Count ways to reach the n’th stair

2.6

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

 
 
 
 
 
 
 

More 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)

We can easily find recursive nature in above problem. The person can reach n’th stair from either (n-1)’th stair or from (n-2)’th stair. Let the total number of ways to reach n’t stair be ‘ways(n)’. The value of ‘ways(n)’ can be written as following.

    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

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

C

// A C program to count number of ways to reach n't stair when
// a person can climb 1, 2, ..m stairs at a time.
#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

# A program to count the number of ways to reach n'th stair

# Recurssive program to find n'th fibonacci number
def fib(n):
	if n <= 1:
		return n
	return fib(n-1) + fib(n-2)

# returns no. of ways to reach s'th stair
def countWays(s):
	return fib(s + 1)

# Driver program

s = 4
print "Number of ways = ",
print countWays(s)

# Contributed by Harshit Agrawal


Output:
Number of ways = 5

The time complexity of the above implementation is exponential (golden ratio raised to power n). It can be optimized to work in O(Logn) time using the previously discussed Fibonacci function optimizations.

Generalization of the above problem
How to count 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.

We can write the recurrence as following.

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

Following is C++ implementation of above recurrence.

C

// A C program to count number of ways to reach n't 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


Output:
Number of ways = 5

The time complexity of above solution is exponential. It can be optimized to O(mn) by using dynamic programming. Following is dynamic programming based solution. We build a table res[] in bottom up manner.

C

// A C program to count number of ways to reach n't 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("Nuber 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("Nuber 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):
	res = [0 for x in range(n)] # Creates list res witth all elements 0
	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 "Nmber of ways =",countWays(s,m)
	
# Contributed by Harshit Agrawal


Output:
Number of ways = 5

Asked in: Accolite,Amazon

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

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