Prerequisite: Van Eck’s sequence Given a positive integer N, the task is to count the occurrences of Nth term in the first N terms of Van Eck’s sequence. Examples:
Input: N = 5 Output: 1 Explanation: First 5 terms of Van Eck’s Sequence 0, 0, 1, 0, 2 Occurrence of 5th term i.e 2 = 1 Input: 11 Output: 5 Explanation: First 11 terms of Van Eck’s Sequence 0, 0, 1, 0, 2, 0, 2, 2, 1, 6, 0, Occurrence of 11th term i.e 0 is 5
- Naive Approach:
- Generate Van Eck’s sequence upto Nth term
- Iterate through the generated sequence and count the occurrence of Nth term.
Below is the implementation of the above approach:
C++
// C++ program to count the occurrence// of nth term in first n terms// of Van Eck's sequence#include <bits/stdc++.h>using namespace std;#define MAX 100000int sequence[MAX + 1];// Utility function to compute// Van Eck's sequencevoid 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 count// the occurrence of nth term// in first n terms of the sequenceint getCount(int n){ // Get nth term of the sequence int nthTerm = sequence[n - 1]; int count = 0; // Count the occurrence of nth term // in first n terms of the sequence for (int i = 0; i < n; i++) { if (sequence[i] == nthTerm) count++; } // Return count return count;}// Driver codeint main(){ // Pre-compute Van Eck's sequence vanEckSequence(); int n = 5; // Print count of the occurrence of nth term // in first n terms of the sequence cout << getCount(n) << endl; n = 11; // Print count of the occurrence of nth term // in first n terms of the sequence cout << getCount(n) << endl; return 0;} |
Java
// Java program to count the occurrence// of nth term in first n terms// of Van Eck's sequenceclass GFG { static int MAX = 100000; static int sequence[] = new int[MAX + 1]; // Utility function to compute // Van Eck's sequence static 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 count // the occurrence of nth term // in first n terms of the sequence static int getCount(int n) { // Get nth term of the sequence int nthTerm = sequence[n - 1]; int count = 0; // Count the occurrence of nth term // in first n terms of the sequence for (int i = 0; i < n; i++) { if (sequence[i] == nthTerm) count++; } // Return count return count; } // Driver code public static void main(String[] args) { // Pre-compute Van Eck's sequence vanEckSequence(); int n = 5; // Print count of the occurrence of nth term // in first n terms of the sequence System.out.println(getCount(n)); n = 11; // Print count of the occurrence of nth term // in first n terms of the sequence System.out.println(getCount(n)); }} |
Python3
# Python3 program to count the occurrence# of nth term in first n terms# of Van Eck's sequenceMAX = 10000sequence = [0]*(MAX + 1);# Utility function to compute# Van Eck's sequencedef vanEckSequence() : # 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 count# the occurrence of nth term# in first n terms of the sequencedef getCount(n) : # Get nth term of the sequence nthTerm = sequence[n - 1]; count = 0; # Count the occurrence of nth term # in first n terms of the sequence for i in range(n) : if (sequence[i] == nthTerm) : count += 1; # Return count return count;# Driver codeif __name__ == "__main__" : # Pre-compute Van Eck's sequence vanEckSequence(); n = 5; # Print count of the occurrence of nth term # in first n terms of the sequence print(getCount(n)); n = 11; # Print count of the occurrence of nth term # in first n terms of the sequence print(getCount(n));# This code is contributed by AnkitRai01 |
C#
// C# program to count the occurrence// of nth term in first n terms// of Van Eck's sequenceusing System;class GFG { static int MAX = 100000; static int[] sequence = new int[MAX + 1]; // Utility function to compute // Van Eck's sequence static 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 count // the occurrence of nth term // in first n terms of the sequence static int getCount(int n) { // Get nth term of the sequence int nthTerm = sequence[n - 1]; int count = 0; // Count the occurrence of nth term // in first n terms of the sequence for (int i = 0; i < n; i++) { if (sequence[i] == nthTerm) count++; } // Return count return count; } // Driver code public static void Main() { // Pre-compute Van Eck's sequence vanEckSequence(); int n = 5; // Print count of the occurrence of nth term // in first n terms of the sequence Console.WriteLine(getCount(n)); n = 11; // Print count of the occurrence of nth term // in first n terms of the sequence Console.WriteLine(getCount(n)); }} |
Javascript
// Define MAXconst MAX = 10000;// Create sequence arraylet sequence = new Array(MAX + 1).fill(0);//Javascript Equivalent// Utility function to compute Van Eck's sequencefunction vanEckSequence() { // Loop to generate sequence for (let i = 0; i <= MAX; i++) { // Check if sequence[i] has occurred previously or is new to sequence for (let 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 count the occurrence // of nth term in first n terms of the sequencefunction getCount(n) { // Get nth term of the sequence let nthTerm = sequence[n - 1]; let count = 0; // Count the occurrence of nth term // in first n terms of the sequence for (let i = 0; i < n; i++) { if (sequence[i] == nthTerm) { count++; } } // Return count return count;}// Driver code // Pre-compute Van Eck's sequence vanEckSequence(); let n = 5; // Print count of the occurrence of nth // term in first n terms of the sequence console.log(getCount(n)); n = 11; // Print count of the occurrence of nth // term in first n terms of the sequence console.log(getCount(n)); |
Output
1 5
- Efficient Approach:
- For a given term in Van Eck’s sequence, its next term indicates the distance between last occurrence of the given term.
