# Nth term of given recurrence relation having each term equal to the product of previous K terms

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.

Algorithm:

1. Initialize an array ans[ ] of size N+1 to store the terms of the recurrence relation.
2. Initialize the first K terms of ans[ ] to the given initial values F[ ].
3. Iterate from K to N:
a. Set ans[i] = 1
b. Iterate from i-K to i-1:
i. Multiply ans[i] with ans[j] to find the current term using the product of previous K terms.
ii. Take the modulus of the product with a given constant mod.
c. Store the current term ans[i] in the ith index of the ans[ ] array.
4. Return the Nth term ans[N] as the answer.

Pseudocode:

function NthTerm(F, K, N):

ans = array of size N+1

for i from 0 to K-1:
ans[i] = F[i]

for i from K to N:
ans[i] = 1

for j from i-K to i-1:
ans[i] *= ans[j]
ans[i] %= mod

return ans[N]

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach   #include #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     // recurrence 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     // recurrence 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     # recurrence 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   // recurrence 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



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 #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 Theorem     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 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 Theorem   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 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 Theorem     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 Theorem   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 q = new List();     // 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

## Javascript



Output

32

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

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Previous
Next