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Count ways to reach Nth Stairs by taking 1 and 2 steps with exactly one 3 step
  • Difficulty Level : Medium
  • Last Updated : 09 Apr, 2021

Given a number N which denotes the number of stairs, the task is to reach the Nth stair by taking 1, 2 steps any number of times and taking a step of 3 exactly once. 
Examples: 
 

Input: N = 4 
Output:
Explanation: 
Since a step of 3 has to be taken compulsorily and only once, 
There are only two possible ways: (1, 3) or (3, 1)
Input: N = 5 
Output:
Explanation: 
Since a step of 3 has to be taken compulsorily and only once, 
There are only 5 possible ways: 
(1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 3), (3, 2) 
 

 

Approach: This problem can be solved using Dynamic Programming. To Reach Nth stair the person should be on (N – 1)th, (N – 2)th or (N – 3)th. So, To reach Nth stair from the base Reach (N – 1)th stair which includes exact only one step of 3, Reach (N – 2)th step with the steps which inclues exactly one step of 3 and Reach the (N – 3)th step without taking any step of 3. 
So, The Recurrence Relation for this problem will be – 
 

includes_3[i] = includes_3[i-1] + includes_3[i-2] + not_includes[i-3] 
 



whereas, the Recurrence relation when the 3 number of steps at a time is not allowed is 
 

not_includes[i] = not_includes[i – 1] + not_includes[i – 2] 
 

Below is the implementation of the above approach.
 

C++




// C++ implementation to find the number
// the number of ways to reach Nth stair
// by taking 1, 2 step at a time and
// 3 Steps at a time exactly once.
 
#include <iostream>
using namespace std;
 
// Function to find the number
// the number of ways to reach Nth stair
int number_of_ways(int n)
{
    // Array including number
    // of ways that includes 3
    int includes_3[n + 1] = {};
 
    // Array including number of
    // ways that doesn't includes 3
    int not_includes_3[n + 1] = {};
 
    // Intially to reach 3 stairs by
    // taking 3 steps can be
    // reached by 1 way
    includes_3[3] = 1;
 
    not_includes_3[1] = 1;
    not_includes_3[2] = 2;
    not_includes_3[3] = 3;
 
    // Loop to find the number
    // the number of ways to reach Nth stair
    for (int i = 4; i <= n; i++) {
        includes_3[i]
            = includes_3[i - 1] + includes_3[i - 2] + not_includes_3[i - 3];
        not_includes_3[i]
            = not_includes_3[i - 1] + not_includes_3[i - 2];
    }
    return includes_3[n];
}
 
// Driver Code
int main()
{
    int n = 7;
 
    cout << number_of_ways(n);
 
    return 0;
}

Java




// Java implementation to find the number
// the number of ways to reach Nth stair
// by taking 1, 2 step at a time and
// 3 Steps at a time exactly once.
class GFG
{
 
// Function to find the number
// the number of ways to reach Nth stair
static int number_of_ways(int n)
{
    // Array including number
    // of ways that includes 3
    int []includes_3 = new int[n + 1];
     
    // Array including number of
    // ways that doesn't includes 3
    int []not_includes_3 = new int[n + 1];
 
    // Intially to reach 3 stairs by
    // taking 3 steps can be
    // reached by 1 way
    includes_3[3] = 1;
 
    not_includes_3[1] = 1;
    not_includes_3[2] = 2;
    not_includes_3[3] = 3;
 
    // Loop to find the number
    // the number of ways to reach Nth stair
    for (int i = 4; i <= n; i++)
    {
        includes_3[i]
            = includes_3[i - 1] + includes_3[i - 2] +
               not_includes_3[i - 3];
        not_includes_3[i]
            = not_includes_3[i - 1] + not_includes_3[i - 2];
    }
    return includes_3[n];
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 7;
 
    System.out.print(number_of_ways(n));
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 implementation to find the number
# the number of ways to reach Nth stair
# by taking 1, 2 step at a time and
# 3 Steps at a time exactly once.
 
# Function to find the number
# the number of ways to reach Nth stair
def number_of_ways(n):
     
    # Array including number
    # of ways that includes 3
    includes_3 = [0]*(n + 1)
 
    # Array including number of
    # ways that doesn't includes 3
    not_includes_3 = [0] * (n + 1)
 
    # Intially to reach 3 stairs by
    # taking 3 steps can be
    # reached by 1 way
    includes_3[3] = 1
 
    not_includes_3[1] = 1
    not_includes_3[2] = 2
    not_includes_3[3] = 3
 
    # Loop to find the number
    # the number of ways to reach Nth stair
    for i in range(4, n + 1):
        includes_3[i] = includes_3[i - 1] + \
                        includes_3[i - 2] + \
                        not_includes_3[i - 3]
        not_includes_3[i] = not_includes_3[i - 1] + \
                           not_includes_3[i - 2]
    return includes_3[n]
 
# Driver Code
n = 7
 
print(number_of_ways(n))
 
# This code is contributed by mohit kumar 29

C#




// C# implementation to find the number
// the number of ways to reach Nth stair
// by taking 1, 2 step at a time and
// 3 Steps at a time exactly once.
using System;
 
class GFG
{
 
// Function to find the number
// the number of ways to reach Nth stair
static int number_of_ways(int n)
{
    // Array including number
    // of ways that includes 3
    int []includes_3 = new int[n + 1];
     
    // Array including number of
    // ways that doesn't includes 3
    int []not_includes_3 = new int[n + 1];
 
    // Intially to reach 3 stairs by
    // taking 3 steps can be
    // reached by 1 way
    includes_3[3] = 1;
 
    not_includes_3[1] = 1;
    not_includes_3[2] = 2;
    not_includes_3[3] = 3;
 
    // Loop to find the number
    // the number of ways to reach Nth stair
    for (int i = 4; i <= n; i++)
    {
        includes_3[i]
            = includes_3[i - 1] + includes_3[i - 2] +
            not_includes_3[i - 3];
        not_includes_3[i]
            = not_includes_3[i - 1] + not_includes_3[i - 2];
    }
    return includes_3[n];
}
 
// Driver Code
public static void Main(String[] args)
{
    int n = 7;
 
    Console.Write(number_of_ways(n));
}
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
 
// JavaScript implementation to find the number
// the number of ways to reach Nth stair
// by taking 1, 2 step at a time and
// 3 Steps at a time exactly once.
 
// Function to find the number
// the number of ways to reach Nth stair
function number_of_ways(n)
{
    // Array including number
    // of ways that includes 3
    let includes_3 = new Uint8Array(n + 1);
 
    // Array including number of
    // ways that doesn't includes 3
    let not_includes_3 = new Uint8Array(n + 1);
 
    // Intially to reach 3 stairs by
    // taking 3 steps can be
    // reached by 1 way
    includes_3[3] = 1;
 
    not_includes_3[1] = 1;
    not_includes_3[2] = 2;
    not_includes_3[3] = 3;
 
    // Loop to find the number
    // the number of ways to reach Nth stair
    for (let i = 4; i <= n; i++) {
        includes_3[i]
            = includes_3[i - 1] + includes_3[i - 2] + not_includes_3[i - 3];
        not_includes_3[i]
            = not_includes_3[i - 1] + not_includes_3[i - 2];
    }
    return includes_3[n];
}
 
// Driver Code
 
    let n = 7;
 
    document.write(number_of_ways(n));
 
 
 
// This code is contributed by Surbhi Tyagi.
 
</script>
Output: 
20

 

Similar Article: Count ways to reach Nth Stair using 1, 2 or 3 steps at a time
 

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