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# Count ways to reach the nth stair using step 1, 2 or 3

A child is running up a staircase with n steps and can hop either 1 step, 2 steps, or 3 steps at a time. Implement a method to count how many possible ways the child can run up the stairs.

Examples:

```Input : 4
Output : 7
Explanation:
Below are the seven ways
1 step + 1 step + 1 step + 1 step
1 step + 2 step + 1 step
2 step + 1 step + 1 step
1 step + 1 step + 2 step
2 step + 2 step
3 step + 1 step
1 step + 3 step

Input : 3
Output : 4
Explanation:
Below are the four ways
1 step + 1 step + 1 step
1 step + 2 step
2 step + 1 step
3 step```

There are two methods to solve this problem:

1. Recursive Method
2. Dynamic Programming

Method 1: Recursive.
There are n stairs, and a person is allowed to jump next stair, skip one stair or skip two stairs. So there are n stairs. So if a person is standing at i-th stair, the person can move to i+1, i+2, i+3-th stair. A recursive function can be formed where at current index i the function is recursively called for i+1, i+2 and i+3 th stair.
There is another way of forming the recursive function. To reach a stair i, a person has to jump either from i-1, i-2 or i-3 th stair or i is the starting stair.

Algorithm:

1. Create a recursive function (count(int n)) which takes only one parameter.
2. Check the base cases. If the value of n is less than 0 then return 0, and if the value of n is equal to zero then return 1 as it is the starting stair.
3. Call the function recursively with values n-1, n-2 and n-3 and sum up the values that are returned, i.e. sum = count(n-1) + count(n-2) + count(n-3)
4. Return the value of the sum.

## C++

 `// C++ Program to find n-th stair using step size` `// 1 or 2 or 3.` `#include ` `using` `namespace` `std;`   `class` `GFG {`   `    ``// Returns count of ways to reach n-th stair` `    ``// using 1 or 2 or 3 steps.` `public``:` `    ``int` `findStep(``int` `n)` `    ``{` `        ``if` `(n == 0)` `            ``return` `1;` `        ``else` `if` `(n < 0)` `            ``return` `0;`   `        ``else` `            ``return` `findStep(n - 3) + findStep(n - 2)` `                   ``+ findStep(n - 1);` `    ``}` `};`   `// Driver code` `int` `main()` `{` `    ``GFG g;` `    ``int` `n = 4;` `    ``cout << g.findStep(n);` `    ``return` `0;` `}`   `// This code is contributed by SoM15242`

## C

 `// Program to find n-th stair using step size` `// 1 or 2 or 3.` `#include `   `// Returns count of ways to reach n-th stair` `// using 1 or 2 or 3 steps.` `int` `findStep(``int` `n)` `{` `    ``if` `(n == 0)` `       ``return` `1;` `    ``else` `if` `(n < 0)` `        ``return` `0;`   `    ``else` `        ``return` `findStep(n - 3) + findStep(n - 2)` `               ``+ findStep(n - 1);` `}`   `// Driver code` `int` `main()` `{` `    ``int` `n = 4;` `    ``printf``(``"%d\n"``, findStep(n));` `    ``return` `0;` `}`

## Java

 `// Program to find n-th stair` `// using step size 1 or 2 or 3.` `import` `java.lang.*;` `import` `java.util.*;`   `public` `class` `GfG {`   `    ``// Returns count of ways to reach` `    ``// n-th stair using 1 or 2 or 3 steps.` `    ``public` `static` `int` `findStep(``int` `n)` `    ``{` `        ``if` `( n == ``0``)` `            ``return` `1``;` `        ``else` `if` `(n < ``0``)` `            ``return` `0``;`   `        ``else` `            ``return` `findStep(n - ``3``) + findStep(n - ``2``)` `                ``+ findStep(n - ``1``);` `    ``}`   `    ``// Driver function` `    ``public` `static` `void` `main(String argc[])` `    ``{` `        ``int` `n = ``4``;` `        ``System.out.println(findStep(n));` `    ``}` `}`   `/* This code is contributed by Sagar Shukla */`

## Python

 `# Python program to find n-th stair` `# using step size 1 or 2 or 3.`   `# Returns count of ways to reach n-th` `# stair using 1 or 2 or 3 steps.`     `def` `findStep(n):` `    ``if` `( n ``=``=` `0` `):` `        ``return` `1` `    ``elif` `(n < ``0``):` `        ``return` `0`   `    ``else``:` `        ``return` `findStep(n ``-` `3``) ``+` `findStep(n ``-` `2``) ``+` `findStep(n ``-` `1``)`     `# Driver code` `n ``=` `4` `print``(findStep(n))`   `# This code is contributed by Nikita Tiwari.`

## C#

 `// Program to find n-th stair` `// using step size 1 or 2 or 3.` `using` `System;`   `public` `class` `GfG {`   `    ``// Returns count of ways to reach` `    ``// n-th stair using 1 or 2 or 3 steps.` `    ``public` `static` `int` `findStep(``int` `n)` `    ``{` `        ``if` `( n == 0)` `            ``return` `1;` `        ``else` `if` `(n < 0)` `            ``return` `0;`   `        ``else` `            ``return` `findStep(n - 3) + findStep(n - 2)` `                ``+ findStep(n - 1);` `    ``}`   `    ``// Driver function` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `n = 4;` `        ``Console.WriteLine(findStep(n));` `    ``}` `}`   `/* This code is contributed by vt_m */`

