Given a positive integer N, the task is to print Nth term of the Van Eck’s sequence.
In mathematics, Van Eck’s sequence is an integer sequence which is defined recursively as follows:
- Let the first term be 0 i.e a0 = 0.
- Then for n >= 0, if there exists an m < n such that
am = an
- take the largest such m and set an+1 = n − m;
- Otherwise an+1 = 0.
- Start with a(1)=0.
First few terms of Van Eck’s Sequence are as follows:
0, 0, 1, 0, 2, 0, 2, 2, 1, 6, 0, 5, 0, 2, 6, 5, 4, 0, 5 …
Example:
Input: N = 5
Output: 2
Input: N = 10
Output: 6
Approach:
As described above we can follow the below steps to generate Van Eck’s sequence:
- Set the first term of the sequence as 0.
-
Then Repeatedly apply:
- If the last term has not occurred yet and is new to the sequence so far then, set the next term as zero.
- Otherwise, the next term is how far back this last term has occurred previously.
- Once the sequence is generated we can get our nth term easily.
Below is the implementation of above approach:
// C++ program to print Nth // term of Van Eck's sequence #include <bits/stdc++.h> using namespace std;
#define MAX 1000 int sequence[MAX];
// Utility function to compute // Van Eck's sequence void vanEckSequence()
{ // Initialize sequence array
for ( int i = 0; i < MAX; i++) {
sequence[i] = 0;
}
// Loop to generate sequence
for ( int i = 0; i < MAX; i++) {
// Check if sequence[i] has occurred
// previously or is new to sequence
for ( int j = i - 1; j >= 0; j--) {
if (sequence[j] == sequence[i]) {
// If occurrence found
// then the next term will be
// how far back this last term
// occurred previously
sequence[i + 1] = i - j;
break ;
}
}
}
} // Utility function to return // Nth term of sequence int getNthTerm( int n)
{ return sequence[n];
} // Driver code int main()
{ // Pre-compute Van Eck's sequence
vanEckSequence();
int n = 6;
// Print nth term of the sequence
cout << getNthTerm(n) << endl;
n = 100;
// Print nth term of the sequence
cout << getNthTerm(n) << endl;
return 0;
} |
// Java program to print Nth // term of Van Eck's sequence class GFG {
static int MAX = 1000 ;
// Array to store terms of sequence
static int sequence[] = new int [MAX];
// Utility function to compute
// Van Eck's sequence
static void vanEckSequence()
{
// Initialize sequence array
for ( int i = 0 ; i < MAX - 1 ; i++) {
sequence[i] = 0 ;
}
// Loop to generate sequence
for ( int i = 0 ; i < MAX - 1 ; i++) {
// Check if sequence[i] has occurred
// previously or is new to sequence
for ( int j = i - 1 ; j >= 0 ; j--) {
if (sequence[j] == sequence[i]) {
// If occurrence found
// then the next term will be
// how far back this last term
// occurred previously
sequence[i + 1 ] = i - j;
break ;
}
}
}
}
// Utility function to return
// Nth term of sequence
static int getNthTerm( int n)
{
return sequence[n];
}
// Driver code
public static void main(String[] args)
{
// Pre-compute Van Eck's sequence
vanEckSequence();
int n = 6 ;
// Print nth term of the sequence
System.out.println(getNthTerm(n));
n = 100 ;
// Print nth term of the sequence
System.out.println(getNthTerm(n));
}
} |
# Python3 program to print Nth # term of Van Eck's sequence MAX = 1000
sequence = [ 0 ] * ( MAX + 1 );
# Utility function to compute # Van Eck's sequence def vanEckSequence() :
# Initialize sequence array
for i in range ( MAX ) :
sequence[i] = 0 ;
# Loop to generate sequence
for i in range ( MAX ) :
# Check if sequence[i] has occurred
# previously or is new to sequence
for j in range (i - 1 , - 1 , - 1 ) :
if (sequence[j] = = sequence[i]) :
# If occurrence found
# then the next term will be
# how far back this last term
# occurred previously
sequence[i + 1 ] = i - j;
break ;
# Utility function to return # Nth term of sequence def getNthTerm(n) :
return sequence[n];
# Driver code if __name__ = = "__main__" :
# Pre-compute Van Eck's sequence
vanEckSequence();
n = 6 ;
# Print nth term of the sequence
print (getNthTerm(n));
n = 100 ;
# Print nth term of the sequence
print (getNthTerm(n));
# This code is contributed by kanugargng |
// C# program to print Nth // term of Van Eck's sequence using System;
class GFG {
static int MAX = 1000;
// Array to store terms of sequence
static int [] sequence = new int [MAX];
// Utility function to compute
// Van Eck's sequence
static void vanEckSequence()
{
// Initialize sequence array
for ( int i = 0; i < MAX - 1; i++) {
sequence[i] = 0;
}
// Loop to generate sequence
for ( int i = 0; i < MAX - 1; i++) {
// Check if sequence[i] has occurred
// previously or is new to sequence
for ( int j = i - 1; j >= 0; j--) {
if (sequence[j] == sequence[i]) {
// If occurrence found
// then the next term will be
// how far back this last term
// occurred previously
sequence[i + 1] = i - j;
break ;
}
}
}
}
// Utility function to return
// Nth term of sequence
static int getNthTerm( int n)
{
return sequence[n];
}
// Driver code
public static void Main()
{
// Pre-compute Van Eck's sequence
vanEckSequence();
int n = 6;
// Print nth term of the sequence
Console.WriteLine(getNthTerm(n));
n = 100;
// Print nth term of the sequence
Console.WriteLine(getNthTerm(n));
}
} |
<script> function getVanEckSequence(n) {
let result = [0];
let dir = {};
for ( let i = 0 ; i < n ; i++) {
const currentData = result[i];
const currentPosition = i + 1;
// If number is already there, then insert the difference of positions.
