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Count ways to reach the n’th stair

  • Difficulty Level : Medium
  • Last Updated : 22 Apr, 2022

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


C#




using System;
public 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;
        Console.WriteLine("Number of ways = "
                           + countWays(s, m));
    }
}
 
// This code is contributed by umadevi9616


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: Memoization

We can use the bottom-up approach of dp to solve this problem as well

For this, we can create an array dp[] and initialize it with -1.

Whenever we see that a subproblem is not solved we can call the recursive method, 

else we stop the recursion if that the subproblem is solved already.

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, int dp[])
{
    if (n <= 1)
        return dp[n] = 1;
   
    if(dp[n] != -1 ){
        return dp[n] ;
    }
    dp[n] = fib(n - 1, dp) + fib(n - 2, dp);
    return  dp[n] ;
}
 
// Returns number of ways to
// reach s'th stair
int countWays(int n)
{
    int dp[n+1] ;
    memset(dp, -1, sizeof dp) ;
    fib(n, dp) ;
    return dp[n] ;
}
 
// Driver C
int main()
{
    int n = 4;
 
    cout << "Number of ways = " << countWays(n);
 
    return 0;
}


Java




// Java program to count number of
// ways to reach Nth stair
class GFG
{
 
  // A simple recursive program to
  // find N'th fibonacci number
  static int fib(int n, int dp[])
  {
    if (n <= 1)
      return dp[n] = 1;
 
    if (dp[n] != -1) {
      return dp[n];
    }
    dp[n] = fib(n - 1, dp) + fib(n - 2, dp);
    return dp[n];
  }
 
  // Returns number of ways to
  // reach s'th stair
  static int countWays(int n)
  {
    int[] dp = new int[n + 1];
    for (int i = 0; i < n + 1; i++) {
      dp[i] = -1;
    }
    fib(n, dp);
    return dp[n];
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int n = 4;
    System.out.println(countWays(n));
  }
}
 
// This code is contributed by Karandeep1234


Python3




# Python program to count number of
# ways to reach Nth stair
 
# A simple recursive program to
# find N'th fibonacci number
def fib(n, dp):
 
    if (n <= 1):
        return 1
   
    if(dp[n] != -1 ):
        return dp[n]
 
    dp[n] = fib(n - 1, dp) + fib(n - 2, dp)
    return  dp[n]
 
# Returns number of ways to
# reach s'th stair
def countWays(n):
 
    dp = [-1 for i in range(n + 1)]
    fib(n, dp)
    return dp[n]
 
# Driver Code
n = 4
 
print("Number of ways = " + str(countWays(n)))
 
# This code is contributed by shinjanpatra


Javascript




<script>
 
// JavaScript program to count number of
// ways to reach Nth stair
 
 
// A simple recursive program to
// find N'th fibonacci number
function fib(n, dp)
{
    if (n <= 1)
        return dp[n] = 1;
   
    if(dp[n] != -1 ){
        return dp[n] ;
    }
    dp[n] = fib(n - 1, dp) + fib(n - 2, dp);
    return  dp[n] ;
}
 
// Returns number of ways to
// reach s'th stair
function countWays(n)
{
    let dp = new Array(n+1).fill(-1) ;
    fib(n, dp) ;
    return dp[n] ;
}
 
// Driver Code
 
let n = 4;
 
document.write("Number of ways = " + countWays(n));
 
// This code is contributed by shinjanpatra.
</script>


Output

Number of ways = 5

Complexity Analysis:

Time Complexity: O(n)

Auxiliary Space: O(n)

Method 3: 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 4: 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 5: 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 containing 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 containing 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 containing 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 containing 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 containing 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.

Method 6: This method uses the technique of Matrix Exponentiation to arrive at the solution.

Approach: The number of ways to reach nth stair (Order matters) is equal to the sum of number of ways to reach (n-1)th stair and (n-2)th stair

This corresponds to the following recurrence relation:

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

f(1) = 1
f(2) = 2

where f(n) indicates the number of ways to reach nth stair

Note: 

f(1) = 1 because there is only 1 way to reach n=1 stair  {1}

f(2) = 2 because there are 2 ways to reach n=2 stairs  {1,1} , {2}

It is a type of linear recurrence relation with constant coefficients and we can solve them using Matrix Exponentiation method which basically finds a transformation matrix for a given recurrence relation and repeatedly applies this transformation to a base vector to arrive at the solution).

