Prerequisite: Van Eck’s sequence

Given a positive integer N, the task is to count the occurrences of N^th term in 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

1. Naive Approach:
   • Generate Van Eck’s sequence upto N^th term
   • Iterate through the generated sequence and count the occurrence of N^th term.

   To serve multiple queries we can pre-compute the Van Eck’s sequence.

   Below is the implementation of 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 100000 int sequence[MAX + 1];    // 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 occured         // 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                 // occured previously                 sequence[i + 1] = i - j;                 break;             }         }     } }    // Utility function to count // the occurrence of nth term // in first n terms of the sequence 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 int 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 sequence    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 occured             // 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                     // occured 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 sequence     MAX = 10000 sequence = [0]*(MAX + 1);     # Utility function to compute  # Van Eck's sequence  def vanEckSequence() :            # Loop to generate sequence      for i in range(MAX) :            # Check if sequence[i] has occured          # 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                  # occured previously                  sequence[i + 1] = i - j;                  break;     # Utility function to count  # the occurrence of nth term  # in first n terms of the sequence  def 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 code  if __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 sequence    using 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 occured             // 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                     // occured 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));     } }

   Output:

   1
   5
   

2. 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 i^th 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 N^th term of the sequence as S_N
     • If S_N+1 is non-zero then increment count
       and do the same for (N- S_N+1)^th term
     • And if S_N+1 is zero then stop.

   Below is the implementation of above approach:
   

   CPP

   
   

   
   
   

   // 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 100000 int sequence[MAX + 1];    // 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 occured         // 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                 // occured previously                 sequence[i + 1] = i - j;                 break;             }         }     } }    // Utility function to count // the occurrence of nth term // in first n terms of the sequence 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 int 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 sequence    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 occured             // 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                     // occured 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 sequence  MAX = 10000 sequence = [0] * (MAX + 1);     # Utility function to compute  # Van Eck's sequence  def vanEckSequence() :        # Loop to generate sequence      for i in range(MAX) :             # Check if sequence[i] has occured          # 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                  # occured previously                  sequence[i + 1] = i - j;                  break;     # Utility function to count  # the occurrence of nth term  # in first n terms of the sequence  def 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 code  if __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 sequence    using 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 occured             // 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                     // occured 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));     } }

   Output:

   1
   5
   

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