Juggler Sequence is a series of integer number in which the first term starts with a positive integer number a and the remaining terms are generated from the immediate previous term using the below recurrence relation :
Juggler Sequence starting with number 3:
5, 11, 36, 6, 2, 1
Juggler Sequence starting with number 9:
9, 27, 140, 11, 36, 6, 2, 1
Given a number N, we have to print the Juggler Sequence for this number as the first term of the sequence.
Examples:
Input: N = 9
Output: 9, 27, 140, 11, 36, 6, 2, 1
We start with 9 and use above formula to get next terms.Input: N = 6
Output: 6, 2, 1
Iterative approach: We have already seen the iterative approach in Set 1 of this problem.
Recursive approach: In this approach, we will recursively traverse starting from N. Follow the steps below for each recursive step
- Output the value of N
- If N has reached 1 end the recursion
- Otherwise, follow the formula based on the number being odd or even and call the recursive function on the newly derived number.
Below is the implementation of the approach:
C++
// C++ code for the above approach#include <bits/stdc++.h>using namespace std;// Recursive function to print// the juggler sequencevoid jum_sequence(int N){ cout << N << " "; if (N <= 1) return; else if (N % 2 == 0) { N = floor(sqrt(N)); jum_sequence(N); } else { N = floor(N * sqrt(N)); jum_sequence(N); }}// Driver codeint main(){ // Juggler sequence starting with 10 jum_sequence(10); return 0;}// This code is contributed by Potta Lokesh |
Java
// Java code for the above approachclass GFG{ // Recursive function to print // the juggler sequence public static void jum_sequence(int N) { System.out.print(N + " "); if (N <= 1) return; else if (N % 2 == 0) { N = (int) (Math.floor(Math.sqrt(N))); jum_sequence(N); } else { N = (int) Math.floor(N * Math.sqrt(N)); jum_sequence(N); } } // Driver code public static void main(String args[]) { // Juggler sequence starting with 10 jum_sequence(10); }}// This code is contributed by Saurabh Jaiswal |
Python3
# Python code to implement the above approach# Recursive function to print# the juggler sequencedef jum_sequence(N): print(N, end =" ") if (N == 1): return elif N & 1 == 0: N = int(pow(N, 0.5)) jum_sequence(N) else: N = int(pow(N, 1.5)) jum_sequence(N)# Juggler sequence starting with 10jum_sequence(10) |
C#
// C# code for the above approachusing System;class GFG{// Recursive function to print// the juggler sequencepublic static void jum_sequence(int N){ Console.Write(N + " "); if (N <= 1) return; else if (N % 2 == 0) { N = (int)(Math.Floor(Math.Sqrt(N))); jum_sequence(N); } else { N = (int)Math.Floor(N * Math.Sqrt(N)); jum_sequence(N); }}// Driver codepublic static void Main() { // Juggler sequence starting with 10 jum_sequence(10);}}// This code is contributed by Saurabh Jaiswal |
Javascript
<script>// Javascript code for the above approach// Recursive function to print// the juggler sequencefunction jum_sequence(N){ document.write(N +" "); if (N <= 1) return; else if (N % 2 == 0) { N = Math.floor(Math.sqrt(N)); jum_sequence(N); } else { N = Math.floor(N * Math.sqrt(N)); jum_sequence(N); }}// Driver code// Juggler sequence starting with 10jum_sequence(10); // This code is contributed by gfgking</script> |
10 3 5 11 36 6 2 1
Time Complexity: O(N)
Auxiliary Space: O(1)
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