## PHP

 ``

## Javascript

 ``

Output

`7`

Working: Complexity Analysis:

• Time Complexity: O(3n).
The time complexity of the above solution is exponential, a close upper bound will be O(3n). From each state, 3 recursive function are called. So the upperbound for n states is O(3n).
• Space Complexity:O(1).
As no extra space is required.

Note: The Time Complexity of the program can be optimized using Dynamic Programming.

Method 2: Dynamic Programming.
The idea is similar, but it can be observed that there are n states but the recursive function is called 3 ^ n times. That means that some states are called repeatedly. So the idea is to store the value of states. This can be done in two ways.

• Top-Down Approach: The first way is to keep the recursive structure intact and just store the value in a HashMap and whenever the function is called again return the value store without computing ().
• Bottom-Up Approach: The second way is to take an extra space of size n and start computing values of states from 1, 2 .. to n, i.e. compute values of i, i+1, i+2 and then use them to calculate the value of i+3.

Algorithm:

1. Create an array of size n + 1 and initialize the first 3 variables with 1, 1, 2. The base cases.
2. Run a loop from 3 to n.
3. For each index i, computer value of ith position as dp[i] = dp[i-1] + dp[i-2] + dp[i-3].
4. Print the value of dp[n], as the Count of the number of ways to reach n th step.

## C++

 `// A C++ program to count number of ways` `// to reach n'th stair when` `#include ` `using` `namespace` `std;`   `// A recursive function used by countWays` `int` `countWays(``int` `n)` `{` `    ``int` `res[n + 1];` `    ``res = 1;` `    ``res = 1;` `    ``res = 2;` `    ``for` `(``int` `i = 3; i <= n; i++)` `        ``res[i] = res[i - 1] + res[i - 2] + res[i - 3];`   `    ``return` `res[n];` `}`   `// Driver program to test above functions` `int` `main()` `{` `    ``int` `n = 4;` `    ``cout << countWays(n);` `    ``return` `0;` `}` `// This code is contributed by shubhamsingh10`

## C

 `// A C program to count number of ways` `// to reach n't stair when` `#include `   `// A recursive function used by countWays` `int` `countWays(``int` `n)` `{` `    ``int` `res[n + 1];` `    ``res = 1;` `    ``res = 1;` `    ``res = 2;` `    ``for` `(``int` `i = 3; i <= n; i++)` `        ``res[i] = res[i - 1] + res[i - 2] + res[i - 3];`   `    ``return` `res[n];` `}`   `// Driver program to test above functions` `int` `main()` `{` `    ``int` `n = 4;` `    ``printf``(``"%d"``, countWays(n));` `    ``return` `0;` `}`

## Java

 `// Program to find n-th stair` `// using step size 1 or 2 or 3.` `import` `java.lang.*;` `import` `java.util.*;`   `public` `class` `GfG {`   `    ``// A recursive function used by countWays` `    ``public` `static` `int` `countWays(``int` `n)` `    ``{` `        ``int``[] res = ``new` `int``[n + ``1``];` `        ``res[``0``] = ``1``;` `        ``res[``1``] = ``1``;` `        ``res[``2``] = ``2``;`   `        ``for` `(``int` `i = ``3``; i <= n; i++)` `            ``res[i] = res[i - ``1``] + res[i - ``2``] + res[i - ``3``];`   `        ``return` `res[n];` `    ``}`   `    ``// Driver function` `    ``public` `static` `void` `main(String argc[])` `    ``{` `        ``int` `n = ``4``;` `        ``System.out.println(countWays(n));` `    ``}` `}`   `/* This code is contributed by Sagar Shukla */`

## Python

 `# Python program to find n-th stair` `# using step size 1 or 2 or 3.`   `# A recursive function used by countWays`     `def` `countWays(n):` `    ``res ``=` `[``0``] ``*` `(n ``+` `2``)` `    ``res[``0``] ``=` `1` `    ``res[``1``] ``=` `1` `    ``res[``2``] ``=` `2`   `    ``for` `i ``in` `range``(``3``, n ``+` `1``):` `        ``res[i] ``=` `res[i ``-` `1``] ``+` `res[i ``-` `2``] ``+` `res[i ``-` `3``]`   `    ``return` `res[n]`     `# Driver code` `n ``=` `4` `print``(countWays(n))`     `# This code is contributed by Nikita Tiwari.`

## C#

 `// Program to find n-th stair` `// using step size 1 or 2 or 3.` `using` `System;`   `public` `class` `GfG {`   `    ``// A recursive function used by countWays` `    ``public` `static` `int` `countWays(``int` `n)` `    ``{` `        ``int``[] res = ``new` `int``[n + 2];` `        ``res = 1;` `        ``res = 1;` `        ``res = 2;`   `        ``for` `(``int` `i = 3; i <= n; i++)` `            ``res[i] = res[i - 1] + res[i - 2] + res[i - 3];`   `        ``return` `res[n];` `    ``}`   `    ``// Driver function` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `n = 4;` `        ``Console.WriteLine(countWays(n));` `    ``}` `}`   `/* This code is contributed by vt_m */`

## PHP

 ``

## Javascript

 ``

Output

`7`
• Working:
```1 -> 1 -> 1 -> 1
1 -> 1 -> 2
1 -> 2 -> 1
1 -> 3
2 -> 1 -> 1
2 -> 2
3 -> 1

So Total ways: 7```
• Complexity Analysis:
• Time Complexity: O(n).
Only one traversal of the array is needed. So Time Complexity is O(n).
• Space Complexity: O(n).
To store the values in a DP, n extra space is needed.

#### Method 3: Matrix Exponentiation (O(logn) Approach)

Matrix Exponentiation is mathematical ways to solve DP problem in better time complexity. Matrix Exponentiation Technique has Transformation matrix of Size K X K and Functional Vector (K X 1) .By taking n-1th power of Transformation matrix and Multiplying It With functional vector Give Resultant Vector say it Res of Size K X 1. First Element of Res will be Answer for given n value. This Approach Will Take O(K^3logn) Time Complexity Which Is Complexity of Finding (n-1) power of Transformation Matrix.

#### Key Terms:

K = No of Terms in which F(n) depend ,from Recurrence Relation We can Say That F(n) depend On F(n-1) and F(n-2). => K =3

F1 =  Vector (1D array) that contain F(n) value of First K terms. Since K=3 =>F1 will have F(n) value of first 2 terms. F1=[1,2,4]

T = Transformation Matrix that is a 2D matrix of Size K X K and  Consist Of All 1 After Diagonal And Rest All Zero except last row. Last Row Will have coefficient Of all K terms in which F(n)  depends In Reverse Order. => T =[ [0 1 0] ,[0 0 1], [1 1 1] ].

#### Algorithms:

```1)Take Input N
2)If N < K then Return Precalculated Answer  //Base Condition
3)construct F1 Vector and T (Transformation Matrix)
4)Take N-1th  power of T by using  Optimal Power(T,N) Methods and assign it in T
5)return (TXF)```

for Optimal Power(T, N) Methods Refer Following Article: https://www.geeksforgeeks.org/write-a-c-program-to-calculate-powxn/

## C++

 `#include ` `#define k 3` `using` `namespace` `std;`   `// Multiply Two Matrix Function` `vector > multiply(vector > A,` `                              ``vector > B)` `{` `    ``// third matrix to store multiplication of Two matrix9*` `    ``vector > C(k + 1, vector<``int``>(k + 1));`   `    ``for` `(``int` `i = 1; i <= k; i++) {` `        ``for` `(``int` `j = 1; j <= k; j++) {` `            ``for` `(``int` `x = 1; x <= k; x++) {` `                ``C[i][j] = (C[i][j] + (A[i][x] * B[x][j]));` `            ``}` `        ``}` `    ``}`   `    ``return` `C;` `}` `// Optimal Way For finding pow(t,n)` `// If n Is Odd then It Will be t*pow(t,n-1)` `// else return pow(t,n/2)*pow(t,n/2)` `vector > ``pow``(vector > t, ``int` `n)` `{` `    ``// base Case` `    ``if` `(n == 1) {` `        ``return` `t;` `    ``}` `    ``// Recurrence Case` `    ``if` `(n & 1) {` `        ``return` `multiply(t, ``pow``(t, n - 1));` `    ``}` `    ``else` `{` `        ``vector > X = ``pow``(t, n / 2);` `        ``return` `multiply(X, X);` `    ``}` `}`   `int` `compute(``int` `n)` `{` `    ``// base Case` `    ``if` `(n == 0)` `        ``return` `1;` `    ``if` `(n == 1)` `        ``return` `1;` `    ``if` `(n == 2)` `        ``return` `2;`   `    ``// Function Vector(indexing 1 )` `    ``// that is [1,2]` `    ``int` `f1[k + 1] = {};` `    ``f1 = 1;` `    ``f1 = 2;` `    ``f1 = 4;`   `    ``// Constructing Transformation Matrix that will be` `    ``/*[[0,1,0],[0,0,1],[3,2,1]]` `     ``*/` `    ``vector > t(k + 1, vector<``int``>(k + 1));` `    ``for` `(``int` `i = 1; i <= k; i++) {` `        ``for` `(``int` `j = 1; j <= k; j++) {` `            ``if` `(i < k) {` `                ``// Store 1 in cell that is next to diagonal` `                ``// of Matrix else Store 0 in cell` `                ``if` `(j == i + 1) {` `                    ``t[i][j] = 1;` `                ``}` `                ``else` `{` `                    ``t[i][j] = 0;` `                ``}` `                ``continue``;` `            ``}` `            ``// Last Row - store the Coefficients in reverse` `            ``// order` `            ``t[i][j] = 1;` `        ``}` `    ``}`   `    ``// Computing T^(n-1) and Setting Transformation matrix T` `    ``// to T^(n-1)` `    ``t = ``pow``(t, n - 1);` `    ``int` `sum = 0;` `    ``// Computing first cell (row=1,col=1) For Resultant` `    ``// Matrix TXF` `    ``for` `(``int` `i = 1; i <= k; i++) {` `        ``sum += t[i] * f1[i];` `    ``}` `    ``return` `sum;` `}` `int` `main()` `{` `    ``int` `n = 4;` `    ``cout << compute(n) << endl;` `    ``n = 5;` `    ``cout << compute(n) << endl;` `    ``n = 10;` `    ``cout << compute(n) << endl;`   `    ``return` `0;` `}`

## Java

 `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG {`   `    ``static` `int` `k = ``3``;`   `    ``// Multiply Two Matrix Function` `    ``static` `int``[][] multiply(``int``[][] A, ``int``[][] B)` `    ``{`   `        ``// Third matrix to store multiplication` `        ``// of Two matrix9*` `        ``int``[][] C = ``new` `int``[k + ``1``][k + ``1``];`   `        ``for` `(``int` `i = ``1``; i <= k; i++) {` `            ``for` `(``int` `j = ``1``; j <= k; j++) {` `                ``for` `(``int` `x = ``1``; x <= k; x++) {` `                    ``C[i][j]` `                        ``= (C[i][j] + (A[i][x] * B[x][j]));` `                ``}` `            ``}` `        ``}` `        ``return` `C;` `    ``}`   `    ``// Optimal Way For finding pow(t,n)` `    ``// If n Is Odd then It Will be t*pow(t,n-1)` `    ``// else return pow(t,n/2)*pow(t,n/2)` `    ``static` `int``[][] pow(``int``[][] t, ``int` `n)` `    ``{`   `        ``// Base Case` `        ``if` `(n == ``1``) {` `            ``return` `t;` `        ``}`   `        ``// Recurrence Case` `        ``if` `((n & ``1``) != ``0``) {` `            ``return` `multiply(t, pow(t, n - ``1``));` `        ``}` `        ``else` `{` `            ``int``[][] X = pow(t, n / ``2``);` `            ``return` `multiply(X, X);` `        ``}` `    ``}`   `    ``static` `int` `compute(``int` `n)` `    ``{`   `        ``// Base Case` `        ``if` `(n == ``0``)` `            ``return` `1``;` `        ``if` `(n == ``1``)` `            ``return` `1``;` `        ``if` `(n == ``2``)` `            ``return` `2``;`   `        ``// Function int(indexing 1 )` `        ``// that is [1,2]` `        ``int` `f1[] = ``new` `int``[k + ``1``];` `        ``f1[``1``] = ``1``;` `        ``f1[``2``] = ``2``;` `        ``f1[``3``] = ``4``;`   `        ``// Constructing Transformation Matrix that will be` `        ``/*[[0,1,0],[0,0,1],[3,2,1]]` `         ``*/` `        ``int``[][] t = ``new` `int``[k + ``1``][k + ``1``];` `        ``for` `(``int` `i = ``1``; i <= k; i++) {` `            ``for` `(``int` `j = ``1``; j <= k; j++) {` `                ``if` `(i < k) {`   `                    ``// Store 1 in cell that is next to` `                    ``// diagonal of Matrix else Store 0 in` `                    ``// cell` `                    ``if` `(j == i + ``1``) {` `                        ``t[i][j] = ``1``;` `                    ``}` `                    ``else` `{` `                        ``t[i][j] = ``0``;` `                    ``}` `                    ``continue``;` `                ``}`   `                ``// Last Row - store the Coefficients` `                ``// in reverse order` `                ``t[i][j] = ``1``;` `            ``}` `        ``}`   `        ``// Computing T^(n-1) and Setting` `        ``// Transformation matrix T to T^(n-1)` `        ``t = pow(t, n - ``1``);` `        ``int` `sum = ``0``;`   `        ``// Computing first cell (row=1,col=1)` `        ``// For Resultant Matrix TXF` `        ``for` `(``int` `i = ``1``; i <= k; i++) {` `            ``sum += t[``1``][i] * f1[i];` `        ``}` `        ``return` `sum;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{`   `        ``// Input` `        ``int` `n = ``4``;` `        ``System.out.println(compute(n));` `        ``n = ``5``;` `        ``System.out.println(compute(n));` `        ``n = ``10``;` `        ``System.out.println(compute(n));` `    ``}` `}`   `// This code is contributed by Shubhamsingh10`

## Python3

 `k ``=` `3`   `# Multiply Two Matrix Function`     `def` `multiply(A, B):`   `    ``# third matrix to store multiplication of Two matrix9*` `    ``C ``=` `[[``0` `for` `x ``in` `range``(k``+``1``)] ``for` `y ``in` `range``(k``+``1``)]`   `    ``for` `i ``in` `range``(``1``, k``+``1``):` `        ``for` `j ``in` `range``(``1``, k``+``1``):` `            ``for` `x ``in` `range``(``1``, k``+``1``):` `                ``C[i][j] ``=` `(C[i][j] ``+` `(A[i][x] ``*` `B[x][j]))`   `    ``return` `C`   `# Optimal Way For finding pow(t,n)` `# If n Is Odd then It Will be t*pow(t,n-1)` `# else return pow(t,n/2)*pow(t,n/2)`     `def` `pow``(t,  n):` `    ``# base Case` `    ``if` `(n ``=``=` `1``):` `        ``return` `t` `    ``# Recurrence Case` `    ``if` `(n & ``1``):` `        ``return` `multiply(t, ``pow``(t, n ``-` `1``))` `    ``else``:` `        ``X ``=` `pow``(t, n ``/``/` `2``)` `    ``return` `multiply(X, X)`     `def` `compute(n):` `    ``# base Case` `    ``if` `(n ``=``=` `0``):` `        ``return` `1` `    ``if` `(n ``=``=` `1``):` `        ``return` `1` `    ``if` `(n ``=``=` `2``):` `        ``return` `2`   `    ``# Function Vector(indexing 1 )` `    ``# that is [1,2]` `    ``f1 ``=` `[``0``]``*``(k ``+` `1``)` `    ``f1[``1``] ``=` `1` `    ``f1[``2``] ``=` `2` `    ``f1[``3``] ``=` `4`   `    ``# Constructing Transformation Matrix that will be` `    ``# [[0,1,0],[0,0,1],[3,2,1]]`   `    ``t ``=` `[[``0` `for` `x ``in` `range``(k``+``1``)] ``for` `y ``in` `range``(k``+``1``)]` `    ``for` `i ``in` `range``(``1``, k``+``1``):` `        ``for` `j ``in` `range``(``1``, k``+``1``):` `            ``if` `(i < k):` `                ``# Store 1 in cell that is next to diagonal of Matrix else Store 0 in` `                ``# cell` `                ``if` `(j ``=``=` `i ``+` `1``):` `                    ``t[i][j] ``=` `1` `                ``else``:` `                    ``t[i][j] ``=` `0` `                ``continue` `            ``# Last Row - store the Coefficients in reverse order` `            ``t[i][j] ``=` `1`   `    ``# Computing T^(n-1) and Setting Transformation matrix T to T^(n-1)` `    ``t ``=` `pow``(t, n ``-` `1``)` `    ``sum` `=` `0` `    ``# Computing first cell (row=1,col=1) For Resultant Matrix TXF` `    ``for` `i ``in` `range``(``1``, k``+``1``):` `        ``sum` `+``=` `t[``1``][i] ``*` `f1[i]` `    ``return` `sum`     `# Driver Code` `n ``=` `4` `print``(compute(n))`   `n ``=` `5` `print``(compute(n))`   `n ``=` `10` `print``(compute(n))`   `# This code is contributed by Shubhamsingh10`

## C#

 `// C# program for the above approach` `using` `System;`   `class` `GFG {`   `    ``static` `int` `k = 3;`   `    ``// Multiply Two Matrix Function` `    ``static` `int``[, ] multiply(``int``[, ] A, ``int``[, ] B)` `    ``{`   `        ``// Third matrix to store multiplication` `        ``// of Two matrix9*` `        ``int``[, ] C = ``new` `int``[k + 1, k + 1];`   `        ``for` `(``int` `i = 1; i <= k; i++) {` `            ``for` `(``int` `j = 1; j <= k; j++) {` `                ``for` `(``int` `x = 1; x <= k; x++) {` `                    ``C[i, j]` `                        ``= (C[i, j] + (A[i, x] * B[x, j]));` `                ``}` `            ``}` `        ``}` `        ``return` `C;` `    ``}`   `    ``// Optimal Way For finding pow(t,n)` `    ``// If n Is Odd then It Will be t*pow(t,n-1)` `    ``// else return pow(t,n/2)*pow(t,n/2)` `    ``static` `int``[, ] pow(``int``[, ] t, ``int` `n)` `    ``{`   `        ``// Base Case` `        ``if` `(n == 1) {` `            ``return` `t;` `        ``}`   `        ``// Recurrence Case` `        ``if` `((n & 1) != 0) {` `            ``return` `multiply(t, pow(t, n - 1));` `        ``}` `        ``else` `{` `            ``int``[, ] X = pow(t, n / 2);` `            ``return` `multiply(X, X);` `        ``}` `    ``}`   `    ``static` `int` `compute(``int` `n)` `    ``{`   `        ``// Base Case` `        ``if` `(n == 0)` `            ``return` `1;` `        ``if` `(n == 1)` `            ``return` `1;` `        ``if` `(n == 2)` `            ``return` `2;`   `        ``// Function int(indexing 1 )` `        ``// that is [1,2]` `        ``int``[] f1 = ``new` `int``[k + 1];` `        ``f1 = 1;` `        ``f1 = 2;` `        ``f1 = 4;`   `        ``// Constructing Transformation Matrix that will be` `        ``/*[[0,1,0],[0,0,1],[3,2,1]]` `         ``*/` `        ``int``[, ] t = ``new` `int``[k + 1, k + 1];` `        ``for` `(``int` `i = 1; i <= k; i++) {` `            ``for` `(``int` `j = 1; j <= k; j++) {` `                ``if` `(i < k) {`   `                    ``// Store 1 in cell that is next to` `                    ``// diagonal of Matrix else Store 0 in` `                    ``// cell` `                    ``if` `(j == i + 1) {` `                        ``t[i, j] = 1;` `                    ``}` `                    ``else` `{` `                        ``t[i, j] = 0;` `                    ``}` `                    ``continue``;` `                ``}`   `                ``// Last Row - store the Coefficients` `                ``// in reverse order` `                ``t[i, j] = 1;` `            ``}` `        ``}`   `        ``// Computing T^(n-1) and Setting` `        ``// Transformation matrix T to T^(n-1)` `        ``t = pow(t, n - 1);` `        ``int` `sum = 0;`   `        ``// Computing first cell (row=1,col=1)` `        ``// For Resultant Matrix TXF` `        ``for` `(``int` `i = 1; i <= k; i++) {` `            ``sum += t[1, i] * f1[i];` `        ``}` `        ``return` `sum;` `    ``}`   `    ``// Driver Code` `    ``static` `public` `void` `Main()` `    ``{`   `        ``// Input` `        ``int` `n = 4;` `        ``Console.WriteLine(compute(n));` `        ``n = 5;` `        ``Console.WriteLine(compute(n));` `        ``n = 10;` `        ``Console.WriteLine(compute(n));` `    ``}` `}`   `// This code is contributed by Shubhamsingh10`

## Javascript

 ``

Output

```7
13
274```
```Explanation:
We Know For This Question
Transformation Matrix M= [[0,1,0],[0,0,1],[1,1,1]]
Functional Vector F1 = [1,2,4]
for n=2 :
ans = (M X F1)
ans = [2,4,7]
ans = 2 //[2,4,7] = First cell value of [2,4,7] i.e 2
for n=3 :
ans = (M X M X F1)  //M^(3-1) X F1 = M X M X F1
ans = (M X [2,4,7])
ans = [4,7,13]
ans = 4
for n = 4 :
ans = (M^(4-1) X F1)
ans = (M X M X M X F1) 
ans = (M X [4,7,13])
ans = [7,13,24]
ans = 7
for n = 5 :
ans = (M^4 X F1)
ans = (M X [7,13,24])
ans = [13,24,44]
ans = 13```

#### Time Complexity:

```O(K^3log(n)) //For Computing pow(t,n-1)
For this question K is 3
So Overall Time Complexity is O(27log(n))=O(logn)```

Auxiliary Space: O(n^2) because extra space of vector have been used

Method 4: Using four variables

The idea is based on the Fibonacci series but here with 3 sums. we will hold the values of the first three stairs in 3 variables and will use the fourth variable to find the number of ways.

## C++

 `// A C++ program to count number of ways` `// to reach nth stair when` `#include ` `using` `namespace` `std;`   `// A recursive function used by countWays` `int` `countWays(``int` `n)` `{` `    ``int` `a = 1, b = 2, c = 4; ``// declaring three variables` `                             ``// and holding the ways` `                             ``// for first three stairs` `    ``int` `d = 0; ``// fourth variable` `    ``if` `(n == 0 || n == 1 || n == 2)` `        ``return` `n;` `    ``if` `(n == 3)` `        ``return` `c;`   `    ``for` `(``int` `i = 4; i <= n; i++) { ``// starting from 4 as` `        ``d = c + b + a; ``// already counted for 3 stairs` `        ``a = b;` `        ``b = c;` `        ``c = d;` `    ``}` `    ``return` `d;` `}`   `// Driver program to test above functions` `int` `main()` `{` `    ``int` `n = 4;` `    ``cout << countWays(n);` `    ``return` `0;` `}` `// This code is contributed by Naveen Shah`

