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 :



title-img 




                        Solution in C++ :

In   C++  :







#include <bits/stdc++.h>
#define MOD 10000000007
using namespace std;
int dp[100001], 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


View More Similar Problems

Tree Coordinates

We consider metric space to be a pair, , where is a set and such that the following conditions hold: where is the distance between points and . Let's define the product of two metric spaces, , to be such that: , where , . So, it follows logically that is also a metric space. We then define squared metric space, , to be the product of a metric space multiplied with itself: . For

View Solution →

Array Pairs

Consider an array of n integers, A = [ a1, a2, . . . . an] . Find and print the total number of (i , j) pairs such that ai * aj <= max(ai, ai+1, . . . aj) where i < j. Input Format The first line contains an integer, n , denoting the number of elements in the array. The second line consists of n space-separated integers describing the respective values of a1, a2 , . . . an .

View Solution →

Self Balancing Tree

An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ

View Solution →

Array and simple queries

Given two numbers N and M. N indicates the number of elements in the array A[](1-indexed) and M indicates number of queries. You need to perform two types of queries on the array A[] . You are given queries. Queries can be of two types, type 1 and type 2. Type 1 queries are represented as 1 i j : Modify the given array by removing elements from i to j and adding them to the front. Ty

View Solution →

Median Updates

The median M of numbers is defined as the middle number after sorting them in order if M is odd. Or it is the average of the middle two numbers if M is even. You start with an empty number list. Then, you can add numbers to the list, or remove existing numbers from it. After each add or remove operation, output the median. Input: The first line is an integer, N , that indicates the number o

View Solution →

Maximum Element

You have an empty sequence, and you will be given N queries. Each query is one of these three types: 1 x -Push the element x into the stack. 2 -Delete the element present at the top of the stack. 3 -Print the maximum element in the stack. Input Format The first line of input contains an integer, N . The next N lines each contain an above mentioned query. (It is guaranteed that each

View Solution →