F(n) = CN-1F(1)
where
C is the transformation matrix
F(1) is the base vector
F(n) is the desired solution

So, for our case the transformation matrix C would be:

0 1
1 1

CN-1 can be calculated using Divide and Conquer technique, in O( (K^3) Log n) where K is dimension of C 

And F(1):

1
2

As an example, For n= 4: 

F(4) = C3F(1)

C3

1 2
2 3

Hence, C3F(1) = 

5
8

C++




#include <bits/stdc++.h>
using namespace std;
typedef vector<vector<long long> > matrix;
 
#define LOOP(i, n) for (int i = 1; i < n; i++)
 
// Computes A*B
// where A,B are square matrices
matrix mul(matrix A, matrix B, long long MOD = 1000000007)
{
    int K = A.size();
    matrix C(K, vector<long long>(K, 0));
    LOOP(i, K)
        LOOP(j, K)
            LOOP(k, K)
                C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % MOD;
    return C;
}
 
// Computes A^n
matrix power(matrix A, long long n)
{
    if (n == 1)
        return A;
    if (n % 2 != 0) {
        // n is odd
        // A^n = A * [ A^(n-1) ]
        A = mul(A, power(A, n - 1));
    }
    else {
        // n is even
        // A^n = [ A^(n/2) ] * [ A^(n/2) ]
        A = power(A, n / 2);
        A = mul(A, A);
    }
    return A;
}
 
long long ways(int n)
{
    vector<long long> F(3);
    F[1] = 1;
    F[2] = 2;
    int K = 2;
    long long MOD = 1000000007;
    // create K*K matrix
    matrix C(K + 1, vector<long long>(K + 1, 0));
    /*
      A matrix with (i+1)th element as 1 and last row
      contains constants
      [
          [0 1 0 0 ... 0]
          [0 0 1 0 ... 0]
          [0 0 0 1 ... 0]
          [. . . . ... .]
          [. . . . ... .]
          [c(k) c(k-1) c(k-2) ... c1]
      ]
    */
    for (int i = 1; i < K; ++i) {
        C[i][i + 1] = 1;
    }
    // Last row is the constants c(k) c(k-1) ... c1
    // f(n) = 1*f(n-1) + 1*f(n-2)
    C[K][1] = 1;
    C[K][2] = 1;
 
    if (n <= 2)
        return F[n];
 
    // f(n) = C^(n-1)*F
    C = power(C, n - 1);
 
    long long result = 0;
 
    // result will be the first row of C^(n-1)*F
    for (int i = 1; i <= K; ++i) {
        result = (result + C[1][i] * F[i]) % MOD;
    }
    return result;
}
 
int main()
{
    int n = 4;
    cout << "Number of ways = " << ways(n) << endl;
}
 
// This code is contributed by darshang631


Java




import java.util.*;
 
class GFG
{
   
    // Computes A*B
    // where A,B are square matrices
    static long[][] mul(long[][] A, long[][] B, long MOD) {
        int K = A.length;
        long[][] C = new long[K][K];
        for (int i = 1; i < K; i++)
            for (int j = 1; j < K; j++)
                for (int k = 1; k < K; k++)
                    C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % MOD;
        return C;
    }
 
    // Computes A^n
    static long[][] power(long[][] A, long n) {
        if (n == 1)
            return A;
        if (n % 2 != 0)
        {
           
            // n is odd
            // A^n = A * [ A^(n-1) ]
            A = mul(A, power(A, n - 1), 1000000007);
        }
      else
      {
            // n is even
            // A^n = [ A^(n/2) ] * [ A^(n/2) ]
            A = power(A, n / 2);
            A = mul(A, A, 1000000007);
        }
        return A;
    }
 
    static long ways(int n) {
        long[] F = new long[3];
        F[1] = 1;
        F[2] = 2;
        int K = 2;
        long MOD = 1000000007;
       
        // create K*K matrix
        long[][] C = new long[K + 1][K + 1];
        /*
         * A matrix with (i+1)th element as 1 and last row contains constants [ [0 1 0 0
         * ... 0] [0 0 1 0 ... 0] [0 0 0 1 ... 0] [. . . . ... .] [. . . . ... .] [c(k)
         * c(k-1) c(k-2) ... c1] ]
         */
        for (int i = 1; i < K; ++i) {
            C[i][i + 1] = 1;
        }
        // Last row is the constants c(k) c(k-1) ... c1
        // f(n) = 1*f(n-1) + 1*f(n-2)
        C[K][1] = 1;
        C[K][2] = 1;
 
        if (n <= 2)
            return F[n];
 
        // f(n) = C^(n-1)*F
        C = power(C, n - 1);
 
        long result = 0;
 
        // result will be the first row of C^(n-1)*F
        for (int i = 1; i <= K; ++i) {
            result = (result + C[1][i] * F[i]) % MOD;
        }
        return result;
    }
 
    public static void main(String[] args) {
        int n = 4;
        System.out.print("Number of ways = " + ways(n) + "\n");
    }
}
 
// This code is contributed by umadevi9616


Python3




# Computes A*B
# where A,B are square matrices
def mul(A, B, MOD):
    K = len(A);
    C = [[0 for i in range(K)] for j in range(K)] ;
    for i in range(1, K):
        for j in range(1, K):
            for k in range(1, K):
                C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % MOD;
    return C;
 
# Computes A^n
def power( A,  n):
    if (n == 1):
        return A;
    if (n % 2 != 0):
 
        # n is odd
        # A^n = A * [ A^(n-1) ]
        A = mul(A, power(A, n - 1), 1000000007);
    else:
        # n is even
        # A^n = [ A^(n/2) ] * [ A^(n/2) ]
        A = power(A, n // 2);
        A = mul(A, A, 1000000007);
     
    return A;
 
def ways(n):
    F = [0 for i in range(3)];
    F[1] = 1;
    F[2] = 2;
    K = 2;
    MOD = 1000000007;
 
    # create K*K matrix
    C = [[0 for i in range(K+1)] for j in range(K+1)];
    '''
     * A matrix with (i+1)th element as 1 and last row contains constants [ [0 1 0 0
     * ... 0] [0 0 1 0 ... 0] [0 0 0 1 ... 0] [. . . . ... .] [. . . . ... .] [c(k)
     * c(k-1) c(k-2) ... c1] ]
     '''
    for i in range(1,K):
        C[i][i + 1] = 1;
     
    # Last row is the constants c(k) c(k-1) ... c1
    # f(n) = 1*f(n-1) + 1*f(n-2)
    C[K][1] = 1;
    C[K][2] = 1;
 
    if (n <= 2):
        return F[n];
 
    # f(n) = C^(n-1)*F
    C = power(C, n - 1);
    result = 0;
 
    # result will be the first row of C^(n-1)*F
    for i in range(1,K+1):
        result = (result + C[1][i] * F[i]) % MOD;
     
    return result;
 
# Driver code
if __name__ == '__main__':
    n = 4;
    print("Number of ways = " , ways(n) , "");
 
# This code is contributed by gauravrajput1


C#




using System;
 
public class GFG {
 
    // Computes A*B
    // where A,B are square matrices
    static long[,] mul(long[,] A, long[,] B, long MOD) {
        int K = A.GetLength(0);
        long[,] C = new long[K,K];
        for (int i = 1; i < K; i++)
            for (int j = 1; j < K; j++)
                for (int k = 1; k < K; k++)
                    C[i,j] = (C[i,j] + A[i,k] * B[k,j]) % MOD;
        return C;
    }
 
    // Computes A^n
    static long[,] power(long[,] A, long n) {
        if (n == 1)
            return A;
        if (n % 2 != 0) {
 
            // n is odd
            // A^n = A * [ A^(n-1) ]
            A = mul(A, power(A, n - 1), 1000000007);
        } else {
            // n is even
            // A^n = [ A^(n/2) ] * [ A^(n/2) ]
            A = power(A, n / 2);
            A = mul(A, A, 1000000007);
        }
        return A;
    }
 
    static long ways(int n) {
        long[] F = new long[3];
        F[1] = 1;
        F[2] = 2;
        int K = 2;
        long MOD = 1000000007;
 
        // create K*K matrix
        long[,] C = new long[K + 1,K + 1];
        /*
         * A matrix with (i+1)th element as 1 and last row contains constants [ [0 1 0 0
         * ... 0] [0 0 1 0 ... 0] [0 0 0 1 ... 0] [. . . . ... .] [. . . . ... .] [c(k)
         * c(k-1) c(k-2) ... c1] ]
         */
        for (int i = 1; i < K; ++i) {
            C[i,i + 1] = 1;
        }
        // Last row is the constants c(k) c(k-1) ... c1
        // f(n) = 1*f(n-1) + 1*f(n-2)
        C[K,1] = 1;
        C[K,2] = 1;
 
        if (n <= 2)
            return F[n];
 
        // f(n) = C^(n-1)*F
        C = power(C, n - 1);
 
        long result = 0;
 
        // result will be the first row of C^(n-1)*F
        for (int i = 1; i <= K; ++i) {
            result = (result + C[1,i] * F[i]) % MOD;
        }
        return result;
    }
 
    public static void Main(String[] args) {
        int n = 4;
        Console.Write("Number of ways = " + ways(n) + "\n");
    }
}
 
// This code is contributed by umadevi9616


Javascript




<script>
    // Computes A*B
    // where A,B are square matrices
     function mul( A,  B , MOD) {
        var K = A.length;
        var C = Array(K).fill().map(()=>Array(K).fill(0));
        for (var i = 1; i < K; i++)
            for (var j = 1; j < K; j++)
                for (var k = 1; k < K; k++)
                    C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % MOD;
        return C;
    }
 
    // Computes A^n
     function power( A , n) {
        if (n == 1)
            return A;
        if (n % 2 != 0) {
 
            // n is odd
            // A^n = A * [ A^(n-1) ]
            A = mul(A, power(A, n - 1), 1000000007);
        } else {
            // n is even
            // A^n = [ A^(n/2) ] * [ A^(n/2) ]
            A = power(A, n / 2);
            A = mul(A, A, 1000000007);
        }
        return A;
    }
 
    function ways(n) {
        var F = Array(3).fill(0);
        F[1] = 1;
        F[2] = 2;
        var K = 2;
        var MOD = 1000000007;
 
        // create K*K matrix
    var  C = Array(K+1).fill().map(()=>Array(K+1).fill(0));
        /*
         * A matrix with (i+1)th element as 1 and last row contains constants [ [0 1 0 0
         * ... 0] [0 0 1 0 ... 0] [0 0 0 1 ... 0] [. . . . ... .] [. . . . ... .] [c(k)
         * c(k-1) c(k-2) ... c1] ]
         */
        for (var i = 1; i < K; ++i) {
            C[i][i + 1] = 1;
        }
        // Last row is the constants c(k) c(k-1) ... c1
        // f(n) = 1*f(n-1) + 1*f(n-2)
        C[K][1] = 1;
        C[K][2] = 1;
 
        if (n <= 2)
            return F[n];
 
        // f(n) = C^(n-1)*F
        C = power(C, n - 1);
 
        var result = 0;
 
        // result will be the first row of C^(n-1)*F
        for (var i = 1; i <= K; ++i) {
            result = (result + C[1][i] * F[i]) % MOD;
        }
        return result;
    }
 
        var n = 4;
        document.write("Number of ways = " + ways(n) + "\n");
 
// This code is contributed by umadevi9616
</script>


Output

Number of ways = 5

Complexity Analysis:

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

Generalization of the Problem:

Given an array A {a1, a2, …., am} containing all valid steps, compute the number of ways to reach nth stair. (Order does matter)

Examples:

Input:
    A = [1,2] 
    n = 4  
Output: 5  
Explanation:
This is the given problem, i.e, count the number of ways to reach n=4 stairs with climbing 1 or 2 stairs at a time
Therefore, ways will be: {1,1,1,1} {1,1,2} {1,2,1} {2,1,1} {2,2} = 5


Input:
    A = [2,4,5]
    n = 9
Output: 5
Explanation:
There are 5 ways to reach n=9 stairs with climbing 2 or 4 or 5 stairs at a time
Therefore, ways will be: {5,4} {4,5} {2,2,5} {2,5,2} {5,2,2} = 5 

 Approach: 

The number of ways to reach nth stair is given by the following recurrence relation

f(n) =  \sum_{i=1}^{i=m} f(n-A_i) 

Let K be the largest element in A.

Step1: Calculate base vector F(1) ( consisting of f(1) …. f(K)  )

It can be done in O(m2K)  time using dynamic programming approach as follows:

Let’s take A = {2,4,5} as an example. To calculate F(1) = { f(1), f(2), f(3), f(4), f(5)  } we will maintain an initially empty array and iteratively append Ai to it and for each Ai we will find the number of ways to reach [Ai-1, to Ai,] 

Hence for A = {2 ,4 ,5}

Let T be the initially empty array

Iteration 1: T = {2}        n = {1,2}        dp = {0,1}         (Number of ways to reach n=1,2 with steps given by T) 
Iteration 2: T = {2,4}        n = {1,2,3,4}    dp = {0,1,0,2}     (Number of ways to reach n=1,2,3,4 with steps given by T)
Iteration 3: T = {2,4,5}    n = {1,2,3,4,5}    dp = {0,1,0,2,1} (Number of ways to reach n=1,2,3,4,5 with steps given by T)

Note: Since some values are already calculated (1,2 for Iteration 2, etc.) we can avoid them in loop

