Length of longest increasing prime subsequence from a given array

Given an array arr[] consisting of N positive integers, the task is to find the length of the longest increasing subsequence consisting of Prime Numbers in the given array.

Examples:

Input: arr[] = {1, 2, 5, 3, 2, 5, 1, 7}
Output: 4
Explanation:
The Longest Increasing Prime Subsequence is {2, 3, 5, 7}.
Therefore, the answer is 4.

Input: arr[] = {6, 11, 7, 13, 9, 25}
Output: 2
Explanation:
The Longest Increasing Prime Subsequence is {11, 13} and {7, 13}.
Therefore, the answer is 2.

Naive Approach: The simplest approach is to generate all possible subsequence of the given array and print the length of the longest subsequence consisting of prime numbers in increasing order. 



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

Efficient Approach: The idea is to use the Dynamic Programming approach to optimize the above approach. This problem is a basic variation of the Longest Increasing Subsequence (LIS) problem. Below are the steps:

  • Initialize an auxiliary array dp[] of size N such that dp[i] will store the length of LIS of prime numbers ending at index i.
  • Below is the recurrence relation for finding the longest increasing Prime Numbers:

    If arr[i] is prime then
          dp[i] = 1 + max(dp[j], for j belongs (0, i – 1)), where 0 < j < i and arr[j] < arr[i];
          dp[i] = 1, if no such j exists
    else if arr[i] is non-prime then
          dp[i]  = 0

  • Using Sieve of Eratosthenes store all the prime numbers to till 105.
  • Iterate a two nested loop over the given array and update the array dp[] according to the above recurrence relation.
  • After all the above steps, the maximum element in the array dp[] is the length of the longest increasing subsequence of Prime Numbers in the given array.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
  
#include <bits/stdc++.h>
using namespace std;
#define N 100005
  
// Function to find the prime numbers
// till 10^5 using Sieve of Eratosthenes
void SieveOfEratosthenes(bool prime[],
                         int p_size)
{
    // False here indicates
    // that it is not prime
    prime[0] = false;
    prime[1] = false;
  
    for (int p = 2; p * p <= p_size; p++) {
  
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p]) {
  
            // Update all multiples of p,
            // set them to non-prime
            for (int i = p * 2;
                 i <= p_size; i += p)
                prime[i] = false;
        }
    }
}
  
// Function which computes the length
// of the LIS of Prime Numbers
int LISPrime(int arr[], int n)
{
    // Create an array of size n
    int lisp[n];
  
    // Create boolean array to
    // mark prime numbers
    bool prime[N + 1];
  
    // Initialize all values to true
    memset(prime, true, sizeof(prime));
  
    // Precompute N primes
    SieveOfEratosthenes(prime, N);
  
    lisp[0] = prime[arr[0]] ? 1 : 0;
  
    // Compute optimized LIS having
    // prime numbers in bottom up manner
    for (int i = 1; i < n; i++) {
        if (!prime[arr[i]]) {
            lisp[i] = 0;
            continue;
        }
  
        lisp[i] = 1;
        for (int j = 0; j < i; j++) {
  
            // Check for LIS and prime
            if (prime[arr[j]]
                && arr[i] > arr[j]
                && lisp[i] < lisp[j] + 1) {
                lisp[i] = lisp[j] + 1;
            }
        }
    }
  
    // Return maximum value in lis[]
    return *max_element(lisp, lisp + n);
}
  
// Driver Code
int main()
{
    // Given array
    int arr[] = { 1, 2, 5, 3, 2, 5, 1, 7 };
  
    // Size of array
    int M = sizeof(arr) / sizeof(arr[0]);
  
    // Function Call
    cout << LISPrime(arr, M);
  
    return 0;
}

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Java

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// Java program for the above approach
import java.util.*;
  
class GFG{
      
static final int N = 100005;
  
// Function to find the prime numbers
// till 10^5 using Sieve of Eratosthenes
static void SieveOfEratosthenes(boolean prime[],
                                int p_size)
{
      
    // False here indicates
    // that it is not prime
    prime[0] = false;
    prime[1] = false;
  
    for(int p = 2; p * p <= p_size; p++) 
    {
  
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p]) 
        {
  
            // Update all multiples of p,
            // set them to non-prime
            for(int i = p * 2;
                    i <= p_size;
                    i += p)
                prime[i] = false;
        }
    }
}
  
// