# Count ways to reach the n’th stair

• Difficulty Level : Medium
• Last Updated : 25 May, 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.

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  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    // 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

 

## Javascript

 

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  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    // 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

 

## Javascript

 

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  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

## C#

 // C# program to implement // the above approach using System;   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() {     int n = 4;     Console.Write("Number of ways = " + countWays(n)); } }   // This code is contributed by sanjoy_62.

## Javascript

 

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  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    // 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

 

## Javascript

 

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  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    // 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

 

Output

Number of ways = 13

Complexity Analysis:

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

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  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)<

## 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

 

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:

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):

As an example, For n= 4:

F(4) = C3F(1)

C3

Hence, C3F(1) =

## C++

 #include  using namespace std; typedef vector > 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

 

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

 

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:

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:

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  using namespace std; typedef vector > 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

 

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)

My Personal Notes arrow_drop_up
Recommended Articles
Page :