# Recursion: Davis' Staircase

### Problem Statement :

```Davis has a number of staircases in his house and he likes to climb each staircase 1, 2, or 3  steps at a time. Being a very precocious child, he wonders how many ways there are to reach the top of the staircase.

Given the respective heights for each of the  staircases in his house, find and print the number of ways he can climb each staircase, module 10^10 + 7  on a new line.

Example

n  =  5

The staircase has 5 steps. Davis can step on the following sequences of steps:

1 1 1 1 1
1 1 1 2
1 1 2 1
1 2 1 1
2 1 1 1
1 2 2
2 2 1
2 1 2
1 1 3
1 3 1
3 1 1
2 3
3 2

Function Description

Complete the stepPerms function using recursion in the editor below.

stepPerms has the following parameter(s):

int n: the number of stairs in the staircase
Returns

int: the number of ways Davis can climb the staircase, modulo 10000000007
Input Format

The first line contains a single integer, s, the number of staircases in his house.
Each of the following s lines contains a single integer, n, the height of staircase i.

Constraints

1  <=  s   <=  5
1  <=  n   <= 36```

### Solution :

```                        ```Solution in C++ :

In   C++  :

#include <bits/stdc++.h>
#define MOD 10000000007
using namespace std;
int dp, n;

int count_paths(int i) {

if(i == 0)
return 1;
if(i < 0)
return 0;
if(dp[i] != -1)
return dp[i];
dp[i] = count_paths(i - 1) % MOD;
dp[i] = (dp[i] + count_paths(i - 2)) % MOD;
dp[i] = (dp[i] + count_paths(i - 3)) % MOD;
return dp[i];
}

int main() {

int t;
cin >> t;
assert(t >=1 and t<= 5);
for(int i = 0; i < t; i++) {
cin >> n;
assert(n >= 1 and n <= 100000);
memset(dp, -1, sizeof dp);
int ans = count_paths(n);
cout << ans << endl;
}

return 0;
}```
```

```                        ```Solution in Java :

In    Java  :

import java.util.*;

class Solution {
final static long _MOD = 1000000007;

public static long countPathsMemoized(int numberOfSteps) {
long[] memo = new long[numberOfSteps + 1];
long numberOfWays = 1;
for(int i = 1; i < numberOfSteps; i++) {
numberOfWays += countPathsMemoized(i, memo);
}
return numberOfWays % _MOD;
}

public static long countPathsMemoized(int numberOfSteps, long[] memo) {
if(numberOfSteps < 3) {
return (numberOfSteps > 0) ? numberOfSteps : 0;
}
if(memo[numberOfSteps] == 0) {
memo[numberOfSteps] = (
countPathsMemoized(numberOfSteps - 1, memo)
+ countPathsMemoized(numberOfSteps - 2, memo)
+ countPathsMemoized(numberOfSteps - 3, memo)
) % _MOD;

}
return memo[numberOfSteps];
}

public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int numberOfStaircases = scanner.nextInt();
while(numberOfStaircases-- > 0) {
int numberOfSteps = scanner.nextInt();
System.out.println(countPathsMemoized(numberOfSteps));
}

scanner.close();
}
}```
```

```                        ```Solution in Python :

In   Python3  :

def stepPerms(n):
f1, f2, f3 = 1, 2, 4
for i in range(n-1):
f1, f2, f3 = f2, f3, f1 + f2 + f3
return f1```
```

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