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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 */



## 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[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 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[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 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[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()     {         int n = 4;         Console.WriteLine(countWays(n));     } }   /* This code is contributed by vt_m */



## 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)[1]

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(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] = 1;     f1[2] = 2;     f1[3] = 4;       // Constructing Transformation Matrix that will be     /*[[0,1,0],[0,0,1],[3,2,1]]      */     vector > t(k + 1, vector(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; } 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] = 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     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)[1]
ans = [2,4,7][1]
ans = 2 //[2,4,7][1] = First cell value of [2,4,7] i.e 2
for n=3 :
ans = (M X M X F1)[1]  //M^(3-1) X F1 = M X M X F1
ans = (M X [2,4,7])[1]
ans = [4,7,13][1]
ans = 4
for n = 4 :
ans = (M^(4-1) X F1)[1]
ans = (M X M X M X F1) [1]
ans = (M X [4,7,13])[1]
ans = [7,13,24][1]
ans = 7
for n = 5 :
ans = (M^4 X F1)[1]
ans = (M X [7,13,24])[1]
ans = [13,24,44][1]
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& 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 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.



## 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)

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