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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 ``#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

 ``
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 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 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;``    ``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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