} 

Java

// Java program to count number of ways 
// to reach n't stair when a person 
// can climb 1, 2, ..m stairs at a time 
  
class GFG { 
    // A recursive function used by countWays 
    static int countWaysUtil(int n, int m) 
    { 
        int res[] = new int[n]; 
        res[0] = 1; 
        res[1] = 1; 
        for (int i = 2; i < n; i++) { 
            res[i] = 0; 
            for (int j = 1; j <= m && j <= i; j++) 
                res[i] += res[i - j]; 
        } 
        return res[n - 1]; 
    } 
  
    // Returns number of ways to reach s'th stair 
    static int countWays(int s, int m) 
    { 
        return countWaysUtil(s + 1, m); 
    } 
  
    // Driver method 
    public static void main(String[] args) 
    { 
        int s = 4, m = 2; 
        System.out.println("Number of ways = "
                           + countWays(s, m)); 
    } 
} 

Python

# A program to count the number of  
# ways to reach n'th stair 
  
# Recursive function used by countWays 
def countWaysUtil(n, m): 
    # Creates list res with all elements 0 
    res = [0 for x in range(n)]  
    res[0], res[1] = 1, 1
      
    for i in range(2, n): 
        j = 1
        while j<= m and j<= i: 
            res[i] = res[i] + res[i-j] 
            j = j + 1 
    return res[n-1] 
  
# Returns number of ways to reach s'th stair 
def countWays(s, m): 
    return countWaysUtil(s + 1, m) 
      
# Driver Program 
s, m = 4, 2
print "Number of ways =", countWays(s, m) 
      
# Contributed by Harshit Agrawal 

C#

// C# program to count number 
// of ways to reach n'th stair when 
// a person can climb 1, 2, ..m 
// stairs at a time 
using System; 
class GFG { 
  
    // A recursive function 
    // used by countWays 
    static int countWaysUtil(int n, int m) 
    { 
        int[] res = new int[n]; 
        res[0] = 1; 
        res[1] = 1; 
        for (int i = 2; i < n; i++) { 
            res[i] = 0; 
            for (int j = 1; j <= m && j <= i; j++) 
                res[i] += res[i - j]; 
        } 
        return res[n - 1]; 
    } 
  
    // Returns number of ways 
    // to reach s'th stair 
    static int countWays(int s, int m) 
    { 
        return countWaysUtil(s + 1, m); 
    } 
  
    // Driver Code 
    public static void Main() 
    { 
        int s = 4, m = 2; 
        Console.WriteLine("Number of ways = "
                          + countWays(s, m)); 
    } 
} 
  
// This code is contributed by anuj_67. 

PHP

<?php 
// A PHP program to count number  
// of ways to reach n't stair when 
// a person can climb 1, 2, ..m  
// stairs at a time 
  
// A recursive function used by countWays 
function countWaysUtil($n, $m) 
{ 
    $res[0] = 1; $res[1] = 1; 
    for ($i = 2; $i < $n; $i++) 
    { 
        $res[$i] = 0; 
        for ($j = 1; $j <= $m && $j <= $i; $j++) 
        $res[$i] += $res[$i - $j]; 
    } 
    return $res[$n - 1]; 
} 
  
// Returns number of ways 
// to reach s'th stair 
function countWays($s, $m) 
{ 
    return countWaysUtil($s + 1, $m); 
} 
  
    // Driver Code 
    $s = 4;  
    $m = 2; 
    echo "Number of ways = ", countWays($s, $m); 
  
// This code is contributed by m_kit  
?> 

Output:

Number of ways = 5

Complexity Analysis:

Time Complexity: O(m*n)
Auxiliary Space: O(n)

Method 3: The third method uses the technique of Sliding Window to arrive at the solution.

Approach: This method efficiently implements the above Dynamic Programming approach.
In this approach for the i^th stair, we keep a window of sum of last m possible stairs from which we can climb to the i^th stair. Instead of running an inner loop, we maintain the result of the inner loop in a temporary variable. We remove the elements of the previous window and add the element of the current window and update the sum.

