Related Articles

# Length of longest common prime subsequence from two given arrays

• Difficulty Level : Hard
• Last Updated : 11 Feb, 2021

Given two arrays arr1[] and arr2[] of length N and M respectively, the task is to find the length of the longest common prime subsequence that can be obtained from the two given arrays.

Examples:

Input: arr1[] = {1, 2, 3, 4, 5, 6, 7, 8, 9}, arr2[] = {2, 5, 6, 3, 7, 9, 8}
Output:
Explanation:
The longest common prime subsequence present in both the arrays is {2, 3, 5, 7}.

Input: arr1[] = {1, 3, 5, 7, 9}, arr2[] = {2, 4, 6, 8, 10}
Output:
Explanation:
In the above arrays, the prime subsequence of arr1[] is {1, 3, 5, 7} and arr2[] is {2}. Therefore, there is no common prime numbers which are present in both the arrays. Hence, the result is 0.

Naive Approach: The simplest idea is to consider all subsequences of arr1[] and check if all numbers in this subsequence are prime and appear in arr2[]. Then find the longest length of these subsequences.

Time Complexity: O(M * 2N
Auxiliary Space: O(N)

Efficient Approach: The idea is to find all the prime numbers from both the arrays and then find the longest common prime subsequence from them using Dynamic Programming. Follow the steps below to solve the problem:

1. Find all the prime numbers between the minimum element of the array and maximum element of the array using Sieve Of Eratosthenes algorithm.
2. Store the sequence of primes from the arrays arr1[] and arr2[].
3. Find the LCS of the two sequences of primes.

Below is the implementation of the above approach:

## C++

 `// CPP implementation of the above approach``#include ``using` `namespace` `std;` `// Function to calculate the LCS``int` `recursion(vector<``int``> arr1,``              ``vector<``int``> arr2, ``int` `i,``              ``int` `j, map,``              ``int``> dp)``{``    ``if` `(i >= arr1.size() or j >= arr2.size())``        ``return` `0;``    ``pair<``int``, ``int``> key = { i, j };``    ``if` `(arr1[i] == arr2[j])``        ``return` `1``               ``+ recursion(arr1, arr2,``                            ``i + 1, j + 1,``                           ``dp);``    ``if` `(dp.find(key) != dp.end())``        ``return` `dp[key];` `    ``else``        ``dp[key] = max(recursion(arr1, arr2,``                                ``i + 1, j, dp),``                      ``recursion(arr1, arr2, i,``                                ``j + 1, dp));``    ``return` `dp[key];``}` `// Function to generate``// all the possible``// prime numbers``vector<``int``> primegenerator(``int` `n)``{``    ``int` `cnt = 0;``    ``vector<``int``> primes(n + 1, ``true``);``    ``int` `p = 2;``    ``while` `(p * p <= n)``    ``{``        ``for` `(``int` `i = p * p; i <= n; i += p)``            ``primes[i] = ``false``;``        ``p += 1;``    ``}``    ``return` `primes;``}` `// Function which returns the``// length of longest common``// prime subsequence``int` `longestCommonSubseq(vector<``int``> arr1,``                        ``vector<``int``> arr2)``{` `    ``// Minimum element of``    ``// both arrays``    ``int` `min1 = *min_element(arr1.begin(),``                            ``arr1.end());``    ``int` `min2 = *min_element(arr2.begin(),``                            ``arr2.end());` `    ``// Maximum element of``    ``// both arrays``    ``int` `max1 = *max_element(arr1.begin(),``                            ``arr1.end());``    ``int` `max2 = *max_element(arr2.begin(),``                            ``arr2.end());` `    ``// Generating all primes within``    ``// the max range of arr1``    ``vector<``int``> a = primegenerator(max1);` `    ``// Generating all primes within``    ``// the max range of arr2``    ``vector<``int``> b = primegenerator(max2);` `    ``vector<``int``> finala;``    ``vector<``int``> finalb;` `    ``// Store precomputed values``    ``map, ``int``> dp;` `    ``// Store all primes in arr1[]``    ``for` `(``int` `i = min1; i <= max1; i++)``    ``{``        ``if` `(find(arr1.begin(), arr1.end(), i)``            ``!= arr1.end()``            ``and a[i] == ``true``)``            ``finala.push_back(i);``    ``}` `    ``// Store all primes of arr2[]``    ``for` `(``int` `i = min2; i <= max2; i++)``    ``{``        ``if` `(find(arr2.begin(), arr2.end(), i)``            ``!= arr2.end()``            ``and b[i] == ``true``)``            ``finalb.push_back(i);``    ``}` `    ``// Calculating the LCS``    ``return` `recursion(finala, finalb, 0, 0, dp);``}` `// Driver Code``int` `main()``{``    ``vector<``int``> arr1 = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };``    ``vector<``int``> arr2 = { 2, 5, 6, 3, 7, 9, 8 };``    ` `    ``// Function Call``    ``cout << longestCommonSubseq(arr1, arr2);``}`

