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Count ways to reach the Nth stair using multiple 1 or 2 steps and a single step 3

  • Difficulty Level : Medium
  • Last Updated : 12 May, 2021

Given an integer N number of stairs, the task is count the number ways to reach the Nth stair by taking 1 or 2 step any number of times but 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. So, there are only two possible ways: (1, 3) or (3, 1)
Input: N = 5 
Output:
Explanation: 
There are five possible ways to reach 5th stair: (2, 3), (3, 2), (1, 1, 3), (1, 3, 1), (3, 1, 1) 
 

 

Approach: This problem can be solved using the concept of Permutations & Combinations. There is a constraint according to which 3 steps at a time have to be taken exactly once. The idea is to count the number of positions that are possible for 3 steps to be taken.
Now, the next task is to find the number of possible two-steps in the movement to Nth stair, which is (N – 3) / 2. In this way, all the possible ways will be covered if the count of two-step is incremented once at a time and for each step, all the possible combinations are calculated.
After Generalising the steps the possible combinations for each possible sequence will be 
 

Ways = (Length of sequence)! * (Count of 2-step)! 
       ------------------------------------------
                  (Count of 1-step)! 

Algorithm: 
 



  • Initialize the length of the sequence to (N – 2).
  • Initialize the count of 1-steps to (N – 3).
  • Initialize the count of 2-steps to 0
  • Initialize the possible count of 2-steps to (N-3)/2.
  • Run a loop until the possible count of 2-steps is equal to the actual count of 2-steps. 
    • Find the possible number of ways with the help of the current length of the sequence, count of 1-steps, count of 2-steps and by above-given formulae.
    • Reduce the length of the sequence by 1.
    • Increase the count of 2-steps by 1

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 or 2 steps at a time and
// 3rd step exactly once
#include <bits/stdc++.h>
using namespace std;
 
int factorial(int n)
{
    // Single line to find factorial
    return (n == 1 || n == 0) ? 1 :
            n * factorial(n - 1);
}
 
// Function to find
// the number of ways
int ways(int n)
{
    // Base Case
    if (n < 3)
    {
        return 0;
    }
 
    // Count of 2-steps
    int c2 = 0;
 
    // Count of 1-steps
    int c1 = n - 3;
 
    // Initial length of sequence
    int l = c1 + 1;
    int s = 0;
 
    // Expected count of 2-steps
    int exp_c2 = c1 / 2;
     
    // Loop to find the ways for
    // every possible sequence
    while (exp_c2 >= c2)
    {
        int f1 = factorial(l);
        int f2 = factorial(c1);
        int f3 = factorial(c2);
        int f4 = (f2 * f3);
 
        s += f1 / f4;
        c2 += 1;
        c1 -= 2;
        l -= 1;
    }
    return s;
}
 
// Driver code
int main()
{
    int n = 7;
    int ans = ways(n);
     
    cout << ans << endl;
    return 0;
}
 
// This code is contributed by coder001

Java




// Java implementation to find the number
// the number of ways to reach Nth stair
// by taking 1 or 2 steps at a time and
// 3rd step exactly once
class GFG {
 
static int factorial(int n)
{
     
    // Single line to find factorial
    return (n == 1 || n == 0) ? 1 :
            n * factorial(n - 1);
}
 
// Function to find
// the number of ways
static int ways(int n)
{
     
    // Base Case
    if (n < 3)
    {
        return 0;
    }
 
    // Count of 2-steps
    int c2 = 0;
 
    // Count of 1-steps
    int c1 = n - 3;
 
    // Intial length of sequence
    int l = c1 + 1;
    int s = 0;
 
    // Expected count of 2-steps
    int exp_c2 = c1 / 2;
 
    // Loop to find the ways for
    // every possible sequence
    while (exp_c2 >= c2)
    {
        int f1 = factorial(l);
        int f2 = factorial(c1);
        int f3 = factorial(c2);
        int f4 = (f2 * f3);
         
        s += f1 / f4;
        c2 += 1;
        c1 -= 2;
         l -= 1;
    }
    return s;
}
 
// Driver Code
public static void main(String args[])
{
    int n = 7;
    int ans = ways(n);
     
