Recaman’s sequence

2.2

Given an integer n. Print first n elements of Recaman’s sequence.

Examples:

Input : n = 6
Output : 0, 1, 3, 6, 2, 7

Input  : n = 17
Output : 0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 
         11, 22, 10, 23, 9, 24, 8

It is basically a function with domain and co-domain as natural numbers and 0. It is recursively defined as below:
Specifically, let a(n) denote the (n+1)-th term. (0 being already there).
The rule says:

a(0) = 0,
if n > 0 and the number is not 
   already included in the sequence,
     a(n) = a(n - 1) - n 
else 
     a(n) = a(n-1) + n. 

Below is simple implementation where we store all n Recaman Sequence numbers in an array. We compute next number using recursive formula mentioned above.

// C++ program to print n-th number in Recaman's 
// sequence
#include <bits/stdc++.h>
using namespace std;

// Prints first n terms of Recaman sequence
int recaman(int n)
{
    // Create an array to store terms
    int arr[n];

    // First term of the sequence is always 0
    arr[0] = 0;
    printf("%d, ", arr[0]);

    // Fill remaining terms using recursive
    // formula.
    for (int i=1; i< n; i++)
    {
        int curr = arr[i-1] - i;
        int j;
        for (j = 0; j < i; j++)
        {
            // If arr[i-1] - i is negative or
            // already exists.
            if ((arr[j] == curr) || curr < 0)
            {
                curr = arr[i-1] + i;
                break;
            }
        }

        arr[i] = curr;
        printf("%d, ", arr[i]);
    }
}

// Driver code
int main()
{
    int n = 17;
    recaman(n);
    return 0;
}

Output:

0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9, 24, 8, 

Time Complexity : O(n2)
Auxiliary Space : O(n)

 

Optimizations :
We can us hashing to store previously computed values and can make this program work in O(n) time.

// C++ program to print n-th number in Recaman's 
// sequence
#include <bits/stdc++.h>
using namespace std;

// Prints first n terms of Recaman sequence
void recaman(int n)
{
    if (n <= 0)
      return;

    // Print first term and store it in a hash 
    printf("%d, ", 0);
    unordered_set<int> s;
    s.insert(0);

    // Print remaining terms using recursive
    // formula.
    int prev = 0;
    for (int i=1; i< n; i++)
    {
        int curr = prev - i;

        // If arr[i-1] - i is negative or
        // already exists.
        if (curr < 0 || s.find(curr) != s.end())
           curr = prev + i;

        s.insert(curr);

        printf("%d, ", curr);
        prev = curr;
    }
}

// Driver code
int main()
{
    int n = 17;
    recaman(n);
    return 0;
}

Output:

0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9, 24, 8, 

Time Complexity : O(n)
Auxiliary Space : O(n)

This article is contributed by Kishlay Verma. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

GATE CS Corner    Company Wise Coding Practice

Recommended Posts:



2.2 Average Difficulty : 2.2/5.0
Based on 11 vote(s)










Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.