- So, for ith term, its previous occurrence will be at i – value (i + 1)th term. For example:
- Also, if the next term in the sequence is 0 then this means that the term has not occurred before. For example:
- Algorithm:
- Let us consider Nth term of the sequence as SN
- If SN+1 is non-zero then increment the count and do the same for (N- SN+1)th term
- And if SN+1 is zero then stop.
Below is the implementation of the above approach:
C++
// C++ program to count the occurrence// of nth term in first n terms// of Van Eck's sequence#include <bits/stdc++.h>using namespace std;#define MAX 100000int sequence[MAX + 1];// Utility function to compute// Van Eck's sequencevoid 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 count// the occurrence of nth term// in first n terms of the sequenceint getCount(int n){ // Initialize count as 1 int count = 1; int i = n - 1; while (sequence[i + 1] != 0) { // Increment count if (i+1)th term // is non-zero count++; // Previous occurrence of sequence[i] // will be it (i - sequence[i+1])th position i = i - sequence[i + 1]; } // Return the count of occurrence return count;}// Driver codeint main(){ // Pre-compute Van Eck's sequence vanEckSequence(); int n = 5; // Print count of the occurrence of nth term // in first n terms of the sequence cout << getCount(n) << endl; n = 11; // Print count of the occurrence of nth term // in first n terms of the sequence cout << getCount(n) << endl; return 0;} |
Java
// Java program to count the occurrence// of nth term in first n terms// of Van Eck's sequenceclass GFG { static int MAX = 100000; static int sequence[] = new int[MAX + 1]; // Utility function to compute // Van Eck's sequence static 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 count // the occurrence of nth term // in first n terms of the sequence static int getCount(int n) { // Initialize count as 1 int count = 1; int i = n - 1; while (sequence[i + 1] != 0) { // Increment count if (i+1)th term // is non-zero count++; // Previous occurrence of sequence[i] // will be it (i - sequence[i+1])th position i = i - sequence[i + 1]; } // Return the count of occurrence return count; } // Driver code public static void main(String[] args) { // Pre-compute Van Eck's sequence vanEckSequence(); int n = 5; // Print count of the occurrence of nth term // in first n terms of the sequence System.out.println(getCount(n)); n = 11; // Print count of the occurrence of nth term // in first n terms of the sequence System.out.println(getCount(n)); }} |
Python3
# Python3 program to count the occurrence# of nth term in first n terms# of Van Eck's sequenceMAX = 10000sequence = [0] * (MAX + 1);# Utility function to compute# Van Eck's sequencedef vanEckSequence() : # 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 count# the occurrence of nth term# in first n terms of the sequencedef getCount(n) : # Initialize count as 1 count = 1; i = n - 1; while (sequence[i + 1] != 0) : # Increment count if (i+1)th term # is non-zero count += 1; # Previous occurrence of sequence[i] # will be it (i - sequence[i+1])th position i = i - sequence[i + 1]; # Return the count of occurrence return count;# Driver codeif __name__ == "__main__" : # Pre-compute Van Eck's sequence vanEckSequence(); n = 5; # Print count of the occurrence of nth term # in first n terms of the sequence print(getCount(n)); n = 11; # Print count of the occurrence of nth term # in first n terms of the sequence print(getCount(n)) ;# This code is contributed by AnkitRai01 |
C#
// C# program to count the occurrence// of nth term in first n terms// of Van Eck's sequenceusing System;class GFG { static int MAX = 100000; static int[] sequence = new int[MAX + 1]; // Utility function to compute // Van Eck's sequence static 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 count // the occurrence of nth term // in first n terms of the sequence static int getCount(int n) { // Initialize count as 1 int count = 1; int i = n - 1; while (sequence[i + 1] != 0) { // Increment count if (i+1)th term // is non-zero count++; // Previous occurrence of sequence[i] // will be it (i - sequence[i+1])th position i = i - sequence[i + 1]; } // Return the count of occurrence return count; } // Driver code public static void Main(string[] args) { // Pre-compute Van Eck's sequence vanEckSequence(); int n = 5; // Print count of the occurrence of nth term // in first n terms of the sequence Console.WriteLine(getCount(n)); n = 11; // Print count of the occurrence of nth term // in first n terms of the sequence Console.WriteLine(getCount(n)); }} |
Javascript
// JS program to count the occurrence// of nth term in first n terms// of Van Eck's sequencelet MAX = 100000;let sequence = new Array(MAX + 1);// Utility function to compute// Van Eck's sequencefunction vanEckSequence(){ // Initialize sequence array for (var i = 0; i < MAX; i++) { sequence[i] = 0; } // Loop to generate sequence for (var i = 0; i < MAX; i++) { // Check if sequence[i] has occurred // previously or is new to sequence for (var 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 count// the occurrence of nth term// in first n terms of the sequencefunction getCount(n){ // Initialize count as 1 let count = 1; let i = n - 1; while (sequence[i + 1] != 0) { // Increment count if (i+1)th term // is non-zero count++; // Previous occurrence of sequence[i] // will be it (i - sequence[i+1])th position i = i - sequence[i + 1]; } // Return the count of occurrence return count;}// Driver code// Pre-compute Van Eck's sequencevanEckSequence();let n = 5;// Print count of the occurrence of nth term// in first n terms of the sequenceconsole.log(getCount(n));n = 11;// Print count of the occurrence of nth term// in first n terms of the sequenceconsole.log(getCount(n));// This code is contributed by phasing17 |
Output
1 5
Time Complexity: O(MAX*MAX)
Auxiliary Space: O(MAX)
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