    System.out.println(ans);
}
}
 
// This code is contributed by rutvik_56

Python3




# Python3 implementation to find the number
# the number of ways to reach Nth stair
# by taking 1 or 2 steps at a time and
# 3rd Step exactly once
 
import math
 
# Function to find
# the number of ways
def ways(n):
     
    # Base Case
    if n < 3:
        return 0
 
    # Count of 2-steps
    c2 = 0
 
    # Count of 1-steps
    c1 = n-3
 
    # Intial length of sequence
    l = c1 + 1
    s = 0
 
    # expected count of 2-steps
    exp_c2 = c1 / 2
     
    # Loop to find the ways for
    # every possible sequence
    while exp_c2 >= c2:
        f1 = math.factorial(l)
        f2 = math.factorial(c1)
        f3 = math.factorial(c2)
        s += f1//(f2 * f3)
        c2 += 1
        c1 -= 2
        l -= 1
    return s
 
# Driver Code
if __name__ == "__main__":
    N = 7
    print(ways(N))

C#




// C# implementation to find the number
// the number of ways to reach Nth stair
// by taking 1 or 2 steps at a time and
// 3rd step exactly once
using System;
class GFG{
     
static int factorial(int n)
{
         
    // Single line to find factorial
    return (n == 1 || n == 0) ? 1 :
            n * factorial(n - 1);
}
     
// Function to find
// the number of ways
static int ways(int n)
{
         
    // Base Case
    if (n < 3)
    {
        return 0;
    }
     
    // Count of 2-steps
    int c2 = 0;
     
    // Count of 1-steps
    int c1 = n - 3;
     
    // Intial length of sequence
    int l = c1 + 1;
    int s = 0;
     
    // Expected count of 2-steps
    int exp_c2 = c1 / 2;
     
    // Loop to find the ways for
    // every possible sequence
    while (exp_c2 >= c2)
    {
        int f1 = factorial(l);
        int f2 = factorial(c1);
        int f3 = factorial(c2);
        int f4 = (f2 * f3);
             
        s += f1 / f4;
        c2 += 1;
        c1 -= 2;
        l -= 1;
    }
    return s;
}
 
// Driver code
static void Main()
{
    int n = 7;
    int ans = ways(n);
         
    Console.WriteLine(ans);
}
}
 
// This code is contributed by divyeshrabadiya07

Javascript




<script>
// JavaScript implementation to find the number
// the number of ways to reach Nth stair
// by taking 1 or 2 steps at a time and
// 3rd step exactly once
 
function factorial(n)
{
 
    // Single line to find factorial
    return (n == 1 || n == 0) ? 1 :
            n * factorial(n - 1);
}
 
// Function to find
// the number of ways
function ways(n)
{
    // Base Case
    if (n < 3)
    {
        return 0;
    }
 
    // Count of 2-steps
    let c2 = 0;
 
    // Count of 1-steps
    let c1 = n - 3;
 
    // Intial length of sequence
    let l = c1 + 1;
    let s = 0;
 
    // Expected count of 2-steps
    let exp_c2 = Math.floor(c1 / 2);
     
    // Loop to find the ways for
    // every possible sequence
    while (exp_c2 >= c2)
    {
        let f1 = factorial(l);
        let f2 = factorial(c1);
        let f3 = factorial(c2);
        let f4 = (f2 * f3);
 
        s += Math.floor(f1 / f4);
        c2 += 1;
        c1 -= 2;
        l -= 1;
    }
    return s;
}
 
// Driver code
    let n = 7;
    let ans = ways(n);
    document.write(ans);
 
// This code is contributed by Surbhi Tyagi.
</script>
Output: 
20

 

Performance Analysis: 
 

  • Time Complexity: As in the above approach, there is one loop to check the possible permutations for each sequence which can go uptil (N-3) which takes O(N) time and to find the possible combinations it will take O(N), Hence the Time Complexity will be O(N2).
  • Space Complexity: As in the above approach, there is no extra space used, Hence the space complexity will be O(1).

Note: The time complexity can be improved up to O(N) by precomputing the factorials for every number till N.
 




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