## Java

 `// A Java program to count number of ways` `// to reach nth stair when` `import` `java.io.*;`   `class` `GFG{` `    `  `// A recursive function used by countWays` `static` `int` `countWays(``int` `n)` `{` `    `  `    ``// Declaring three variables` `    ``// and holding the ways` `    ``// for first three stairs` `    ``int` `a = ``1``, b = ``2``, c = ``4``; ` `                             `  `    ``// Fourth variable                         ` `    ``int` `d = ``0``; ` `    ``if` `(n == ``0` `|| n == ``1` `|| n == ``2``)` `        ``return` `n;` `    ``if` `(n == ``3``)` `        ``return` `c;`   `    ``for``(``int` `i = ``4``; i <= n; i++)` `    ``{ ` `        ``// Starting from 4 as` `        ``// already counted for 3 stairs` `        ``d = c + b + a; ` `        ``a = b;` `        ``b = c;` `        ``c = d;` `    ``}` `    ``return` `d;` `}`   `// Driver code` `public` `static` `void` `main(String[] args) ` `{` `    ``int` `n = ``4``;` `    `  `    ``System.out.println(countWays(n));` `}` `}`   `// This code is contributed by shivanisinghss2110`

## Python3

 `# A Python program to count number of ways` `# to reach nth stair when` `# A recursive function used by countWays` `def` `countWays(n):` `  `  `    ``# declaring three variables` `    ``# and holding the ways` `    ``# for first three stairs` `    ``a ``=` `1` `    ``b ``=` `2` `    ``c ``=` `4`   `    ``d ``=` `0` `# fourth variable` `    ``if` `(n ``=``=` `0` `or` `n ``=``=` `1` `or` `n ``=``=` `2``):` `        ``return` `n` `    ``if` `(n ``=``=` `3``):` `        ``return` `c` `        `  `    ``for` `i ``in` `range``(``4``,n``+``1``): ` `      `  `        ``# starting from 4 as` `        ``d ``=` `c ``+` `b ``+` `a ``# already counted for 3 stairs` `        ``a ``=` `b` `        ``b ``=` `c` `        ``c ``=` `d` `    ``return` `d`     `# Driver program to test above functions` `n ``=` `4` `print``(countWays(n))`   `# This code is contributed by shivanisinghss2110`

## C#

 `// A C# program to count number of ways` `// to reach nth stair when` `using` `System;`   `class` `GFG{` `    `  `// A recursive function used by countWays` `static` `int` `countWays(``int` `n)` `{` `    `  `    ``// Declaring three variables` `    ``// and holding the ways` `    ``// for first three stairs` `    ``int` `a = 1, b = 2, c = 4; ` `                             `  `    ``// Fourth variable                         ` `    ``int` `d = 0; ` `    ``if` `(n == 0 || n == 1 || n == 2)` `        ``return` `n;` `    ``if` `(n == 3)` `        ``return` `c;`   `    ``for``(``int` `i = 4; i <= n; i++)` `    ``{ ` `        ``// Starting from 4 as` `        ``// already counted for 3 stairs` `        ``d = c + b + a; ` `        ``a = b;` `        ``b = c;` `        ``c = d;` `    ``}` `    ``return` `d;` `}`   `// Driver code` `public` `static` `void` `Main(String[] args) ` `{` `    ``int` `n = 4;` `    `  `    ``Console.Write(countWays(n));` `}` `}`   `// This code is contributed by shivanisinghss2110`

## Javascript

 ``

Output

`7`

Time Complexity: O(n)
Auxiliary Space: O(1), since no extra space has been taken.

Method 5: DP using memoization(Top down approach)

We can avoid the repeated work done in method 1(recursion) by storing the number of ways calculated so far.

We just need to store all the values in an array.

## C++

 `// C++ Program to find n-th stair using step size` `// 1 or 2 or 3.` `#include ` `using` `namespace` `std;`   `class` `GFG {` `private``:` `    ``int` `findStepHelper(``int` `n, vector<``int``>& dp)` `    ``{` `        ``// Base Case` `        ``if` `(n == 0)` `            ``return` `1;` `        ``else` `if` `(n < 0)` `            ``return` `0;` `        ``// If subproblems are already calculated` `        ``//then return it` `        ``if` `(dp[n] != -1) {` `            ``return` `dp[n];` `        ``}`   `       ``// store the subproblems in the vector` `        ``return` `dp[n] = findStepHelper(n - 3, dp)` `                       ``+ findStepHelper(n - 2, dp)` `                       ``+ findStepHelper(n - 1, dp);` `    ``}`   `    ``// Returns count of ways to reach n-th stair` `    ``// using 1 or 2 or 3 steps.` `public``:` `    ``int` `findStep(``int` `n)` `    ``{` `        ``vector<``int``> dp(n + 1, -1);` `        ``return` `findStepHelper(n, dp);` `    ``}` `};`   `// Driver code` `int` `main()` `{` `    ``GFG g;` `    ``int` `n = 4;`   `    ``cout << g.findStep(n);` `    ``return` `0;` `}`