## Java

 `// JAVA implementation of the above approach``import` `java.util.*;``import` `java.io.*;``import` `java.math.*;``public` `class` `GFG``{` `  ``// Function to calculate the LCS``  ``static` `int` `recursion(ArrayList arr1,``                       ``ArrayList arr2, ``int` `i,``                       ``int` `j, Map,``                       ``Integer> dp)``  ``{``    ``if` `(i >= arr1.size() || j >= arr2.size())``      ``return` `0``;``    ``ArrayList key = ``new` `ArrayList<>();``    ``key.add(i);``    ``key.add(j);``    ``if` `(arr1.get(i) == arr2.get(j))``      ``return` `1` `+ recursion(arr1, arr2,``                           ``i + ``1``, j + ``1``,``                           ``dp);``    ``if` `(dp.get(key) != dp.get(dp.size()-``1``))``      ``return` `dp.get(key);` `    ``else``      ``dp.put(key,Math.max(recursion(arr1, arr2,``                                    ``i + ``1``, j, dp),``                          ``recursion(arr1, arr2, i,``                                    ``j + ``1``, dp)));``    ``return` `dp.get(key);``  ``}` `  ``// Function to generate``  ``// all the possible``  ``// prime numbers``  ``static` `ArrayList primegenerator(``int` `n)``  ``{``    ``int` `cnt = ``0``;``    ``ArrayList primes = ``new` `ArrayList<>();``    ``for``(``int` `i = ``0``; i < n + ``1``; i++)``      ``primes.add(``true``);``    ``int` `p = ``2``;``    ``while` `(p * p <= n)``    ``{``      ``for` `(``int` `i = p * p; i <= n; i += p)``        ``primes.set(i,``false``);``      ``p += ``1``;``    ``}``    ``return` `primes;``  ``}` `  ``// Function that returns the Minimum element of an ArrayList``  ``static` `int` `min_element(ArrayList al)``  ``{``    ``int` `min = Integer.MAX_VALUE;``    ``for``(``int` `i = ``0``; i < al.size(); i++)``    ``{``      ``min=Math.min(min, al.get(i));``    ``}``    ``return` `min;``  ``}` `  ``// Function that returns the Minimum element of an ArrayList``  ``static` `int` `max_element(ArrayList al)``  ``{``    ``int` `max = Integer.MIN_VALUE;``    ``for``(``int` `i = ``0``; i < al.size(); i++)``    ``{``      ``max = Math.max(max, al.get(i));``    ``}``    ``return` `max;``  ``}` `  ``// Function which returns the``  ``// length of longest common``  ``// prime subsequence``  ``static` `int` `longestCommonSubseq(ArrayList arr1,``                                 ``ArrayList arr2)``  ``{` `    ``// Minimum element of``    ``// both arrays``    ``int` `min1 = min_element(arr1);``    ``int` `min2 = min_element(arr2);` `    ``// Maximum element of``    ``// both arrays``    ``int` `max1 = max_element(arr1);``    ``int` `max2 = max_element(arr2);` `    ``// Generating all primes within``    ``// the max range of arr1``    ``ArrayList a = primegenerator(max1);` `    ``// Generating all primes within``    ``// the max range of arr2``    ``ArrayList b = primegenerator(max2);``    ``ArrayList finala = ``new` `ArrayList<>();``    ``ArrayList finalb = ``new` `ArrayList<>();` `    ``// Store precomputed values``    ``Map,Integer> dp =``      ``new` `HashMap ,Integer> ();` `    ``// Store all primes in arr1[]``    ``for` `(``int` `i = min1; i <= max1; i++)``    ``{``      ``if` `(arr1.contains(i)``          ``&& a.get(i) == ``true``)``        ``finala.add(i);``    ``}` `    ``// Store all primes of arr2[]``    ``for` `(``int` `i = min2; i <= max2; i++)``    ``{``      ``if` `(arr2.contains(i)``          ``&& b.get(i) == ``true``)``        ``finalb.add(i);``    ``}` `    ``// Calculating the LCS``    ``return` `recursion(finala, finalb, ``0``, ``0``, dp);``  ``}` `  ``// Driver Code``  ``public` `static` `void` `main(String args[])``  ``{``    ``int` `a1[] = { ``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7``, ``8``, ``9` `};``    ``int` `a2[] = { ``2``, ``5``, ``6``, ``3``, ``7``, ``9``, ``8` `};` `    ``// Converting into list``    ``ArrayList arr1 = ``new` `ArrayList();``    ``for``(``int` `i = ``0``; i < a1.length; i++)``      ``arr1.add(a1[i]);` `    ``ArrayList arr2 = ``new` `ArrayList();``    ``for``(``int` `i = ``0``; i < a2.length; i++)``      ``arr2.add(a2[i]);` `    ``// Function Call``    ``System.out.println(longestCommonSubseq(arr1, arr2));``  ``}``}` `// This code is contributed by jyoti369`