if (dir[currentData]) {
result.push(currentPosition - dir[currentData]);
} else {
result.push(0);
}
// Update with the latest position
dir[currentData] = currentPosition;
}
return result;
} console.log(getVanEckSequence(100)); // This code is contributed by Devesh </script> |
2 23
Time Complexity: O(MAX2)
Auxiliary Space: O(MAX)
Method 2: Using lambda functions
There is no need to store the entire sequence to get the value of nth term. We can recursively build the series upto nth term and find only the value of nth term using the lambda expression.
Below is the implementation of the above approach using lambda function:
#include <iostream> #include <vector> using namespace std;
int vanEck( int n) {
vector< int > s(n + 1, 0);
int l = 0;
for ( int i = 1; i <= n; i++) {
if (l == 0 || s[l] != 0) {
s[i] = 0;
l = (s[l] == 0) ? (i - 1) : (i - s[l]);
} else {
s[i] = 0;
l = i - 1;
}
}
return l;
} int main() {
int n = 6;
cout << vanEck(n) << endl;
n = 100;
cout << vanEck(n) << endl;
return 0;
} |
public class VanEckSequence {
public static int vanEck( int n) {
int [] s = new int [n + 1 ];
int l = 0 ;
for ( int i = 1 ; i <= n; i++) {
if (l == 0 || s[l] != 0 ) {
s[i] = 0 ;
l = s[l] == 0 ? i - 1 : i - s[l];
} else {
s[i] = 0 ;
l = i - 1 ;
}
}
return l;
}
public static void main(String[] args) {
int n = 6 ;
System.out.println(vanEck(n));
n = 100 ;
System.out.println(vanEck(n));
}
} |
# Python3 program to find # Nth term of Van Eck's sequence # using the lambda expression # Lambda function f = lambda n, l = 0 , * s: f(n - 1 , l in s and ~s.index(l), l, * s) \
if n else - l
# The above lambda function recursively # build the sequence and store it in tuple s # The expression l in s and ~s.index(l) # returns False if l is not present in tuple s # otherwise, returns the negation of the value of # the index of l in tuple s # and is appended to tuple s # Thus, tuple s store negation of all the elements # of the sequence in reverse order # At the end, when n reaches 0, function converts # the nth term back to its actual value # and returns it. # Driver code n = 6
# Get Nth term of the sequence print (f(n))
n = 100
# Get Nth term of the sequence print (f(n))
|
using System;
public class VanEckSequence
{ public static int VanEck( int n)
{
int [] s = new int [n + 1];
int l = 0;
for ( int i = 1; i <= n; i++)
{
if (l == 0 || s[l] != 0)
{
s[i] = 0;
l = s[l] == 0 ? i - 1 : i - s[l];
}
else
{
s[i] = 0;
l = i - 1;
}
}
return l;
}
public static void Main( string [] args)
{
int n = 6;
Console.WriteLine(VanEck(n));
n = 100;
Console.WriteLine(VanEck(n));
}
} |
// Lambda function const f = (n, l = 0, ...s) => // Recursively build the sequence and store it in array s
n ? f(n-1, s.includes(l) ? ~s.indexOf(l) : 0, l, ...s) :
// Convert the nth term back to its actual value and return it
-l;
// The expression s.includes(l) ? ~s.indexOf(l) : 0 // returns 0 if l is not present in array s // otherwise, returns the negation of the value of // the index of l in array s // and is appended to array s // Thus, array s store negation of all the elements // of the sequence in reverse order // Driver code let x= 6; // Get Nth term of the sequence console.log(f(x)); x = 100; // Get Nth term of the sequence console.log(f(x)); |
2 23
Time Complexity: O(n)
Auxiliary Space: O(n)
Reference: https://codegolf.stackexchange.com/questions/186654/nth-term-of-van-eck-sequence