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Nth term of given recurrence relation having each term equal to the product of previous K terms
  • Last Updated : 24 Mar, 2021

Given two positive integers N and K and an array F[] consisting of K positive integers. The Nth term of the recurrence relation is given by: 

FN = FN – 1 * FN – 2 * FN – 3 *…….* FN – K

 The task is to find the Nth term of the given recurrence relation. As the Nth term can be very large, print the Nth term modulo 109 + 7.

Examples: 

Input: N = 5, K = 2, F = {1, 2} 
Output: 32 
Explanation: 
The sequence for above input is 1, 2, 2, 4, 8, 32, 256, ……. 
Each term is the product of its two previous terms. 
Therefore the Nth term is 32. 



Input: N = 5, K = 3, F = {1, 2, 3} 
Output: 648 
Explanation: 
The sequence for above input is: 1, 2, 3, 6, 36, 648, 139968, ……. 
Each term is the product of its three previous terms. 
Therefore the Nth term is 648. 

Naive Approach: The idea is to generate all the N terms of the given sequence using the recurrence relation and print the Nth term obtained as the required answer.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
#define int long long int
using namespace std;
 
int mod = 1e9 + 7;
 
// Function to find the nth term
void NthTerm(int F[], int K, int N)
{
    // Stores the terms of
    // reccurrence relation
    int ans[N + 1] = { 0 };
 
    // Initialize first K terms
    for (int i = 0; i < K; i++)
        ans[i] = F[i];
 
    // Find all terms from Kth term
    // to the Nth term
    for (int i = K; i <= N; i++) {
 
        ans[i] = 1;
 
        for (int j = i - K; j < i; j++) {
 
            // Current term is product of
            // previous K terms
            ans[i] *= ans[j];
            ans[i] %= mod;
        }
    }
 
    // Print the Nth term
    cout << ans[N] << endl;
}
 
// Driver Code
int32_t main()
{
    // Given N, K and F[]
    int F[] = { 1, 2 };
    int K = 2;
    int N = 5;
 
    // Function Call
    NthTerm(F, K, N);
 
    return 0;
}

Java




// Java program for the above approach
class GFG{
 
static int mod = (int)(1e9 + 7);
 
// Function to find the nth term
static void NthTerm(int F[], int K, int N)
{
     
    // Stores the terms of
    // reccurrence relation
    int ans[] = new int[N + 1];
 
    // Initialize first K terms
    for(int i = 0; i < K; i++)
        ans[i] = F[i];
 
    // Find all terms from Kth term
    // to the Nth term
    for(int i = K; i <= N; i++)
    {
        ans[i] = 1;
 
        for(int j = i - K; j < i; j++)
        {
             
            // Current term is product of
            // previous K terms
            ans[i] *= ans[j];
            ans[i] %= mod;
        }
    }
 
    // Print the Nth term
    System.out.print(ans[N] + "\n");
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given N, K and F[]
    int F[] = { 1, 2 };
    int K = 2;
    int N = 5;
 
    // Function call
    NthTerm(F, K, N);
}
}
 
// This code is contributed by Amit Katiyar

Python3




# Python3 program for the above approach
mod = 1e9 + 7
 
# Function to find the nth term
def NthTerm(F, K, N):
     
    # Stores the terms of
    # reccurrence relation
    ans = [0] * (N + 1)
 
    # Initialize first K terms
    for i in range(K):
        ans[i] = F[i]
 
    # Find all terms from Kth term
    # to the Nth term
    for i in range(K, N + 1):
        ans[i] = 1
 
        for j in range(i - K, i):
 
            # Current term is product of
            # previous K terms
            ans[i] *= ans[j]
            ans[i] %= mod
 
    # Print the Nth term
    print(ans[N])
 
# Driver Code
if __name__ == '__main__':
     
    # Given N, K and F[]
    F = [1, 2]
    K = 2
    N = 5
 
    # Function Call
    NthTerm(F, K, N)
 
# This code is contributed by mohit kumar 29

C#




// C# program for
// the above approach
using System;
class GFG{
 
static int mod = (int)(1e9 + 7);
 
// Function to find the
// nth term
static void NthTerm(int []F,
                    int K, int N)
{
  // Stores the terms of
  // reccurrence relation
  int []ans = new int[N + 1];
 
  // Initialize first K terms
  for(int i = 0; i < K; i++)
    ans[i] = F[i];
 
  // Find all terms from Kth
  // term to the Nth term
  for(int i = K; i <= N; i++)
  {
    ans[i] = 1;
 
    for(int j = i - K; j < i; j++)
    {
      // Current term is product of
      // previous K terms
      ans[i] *= ans[j];
      ans[i] %= mod;
    }
  }
 
  // Print the Nth term
  Console.Write(ans[N] + "\n");
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given N, K and F[]
  int []F = {1, 2};
  int K = 2;
  int N = 5;
 
  // Function call
  NthTerm(F, K, N);
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
// JavaScript program for the above approach
 
let mod = 1e9 + 7;
 
// Function to find the nth term
function NthTerm(F, K, N)
{
    // Stores the terms of
    // reccurrence relation
    let ans = new Uint8Array(N + 1);
 
    // Initialize first K terms
    for (let i = 0; i < K; i++)
        ans[i] = F[i];
 
    // Find all terms from Kth term
    // to the Nth term
    for (let i = K; i <= N; i++) {
 
        ans[i] = 1;
 
        for (let j = i - K; j < i; j++) {
 
            // Current term is product of
            // previous K terms
            ans[i] *= ans[j];
            ans[i] %= mod;
        }
    }
 
    // Print the Nth term
    document.write(ans[N] + "<br>");
}
 
// Driver Code
 
    // Given N, K and F[]
    let F = [ 1, 2 ];
    let K = 2;
    let N = 5;
 
    // Function Call
    NthTerm(F, K, N);
 
 
// This code is contributed by Surbhi Tyagi.
 
</script>
Output
32

Time Complexity: O(N*K) 
Auxiliary Space: O(N)

Efficient Approach: The idea is to use the deque Data Structure to find the next term using the last K terms. Below are the steps:

  • Initialize an empty deque say dq.
  • Calculate the product of first K terms and since it is equal to the (K + 1)th term of the recurrence relation, insert it at the end of dq.
  • Iterate over the range [K + 2, N] and follow the steps below: 
    • Let the last element of deque be L and the front element of deque be F.
    • Now, calculate the ith term using the formula for ith term = (L * L) / F.
    • Since L is the product of elements from (i – 1 – K) to (i – 2). Therefore, to find the ith term, choose the product of elements from (i – K) to (i – 1), and multiply (i – 1)th term (i.e., L) to the product of elements from (i – 1 – K) to (i – 2), to get the product of elements.
    • Now, divide this product (L * L) by (i – 1 – K)th term which is F in this case.
    • Now, insert the ith term to the back of the deque.
    • Pop one element from the front of the deque.
  • After completing the above steps, print the last element of the deque.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
#define int long long int
using namespace std;
int mod = 1e9 + 7;
 
// Function to calculate (x ^ y) % p
// fast exponentiation ( O(log y)
int power(int x, int y, int p)
{
    // Store the result
    int res = 1;
    x = x % p;
 
    // Till y is greater than 0
    while (y > 0) {
 
        // If y is odd
        if (y & 1)
            res = (res * x) % p;
 
        // Right shift by 1
        y = y >> 1;
        x = (x * x) % p;
    }
 
    // Print the resultant value
    return res;
}
 
// Function to find mod inverse
int modInverse(int n, int p)
{
    // Using Fermat Little Theorm
    return power(n, p - 2, p);
}
 
// Function to find Nth term of the
// given recurrence relation
void NthTerm(int F[], int K, int N)
{
    // Doubly ended queue
    deque<int> q;
 
    // Stores the product of 1st K terms
    int product = 1;
 
    for (int i = 0; i < K; i++) {
 
        product *= F[i];
        product %= mod;
        q.push_back(F[i]);
    }
 
    // Push (K + 1)th term to Dequeue
    q.push_back(product);
 
    for (int i = K + 1; i <= N; i++) {
 
        // First and the last element
        // of the dequeue
        int f = *q.begin();
        int e = *q.rbegin();
 
        // Calculating the ith term
        int next_term
            = ((e % mod * e % mod) % mod
               * (modInverse(f, mod)))
              % mod;
        // Add current term to end
        // of Dequeue
        q.push_back(next_term);
 
        // Remove the first number
        // from dequeue
        q.pop_front();
    }
 
    // Print the Nth term
    cout << *q.rbegin() << endl;
}
 
// Driver Code
int32_t main()
{
    // Given N, K and F[]
    int F[] = { 1, 2 };
    int K = 2;
    int N = 5;
 
    // Function Call
    NthTerm(F, K, N);
    return 0;
}

Java




// Java program for the
// above approach
import java.util.*;
class GFG{
   
static long mod = 1000000007;
 
// Function to calculate
// (x ^ y) % p fast
// exponentiation ( O(log y)
static long power(long x,
                  long y, long p)
{
  // Store the result
  long res = 1;
  x = x % p;
 
  // Till y is
  // greater than 0
  while (y > 0)
  {
    // If y is odd
    if (y % 2 == 1)
      res = (res * x) % p;
 
    // Right shift by 1
    y = y >> 1;
    x = (x * x) % p;
  }
 
  // Print the resultant value
  return res;
}
     
// Function to find mod
// inverse
static long modInverse(long n,
                       long p)
{
  // Using Fermat Little Theorm
  return power(n, p - 2, p);
}
 
// Function to find Nth term
// of the given recurrence
// relation
static void NthTerm(long F[],
                    long K, long N)
{
  // Doubly ended queue
  Vector<Long> q = new Vector<>();
 