After all iterations, the dp array would be: [0,1,0,2,1] 

Thus, base vector F(1) for A = [2,4,5] is: 

0
1
0
2
1

Now that we have the base vector F(1), calculation of C (Transformation matrix) is easy

Step 2: Calculate C, the transformation matrix

It is a matrix having elements Ai,i+1= 1 and last row contains constants

Now constants can be determined by the presence of that element in A

So for A = [2,4,5] constants will be c = [1,1,0,1,0]   (Ci = 1 if (K-i+1) is present in A, or else 0  where 1 <= i <= K )

Thus, Transformation matrix C for A =[2,4,5] is:

0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
1 1 0 1 0

Step 3: Calculate F(n)

To calculate F(n), following formula is used:

F(n) = Cn-1F(1)

Now that we have C and F(1) we can use Divide and Conquer technique to find Cn-1 and hence the desired output

C++




#include <bits/stdc++.h>
using namespace std;
typedef vector<vector<long long> > matrix;
 
#define LOOP(i, n) for (int i = 1; i < n; i++)
 
// Computes A*B when A,B are square matrices of equal
// dimensions)
matrix mul(matrix A, matrix B, long long MOD = 1000000007)
{
    int K = A.size();
    matrix C(K, vector<long long>(K, 0));
    LOOP(i, K)
        LOOP(j, K)
            LOOP(k, K)
                C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % MOD;
    return C;
}
 
matrix power(matrix A, long long n)
{
    if (n == 1)
        return A;
    if (n % 2 != 0) {
        // n is odd
        // A^n = A * [ A^(n-1) ]
        A = mul(A, power(A, n - 1));
    }
    else {
        // n is even
        // A^n = [ A^(n/2) ] * [ A^(n/2) ]
        A = power(A, n / 2);
        A = mul(A, A);
    }
    return A;
}
 
vector<long long> initialize(vector<long long> A)
{
    // Initializes the base vector F(1)
   
    int K = A[A.size() - 1]; // Assuming A is sorted
    vector<long long> F(K + 1, 0);
    vector<long long> dp(K + 1);
    dp[0] = 0;
    dp[A[1]] = 1; // There is one and only one way to reach
                  // first element
    F[A[1]] = 1;
    for (int i = 2; i < A.size(); ++i) {
        // Loop through A[i-1] to A[i] and fill the dp array
        for (int j = A[i - 1] + 1; j <= A[i]; ++j) {
           
            // dp[j] = dp[j-A[0]] + .. + dp[j-A[i-1]]
            for (int k = 1; k < i; ++k) {
                dp[j] += dp[j - A[k]];
            }
        }
        // There will be one more way to reach A[i]
        dp[A[i]] += 1;
        F[A[i]] = dp[A[i]];
    }
    return F;
}
long long ways(vector<long long> A, int n)
{
    int K = A[A.size() - 1]; // Assuming A is sorted
    vector<long long> F = initialize(A); // O(m^2*K)
    int MOD = 1000000007;
    // create K*K matrix
    matrix C(K + 1, vector<long long>(K + 1, 0));
    /*
    A matrix with (i+1)th element as 1 and last row contains
    constants
    [
        [0 1 0 0 ... 0]
        [0 0 1 0 ... 0]
        [0 0 0 1 ... 0]
        [. . . . ... .]
        [. . . . ... .]
        [c(k) c(k-1) c(k-2) ... c1]
    ]
    */
    for (int i = 1; i < K; ++i) {
        C[i][i + 1] = 1;
    }
    // Last row is the constants c(k) c(k-1) ... c1
    // f(n) = 1*f(n-1) + 1*f(n-2)
    for (int i = 1; i < A.size(); ++i) {
        C[K][K - A[i] + 1] = 1;
    }
 
    if (n <= K)
        return F[n];
    // F(n) = C^(n-1)*F
    C = power(C, n - 1); // O(k^3Log(n))
 
    long long result = 0;
 
    // result will be the first row of C^(n-1)*F
    for (int i = 1; i <= K; ++i) {
        result = (result + C[1][i] * F[i]) % MOD;
    }
    return result;
}
 
int main()
{
    int n = 9;
    vector<long long> A = {
        0, 2, 4, 5
    }; // 0 is just because we are using 1 based indexing
    cout << "Number of ways = " << ways(A, n) << endl;
}
 