## Java

 `/*package whatever //do not write package name here */` `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG` `{`   `  ``// Java Program to find n-th stair using step size` `  ``// 1 or 2 or 3.` `  ``static` `class` `gfg {`   `    ``private` `int` `findStepHelper(``int` `n, ``int``[] dp)` `    ``{` `      `  `      ``// Base Case` `      ``if` `(n == ``0``)` `        ``return` `1``;` `      ``else` `if` `(n < ``0``)` `        ``return` `0``;` `      `  `      ``// If subproblems are already calculated` `      ``//then return it` `      ``if` `(dp[n] != -``1``) {` `        ``return` `dp[n];` `      ``}`   `      ``// store the subproblems in the vector` `      ``return` `dp[n] = findStepHelper(n - ``3``, dp)` `        ``+ findStepHelper(n - ``2``, dp)` `        ``+ findStepHelper(n - ``1``, dp);` `    ``}`   `    ``// Returns count of ways to reach n-th stair` `    ``// using 1 or 2 or 3 steps.` `    ``public` `int` `findStep(``int` `n)` `    ``{` `      ``int``[] dp = ``new` `int``[n + ``1``];` `      ``Arrays.fill(dp,-``1``);` `      ``return` `findStepHelper(n, dp);` `    ``}` `  ``};`   `  ``/* Driver program to test above function*/` `  ``public` `static` `void` `main(String args[])` `  ``{` `    ``gfg g = ``new` `gfg();` `    ``int` `n = ``4``;`   `    ``System.out.println(g.findStep(n));` `  ``}` `}`   `// This code is contributed by shinjanpatra`

## Python3

 `# Python Program to find n-th stair using step size` `# 1 or 2 or 3.` `class` `GFG:`   `    ``def` `findStepHelper(``self``, n, dp):` `      `  `        ``# Base Case` `        ``if` `(n ``=``=` `0``):` `            ``return` `1` `        ``elif` `(n < ``0``):` `            ``return` `0` `            `  `        ``# If subproblems are already calculated` `        ``#then return it` `        ``if` `(dp[n] !``=` `-``1``):` `            ``return` `dp[n]`   `        ``# store the subproblems in the vector` `        ``dp[n] ``=` `self``.findStepHelper(n ``-` `3``, dp) ``+` `self``.findStepHelper(n ``-` `2``, dp) ``+` `self``.findStepHelper(n ``-` `1``, dp)` `        `  `        ``return` `dp[n]`   `    ``# Returns count of ways to reach n-th stair` `    ``# using 1 or 2 or 3 steps.` `    ``def` `findStep(``self``, n):`   `        ``dp ``=` `[``-``1` `for` `i ``in` `range``(n ``+` `1``)]` `        ``return` `self``.findStepHelper(n, dp)`   `# Driver code` `g ``=` `GFG()` `n ``=` `4`   `print``(g.findStep(n))`   `# This code is contributed by shinjanpatra.`

## Javascript

 ``

## C#

 `using` `System;` `class` `GFG {` `    ``static` `int` `findStepHelper(``int` `n, ``int``[] dp)` `    ``{` `        ``// Base Case` `        ``if` `(n == 0)` `            ``return` `1;` `        ``else` `if` `(n < 0)` `            ``return` `0;` `        ``// If subproblems are already calculated` `        ``// then return it` `        ``if` `(dp[n] != -1) {` `            ``return` `dp[n];` `        ``}`   `        ``// store the subproblems in the vector` `        ``return` `dp[n] = findStepHelper(n - 3, dp)` `                       ``+ findStepHelper(n - 2, dp)` `                       ``+ findStepHelper(n - 1, dp);` `    ``}` `    ``static` `int` `findStep(``int` `n)` `    ``{` `        ``int``[] dp = ``new` `int``[n + 1];` `        ``for` `(``int` `i = 0; i <= n; i++) {` `            ``dp[i] = -1;` `        ``}` `        ``return` `findStepHelper(n, dp);` `    ``}` `    ``static` `void` `Main()` `    ``{` `        ``int` `n = 4;` `        ``Console.Write(findStep(n));` `    ``}` `}`

Output

`7`

Complexity Analysis:

• Time Complexity: O(n). Only one traversal of the array is needed. So Time Complexity is O(n).
• Space Complexity: O(n). To store the values in a DP, n extra space is needed. Also, stack space for recursion is needed which is again O(n)

My Personal Notes arrow_drop_up