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Count ways to reach the n’th stair
  • Difficulty Level : Medium
  • Last Updated : 26 May, 2021

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.
 

stairs

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) 

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 nth 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-1th stair and n-2th stair and add them to give the answer for the nth 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
?>

Javascript




<script>
// A Javascript 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
let s = 4;
document.write("Number of ways = " + countWays(s));
 
// This code is contributed
// by _saurabh_jaiswal
</script>

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 nth 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 << "Number 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("Number 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

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; $m = 2;
echo "Number of ways = ",
      countWays($s, $m);
     
// This code is contributed
// by akt_mit
?>

Javascript




<script>
// A Javascript 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;
    let res = 0;
    for (let 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
let s = 4;
let m = 2;
document.write("Number of ways = " + countWays(s, m));
     
// This code is contributed by _saurabh_jaiswal
</script>

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 ith index of the array will contain the number of ways required to reach the ith 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
?>

Javascript




<script>
 
// A Javascript 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)
{
    let res = [];
    res[0] = 1;
    res[1] = 1;
    for (let i = 2; i < n; i++)
    {
        res[i] = 0;
        for (let 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
    let s = 4;
    let m = 2;
    document.write("Number of ways = " + countWays(s, m));
 
// This code is contributed by _saurabh_jaiswal
 
</script>

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 ith stair, we keep a window of sum of last m possible stairs from which we can climb to the ith 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

Javascript




<script>
 
// Javascript program to count number of
// ways to reach n't stair when a
// person can climb 1, 2, ..m
// stairs at a time   
 
// Returns number of ways
// to reach s'th stair
    function countWays(n , m)
    {
        var res = Array(n + 1).fill(0);
        var temp = 0;
        res[0] = 1;
 
        for (i = 1; i <= n; i++) {
            var s = i - m - 1;
            var e = i - 1;
            if (s >= 0) {
                temp -= res[s];
            }
            temp += res[e];
            res[i] = temp;
        }
        return res[n];
    }
 
    // Driver Code
     
        var n = 5, m = 3;
        document.write("Number of ways = " +
        countWays(n, m));
 
// This code contributed by Rajput-Ji
 
</script>

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

Method 4: The fourth method uses simple mathematics but this is only applicable for this problem if (Order does not matter) while counting steps.

Approach: In This method we simply count the number of sets having 2.

C++




#include <iostream>
using namespace std;
 
int main() {
    int n;
    n=5;
 
    // Here n/2 is done to count the number 2's in n
    // 1 is added for case where there is no 2.
    // eg: if n=4 ans will be 3.
    // {1,1,1,1} set having no 2.
    // {1,1,2} ans {2,2} (n/2) sets contaning 2.
 
    cout<<"Number of ways when order of steps does not matter is : "<<1+(n/2)<<endl;   
   
    return 0;
}

Java




import java.util.*;
 
class GFG{
 
public static void main(String[] args)
{
    int n;
    n = 5;
     
    // Here n/2 is done to count the number 2's
    // in n 1 is added for case where there is no 2.
    // eg: if n=4 ans will be 3.
    // {1,1,1,1} set having no 2.
    // {1,1,2} ans {2,2} (n/2) sets contaning 2.
    System.out.print("Number of ways when order of steps " +
                     "does not matter is : " + (1 + (n / 2)));
}
}
 
// This code is contributed by todaysgaurav

Python3




n = 5
 
# Here n/2 is done to count the number 2's in n
# 1 is added for case where there is no 2.
# eg: if n=4 ans will be 3.
# {1,1,1,1} set having no 2.
# {1,1,2} ans {2,2} (n/2) sets contaning 2.
print("Number of ways when order "
      "of steps does not matter is : ", 1 + (n // 2)) 
 
# This code is contributed by rohitsingh07052

C#




using System;
 
class GFG{
static public void Main()
{
    int n;
    n = 5;
     
    // Here n/2 is done to count the number 2's
    // in n 1 is added for case where there is no 2.
    // eg: if n=4 ans will be 3.
    // {1,1,1,1} set having no 2.
    // {1,1,2} ans {2,2} (n/2) sets contaning 2.
    Console.WriteLine("Number of ways when order of steps " +
                      "does not matter is : " + (1 + (n / 2)));
 
}
}
 
// This code is contributed by Ankita saini

Javascript




<script>
 
var n;
n = 5;
 
// Here n/2 is done to count the number 2's in n
// 1 is added for case where there is no 2.
// eg: if n=4 ans will be 3.
// {1,1,1,1} set having no 2.
// {1,1,2} ans {2,2} (n/2) sets contaning 2.
document.write("Number of ways when order " +
               "of steps does not matter is : ",
               parseInt(1 + (n / 2)));   
 
// This code is contributed by Ankita saini
 
</script>

Output:

Number of ways when order of steps does not matter is : 3

Complexity Analysis:

  • Time Complexity : O(1)
  • Space Complexity : O(1)

Note: This Method is only applicable for the question Count ways to N’th Stair(Order does not matter) .

Order does not matter means for n = 4  {1 2 1}  ,{2 1 1}  , {1 1 2} are considered same.

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