Below code implements the above idea

C++

// A C++ program to count the number of ways 
// to reach n'th stair when user 
// climb 1 .. m stairs at a time. 
// Contributor: rsampaths16 
#include <iostream> 
using namespace std; 
  
// Returns number of ways 
// to reach s'th stair 
int countWays(int n, int m) 
{ 
    int res[n + 1]; 
    int temp = 0; 
    res[0] = 1; 
    for (int i = 1; i <= n; i++) 
    { 
        int s = i - m - 1; 
        int e = i - 1; 
        if (s >= 0) 
        { 
            temp -= res[s]; 
        } 
        temp += res[e]; 
        res[i] = temp; 
    } 
    return res[n]; 
} 
  
// Driver Code 
int main() 
{ 
    int n = 5, m = 3; 
    cout << "Number of ways = " 
         << countWays(n, m); 
    return 0; 
} 
  
// This code is contributed by shubhamsingh10 

C

// A C program to count the number of ways 
// to reach n'th stair when user 
// climb 1 .. m stairs at a time. 
// Contributor: rsampaths16 
#include <stdio.h> 
  
// Returns number of ways 
// to reach s'th stair 
int countWays(int n, int m) 
{ 
    int res[n + 1]; 
    int temp = 0; 
    res[0] = 1; 
    for (int i = 1; i <= n; i++) { 
        int s = i - m - 1; 
        int e = i - 1; 
        if (s >= 0) { 
            temp -= res[s]; 
        } 
        temp += res[e]; 
        res[i] = temp; 
    } 
    return res[n]; 
} 
  
// Driver Code 
int main() 
{ 
    int n = 5, m = 3; 
    printf("Number of ways = %d", 
           countWays(n, m)); 
    return 0; 
} 

Java

// Java program to count number of  
// ways to reach n't stair when a 
// person can climb 1, 2, ..m  
// stairs at a time 
class GFG{ 
      
// Returns number of ways  
// to reach s'th stair 
static int countWays(int n, int m)  
{ 
    int res[] = new int[n + 1];  
    int temp = 0;  
    res[0] = 1;  
      
    for(int i = 1; i <= n; i++)  
    {  
       int s = i - m - 1;  
       int e = i - 1;  
       if (s >= 0)  
       {  
           temp -= res[s];  
       }  
       temp += res[e];  
       res[i] = temp;  
    }  
    return res[n]; 
} 
      
// Driver Code 
public static void main(String[] args)  
{ 
    int n = 5, m = 3; 
    System.out.println("Number of ways = " + 
                       countWays(n, m)); 
} 
} 
  
// This code is contributed by equbalzeeshan 

Python3

# Python3 program to count the number  
# of ways to reach n'th stair when  
# user climb 1 .. m stairs at a time.  
  
# Function to count number of ways  
# to reach s'th stair  
def countWays(n, m):  
      
    temp = 0
    res = [1]  
      
    for i in range(1, n + 1): 
        s = i - m - 1
        e = i - 1
        if (s >= 0): 
            temp -= res[s]  
        temp += res[e]  
        res.append(temp)  
          
    return res[n]  
  
# Driver Code 
n = 5
m = 3
  
print('Number of ways =', countWays(n, m))  
  
# This code is contributed by 31ajaydandge 

C#

// C# program to count number of 
// ways to reach n'th stair when  
// a person can climb 1, 2, ..m  
// stairs at a time  
using System;  
class GFG{  
      
// Returns number of ways  
// to reach s'th stair 
static int countWays(int n, int m)  
{  
    int[] res = new int[n + 1];  
    int temp = 0;  
    res[0] = 1;  
      
    for(int i = 1; i <= n; i++)  
    {  
       int s = i - m - 1;  
       int e = i - 1;  
       if (s >= 0)  
       {  
           temp -= res[s];  
       }  
       temp += res[e];  
       res[i] = temp;  
    }  
    return res[n]; 
} 
  
// Driver Code  
public static void Main()  
{  
    int n = 5, m = 3;  
    Console.WriteLine("Number of ways = " + 
                      countWays(n, m));  
}  
} 
  
// This code is contributed by equbalzeeshan 

Output:

Number of ways = 13

Complexity Analysis:

Time Complexity: O(n)
Auxiliary Space: O(n)

This article is contributed by Abhishek. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

Count ways to reach the n’th stair

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

C

Java

Python

C#

PHP

C++

C

Java

Python

C#

PHP

C++

C

Java

Python

C#

PHP

C++

C

Java

Python3

C#

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