## Python3

 `# Python implementation of the above approach` `# Function to calculate the LCS` `def` `recursion(arr1, arr2, i, j, dp):``    ``if` `i >``=` `len``(arr1) ``or` `j >``=` `len``(arr2):``        ``return` `0``    ``key ``=` `(i, j)``    ``if` `arr1[i] ``=``=` `arr2[j]:``        ``return` `1` `+` `recursion(arr1, arr2,``                             ``i ``+` `1``, j ``+` `1``, dp)``    ``if` `key ``in` `dp:``        ``return` `dp[key]``    ``else``:``        ``dp[key] ``=` `max``(recursion(arr1, arr2,``                                ``i ``+` `1``, j, dp),``                      ``recursion(arr1, arr2,``                                ``i, j ``+` `1``, dp))``    ``return` `dp[key]` `# Function to generate``# all the possible``# prime numbers`  `def` `primegenerator(n):``    ``cnt ``=` `0``    ``primes ``=` `[``True` `for` `_ ``in` `range``(n ``+` `1``)]``    ``p ``=` `2``    ``while` `p ``*` `p <``=` `n:``        ``for` `i ``in` `range``(p ``*` `p, n ``+` `1``, p):``            ``primes[i] ``=` `False``        ``p ``+``=` `1``    ``return` `primes` `# Function which returns the``# length of longest common``# prime subsequence`  `def` `longestCommonSubseq(arr1, arr2):` `    ``# Minimum element of``    ``# both arrays``    ``min1 ``=` `min``(arr1)``    ``min2 ``=` `min``(arr2)` `    ``# Maximum element of``    ``# both arrays``    ``max1 ``=` `max``(arr1)``    ``max2 ``=` `max``(arr2)` `    ``# Generating all primes within``    ``# the max range of arr1``    ``a ``=` `primegenerator(max1)` `    ``# Generating all primes within``    ``# the max range of arr2``    ``b ``=` `primegenerator(max2)` `    ``finala ``=` `[]``    ``finalb ``=` `[]` `    ``# Store precomputed values``    ``dp ``=` `dict``()` `    ``# Store all primes in arr1[]``    ``for` `i ``in` `range``(min1, max1 ``+` `1``):``        ``if` `i ``in` `arr1 ``and` `a[i] ``=``=` `True``:``            ``finala.append(i)` `    ``# Store all primes of arr2[]``    ``for` `i ``in` `range``(min2, max2 ``+` `1``):``        ``if` `i ``in` `arr2 ``and` `b[i] ``=``=` `True``:``            ``finalb.append(i)` `    ``# Calculating the LCS``    ``return` `recursion(finala, finalb,``                     ``0``, ``0``, dp)`  `# Driver Code``arr1 ``=` `[``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7``, ``8``, ``9``]``arr2 ``=` `[``2``, ``5``, ``6``, ``3``, ``7``, ``9``, ``8``]` `# Function Call``print``(longestCommonSubseq(arr1, arr2))`
Output
`4`

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up