  // Stores the product of 1st K terms
  long product = 1;
 
  for (int i = 0; i < K; i++)
  {
    product *= F[i];
    product %= mod;
    q.add(F[i]);
  }
 
  // Push (K + 1)th
  // term to Dequeue
  q.add(product);
 
  for (long i = K + 1; i <= N; i++)
  {
    // First and the last element
    // of the dequeue
    long f = q.get(0);
    long e = q.get(q.size() - 1);
 
    // Calculating the ith term
    long next_term = ((e % mod * e % mod) % mod *
                      (modInverse(f, mod))) % mod;
 
    // Add current term to end
    // of Dequeue
    q.add(next_term);
 
    // Remove the first number
    // from dequeue
    q.remove(0);
  }
 
  // Print the Nth term
  System.out.print(q.get(q.size() - 1) + "\n");
}
 
// Driver Code
public static void main(String[] args)
{
  // Given N, K and F[]
  long F[] = {1, 2};
  long K = 2;
  long N = 5;
 
  // Function Call
  NthTerm(F, K, N);
}
}
 
// This code is contributed by shikhasingrajput

Python3




# Python3 program for the
# above approach
mod = 1000000007
 
# Function to calculate
# (x ^ y) % p fast
# exponentiation ( O(log y)
def power(x, y, p):
   
    # Store the result
    res = 1
    x = x % p
 
    # Till y is
    # greater than 0
    while (y > 0):
       
        # If y is odd
        if (y % 2 == 1):
            res = (res * x) % p
 
        # Right shift by 1
        y = y >> 1
        x = (x * x) % p
 
    # Print the resultant value
    return res
 
# Function to find mod
# inverse
def modInverse(n, p):
   
    # Using Fermat Little Theorm
    return power(n, p - 2, p);
 
 
# Function to find Nth term
# of the given recurrence
# relation
def NthTerm(F, K, N):
   
    # Doubly ended queue
    q = []
 
    # Stores the product of
    # 1st K terms
    product = 1
 
    for i in range(K):
        product *= F[i]
        product %= mod
        q.append(F[i])
 
    # Push (K + 1)th
    # term to Dequeue
    q.append(product)
 
    for i in range(K + 1, N + 1):
       
        # First and the last element
        # of the dequeue
        f = q[0]
        e = q[len(q) - 1]
 
        # Calculating the ith term
        next_term = ((e % mod * e % mod) %
             mod * (modInverse(f, mod))) % mod
 
        # Add current term to end
        # of Dequeue
        q.append(next_term)
 
        # Remove the first number
        # from dequeue
        q.remove(q[0])
 
    # Print the Nth term
    print(q[len(q) - 1], end = "")
 
# Driver Code
if __name__ == '__main__':
   
    # Given N, K and F
    F = [1, 2]
    K = 2
    N = 5
 
    # Function Call
    NthTerm(F, K, N)
 
# This code is contributed by Princi Singh

C#




// C# program for the
// above approach
using System;
using System.Collections.Generic;
 
class GFG{
   
static long mod = 1000000007;
   
// Function to calculate
// (x ^ y) % p fast
// exponentiation ( O(log y)
static long power(long x, long y,
                  long p)
{
   
  // Store the result
  long res = 1;
  x = x % p;
 
  // Till y is
  // greater than 0
  while (y > 0)
  {
     
    // If y is odd
    if (y % 2 == 1)
      res = (res * x) % p;
 
    // Right shift by 1
    y = y >> 1;
    x = (x * x) % p;
  }
 
  // Print the resultant value
  return res;
}
     
// Function to find mod
// inverse
static long modInverse(long n,
                       long p)
{
   
  // Using Fermat Little Theorm
  return power(n, p - 2, p);
}
 
// Function to find Nth term
// of the given recurrence
// relation
static void NthTerm(long []F,
                    long K, long N)
{
   
  // Doubly ended queue
  List<long> q = new List<long>();
 
  // Stores the product of 1st K terms
  long product = 1;
 
  for(int i = 0; i < K; i++)
  {
    product *= F[i];
    product %= mod;
    q.Add(F[i]);
  }
 
  // Push (K + 1)th
  // term to Dequeue
  q.Add(product);
 
  for(long i = K + 1; i <= N; i++)
  {
     
    // First and the last element
    // of the dequeue
    long f = q[0];
    long e = q[q.Count - 1];
 
    // Calculating the ith term
    long next_term = ((e % mod * e % mod) % mod *
                    (modInverse(f, mod))) % mod;
     
    // Add current term to end
    // of Dequeue
    q.Add(next_term);
 
    // Remove the first number
    // from dequeue
    q.RemoveAt(0);
  }
   
  // Print the Nth term
  Console.Write(q[q.Count - 1] + "\n");
}
 
// Driver Code
public static void Main(String[] args)
{
   
  // Given N, K and F[]
  long []F = {1, 2};
  long K = 2;
  long N = 5;
 
  // Function Call
  NthTerm(F, K, N);
}
}
 
// This code is contributed by Rajput-Ji
Output
32

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

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