// This code is contributed by darshang631


Java




import java.util.*;
 
class GFG{
 
  // Computes A*B when A,B are square matrices of equal
  // dimensions)
  static int[][] mul(int[][] A, int[][] B,int MOD)
  {
    int K = A.length;
    int[][] C = new int[K][K];
    for (int i = 1; i < K; i++)
      for (int j = 1; j < K; j++)
        for (int k = 1; k < K; k++)
          C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % MOD;
    return C;
  }
 
  static int[][] power(int[][] A, long n)
  {
    if (n == 1)
      return A;
    if (n % 2 != 0)
    {
 
      // n is odd
      // A^n = A * [ A^(n-1) ]
      A = mul(A, power(A, n - 1), 1000000007);
    }
    else {
      // n is even
      // A^n = [ A^(n/2) ] * [ A^(n/2) ]
      A = power(A, n / 2);
      A = mul(A, A, 1000000007);
    }
    return A;
  }
 
  static int[] initialize(int[] A)
  {
    // Initializes the base vector F(1)
 
    int K = A[A.length - 1]; // Assuming A is sorted
    int[] F = new int[K+1];
    int[] dp = new int[K+1];
    dp[0] = 0;
    dp[A[1]] = 1; // There is one and only one way to reach
    // first element
    F[A[1]] = 1;
    for (int i = 2; i < A.length; ++i)
    {
 
      // Loop through A[i-1] to A[i] and fill the dp array
      for (int j = A[i - 1] + 1; j <= A[i]; ++j) {
 
        // dp[j] = dp[j-A[0]] + .. + dp[j-A[i-1]]
        for (int k = 1; k < i; ++k) {
          dp[j] += dp[j - A[k]];
        }
      }
 
      // There will be one more way to reach A[i]
      dp[A[i]] += 1;
      F[A[i]] = dp[A[i]];
    }
    return F;
  }
  static int ways(int[] A, int n)
  {
    int K = A[A.length - 1]; // Assuming A is sorted
    int[] F = initialize(A); // O(m^2*K)
    int MOD = 1000000007;
 
    // create K*K matrix
    int[][] C = new int[K + 1][K + 1];
 
    /*
    A matrix with (i+1)th element as 1 and last row contains
    constants
    [
        [0 1 0 0 ... 0]
        [0 0 1 0 ... 0]
        [0 0 0 1 ... 0]
        [. . . . ... .]
        [. . . . ... .]
        [c(k) c(k-1) c(k-2) ... c1]
    ]
    */
    for (int i = 1; i < K; ++i) {
      C[i][i + 1] = 1;
    }
 
    // Last row is the constants c(k) c(k-1) ... c1
    // f(n) = 1*f(n-1) + 1*f(n-2)
    for (int i = 1; i < A.length; ++i) {
      C[K][K - A[i] + 1] = 1;
    }
 
    if (n <= K)
      return F[n];
    // F(n) = C^(n-1)*F
    C = power(C, n - 1); // O(k^3Log(n))
 
    int result = 0;
 
    // result will be the first row of C^(n-1)*F
    for (int i = 1; i <= K; ++i) {
      result = (result + C[1][i] * F[i]) % MOD;
    }
    return result;
  }
 
  public static void main(String[] args)
  {
    int n = 9;
    int[] A = {0, 2, 4, 5};
 
    // 0 is just because we are using 1 based indexing
    System.out.print("Number of ways = " +  ways(A, n) +"\n");
  }
}
 
// This code is contributed by umadevi9616


Python3




# Computes A*B when A,B are square matrices of equal
# dimensions)
def mul(A, B, MOD):
    K = len(A);
    C = [[0 for i in range(K)] for j in range(K)] ;
    for i in range(1, K):
        for j in range(1, K):
            for k in range(1, K):
                C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % MOD;
    return C;
 
def power(A, n):
    if (n == 1):
        return A;
    if (n % 2 != 0):
 
        # n is odd
        # A^n = A * [ A^(n-1) ]
        A = mul(A, power(A, n - 1), 1000000007);
    else:
        # n is even
        # A^n = [ A^(n/2) ] * [ A^(n/2) ]
        A = power(A, n // 2);
        A = mul(A, A, 1000000007);
     
    return A;
 
def initialize(A):
    # Initializes the base vector F(1)
 
    K = A[len(A)-1]; # Assuming A is sorted
    F = [0 for i in range(K+1)];
    dp = [0 for i in range(K+1)];
    dp[0] = 0;
    dp[A[1]] = 1; # There is one and only one way to reach
    # first element
    F[A[1]] = 1;
    for i in range(2,len(A)):
 
        # Loop through A[i-1] to A[i] and fill the dp array
        for j in range(A[i - 1] + 1,A[i]+1):
 
            # dp[j] = dp[j-A[0]] + .. + dp[j-A[i-1]]
            for k in range(1,i):
                dp[j] += dp[j - A[k]];
             
        # There will be one more way to reach A[i]
        dp[A[i]] += 1;
        F[A[i]] = dp[A[i]];
     
    return F;
 
def ways(A, n):
    K = A[len(A) - 1]; # Assuming A is sorted
    F = initialize(A); # O(m^2*K)
    MOD = 1000000007;
 
    # create K*K matrix
    C = [[0 for i in range(K+1)] for j in range(K+1)];
 
    '''
     * A matrix with (i+1)th element as 1 and last row contains constants [ [0 1 0 0
     * ... 0] [0 0 1 0 ... 0] [0 0 0 1 ... 0] [. . . . ... .] [. . . . ... .] [c(k)
     * c(k-1) c(k-2) ... c1] ]
     '''
    for i in range(1,K):
        C[i][i + 1] = 1;
     
 
    # Last row is the constants c(k) c(k-1) ... c1
    # f(n) = 1*f(n-1) + 1*f(n-2)
    for i in range(1, len(A)):
        C[K][K - A[i] + 1] = 1;
     
    if (n <= K):
        return F[n];
    # F(n) = C^(n-1)*F
    C = power(C, n - 1); # O(k^3Log(n))
 
    result = 0;
 
    # result will be the first row of C^(n-1)*F
    for i in range(1, K+1):
        result = (result + C[1][i] * F[i]) % MOD;
     
    return result;
 
# Driver code
if __name__ == '__main__':
    n = 9;
    A = [ 0, 2, 4, 5] ;
 
    # 0 is just because we are using 1 based indexing
    print("Number of ways = " ,ways(A, n));
 
# This code is contributed by gauravrajput1


C#




using System;
 
public class GFG {
 
    // Computes A*B when A,B are square matrices of equal
    // dimensions)
    static int[,] mul(int[,] A, int[,] B, int MOD) {
        int K = A.GetLength(0);
        int[,] C = new int[K,K];
        for (int i = 1; i < K; i++)
            for (int j = 1; j < K; j++)
                for (int k = 1; k < K; k++)
                    C[i,j] = (C[i,j] + A[i,k] * B[k,j]) % MOD;
        return C;
    }
 
    static int[,] power(int[,] A, long n) {
        if (n == 1)
            return A;
        if (n % 2 != 0) {
 
            // n is odd
            // A^n = A * [ A^(n-1) ]
            A = mul(A, power(A, n - 1), 1000000007);
        } else {
            // n is even
            // A^n = [ A^(n/2) ] * [ A^(n/2) ]
            A = power(A, n / 2);
            A = mul(A, A, 1000000007);
        }
        return A;
    }
 
    static int[] initialize(int[] A) {
        // Initializes the base vector F(1)
 
        int K = A[A.Length - 1]; // Assuming A is sorted
        int[] F = new int[K + 1];
        int[] dp = new int[K + 1];
        dp[0] = 0;
        dp[A[1]] = 1; // There is one and only one way to reach
        // first element
        F[A[1]] = 1;
        for (int i = 2; i < A.Length; ++i) {
 
            // Loop through A[i-1] to A[i] and fill the dp array
            for (int j = A[i - 1] + 1; j <= A[i]; ++j) {
 
                // dp[j] = dp[j-A[0]] + .. + dp[j-A[i-1]]
                for (int k = 1; k < i; ++k) {
                    dp[j] += dp[j - A[k]];
                }
            }
 
            // There will be one more way to reach A[i]
            dp[A[i]] += 1;
            F[A[i]] = dp[A[i]];
        }
        return F;
    }
 
    static int ways(int[] A, int n) {
        int K = A[A.Length - 1]; // Assuming A is sorted
        int[] F = initialize(A); // O(m^2*K)
        int MOD = 1000000007;
 
        // create K*K matrix
        int[,] C = new int[K + 1,K + 1];
 
        /*
         * A matrix with (i+1)th element as 1 and last row contains constants [ [0 1 0 0
         * ... 0] [0 0 1 0 ... 0] [0 0 0 1 ... 0] [. . . . ... .] [. . . . ... .] [c(k)
         * c(k-1) c(k-2) ... c1] ]
         */
        for (int i = 1; i < K; ++i) {
            C[i,i + 1] = 1;
        }
 
        // Last row is the constants c(k) c(k-1) ... c1
        // f(n) = 1*f(n-1) + 1*f(n-2)
        for (int i = 1; i < A.GetLength(0); ++i) {
            C[K,K - A[i] + 1] = 1;
        }
 
        if (n <= K)
            return F[n];
        // F(n) = C^(n-1)*F
        C = power(C, n - 1); // O(k^3Log(n))
 
        int result = 0;
 
        // result will be the first row of C^(n-1)*F
        for (int i = 1; i <= K; ++i) {
            result = (result + C[1,i] * F[i]) % MOD;
        }
        return result;
    }
 
    public static void Main(String[] args) {
        int n = 9;
        int[] A = { 0, 2, 4, 5 };
 
        // 0 is just because we are using 1 based indexing
        Console.Write("Number of ways = " + ways(A, n) + "\n");
    }
}
 
// This code is contributed by gauravrajput1


Javascript




<script>
    // Computes A*B when A,B are square matrices of equal
    // dimensions)
     function mul(A,  B , MOD) {
        var K = A.length;
        var  C = Array(K).fill().map(()=>Array(K).fill(0));
        for (var i = 1; i < K; i++)
            for (var j = 1; j < K; j++)
                for (var k = 1; k < K; k++)
                    C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % MOD;
        return C;
    }
 
     function power(A , n) {
        if (n == 1)
            return A;
        if (n % 2 != 0) {
 
            // n is odd
            // A^n = A * [ A^(n-1) ]
            A = mul(A, power(A, n - 1), 1000000007);
        } else {
            // n is even
            // A^n = [ A^(n/2) ] * [ A^(n/2) ]
            A = power(A, parseInt(n / 2));
            A = mul(A, A, 1000000007);
        }
        return A;
    }
 
     function initialize(A) {
        // Initializes the base vector F(1)
 
        var K = A[A.length - 1]; // Assuming A is sorted
        var F = Array(K+1).fill(0);
        var dp = Array(K+1).fill(0);
        dp[0] = 0;
        dp[A[1]] = 1; // There is one and only one way to reach
        // first element
        F[A[1]] = 1;
        for (var i = 2; i < A.length; ++i) {
 
            // Loop through A[i-1] to A[i] and fill the dp array
            for (var j = A[i - 1] + 1; j <= A[i]; ++j) {
 
                // dp[j] = dp[j-A[0]] + .. + dp[j-A[i-1]]
                for (var k = 1; k < i; ++k) {
                    dp[j] += dp[j - A[k]];
                }
            }
 
            // There will be one more way to reach A[i]
            dp[A[i]] += 1;
            F[A[i]] = dp[A[i]];
        }
        return F;
    }
 
    function ways(A , n) {
        var K = A[A.length - 1]; // Assuming A is sorted
        var F = initialize(A); // O(m^2*K)
        var MOD = 1000000007;
 
        // create K*K matrix
        var C = Array(K+1).fill().map(()=>Array(K+1).fill(0));
 
        /*
         * A matrix with (i+1)th element as 1 and last row contains constants [ [0 1 0 0
         * ... 0] [0 0 1 0 ... 0] [0 0 0 1 ... 0] [. . . . ... .] [. . . . ... .] [c(k)
         * c(k-1) c(k-2) ... c1] ]
         */
        for (var i = 1; i < K; ++i) {
            C[i][i + 1] = 1;
        }
 
        // Last row is the constants c(k) c(k-1) ... c1
        // f(n) = 1*f(n-1) + 1*f(n-2)
        for (var i = 1; i < A.length; ++i) {
            C[K][K - A[i] + 1] = 1;
        }
 
        if (n <= K)
            return F[n];
        // F(n) = C^(n-1)*F
        C = power(C, n - 1); // O(k^3Log(n))
 
        var result = 0;
 
        // result will be the first row of C^(n-1)*F
        for (var i = 1; i <= K; ++i) {
            result = (result + C[1][i] * F[i]) % MOD;
        }
        return result;
    }
 
     
        var n = 9;
        var A = [ 0, 2, 4, 5 ];
 
        // 0 is just because we are using 1 based indexing
        document.write("Number of ways = " + ways(A, n) + "\n");
 
// This code is contributed by gauravrajput1
</script>


Output

Number of ways = 5

Complexity Analysis:

Time Complexity: O( m2K + K3Logn )
            where
                m is the size of Array A
                K is the largest element in A
                n denotes the stair number (nth stair)
Space Complexity: O(K2)

Note: 

This approach is ideal when n is too large for iteration 

For Example: Consider this approach when (1 ≤ n ≤ 109) and (1 ≤ m,k ≤ 102)


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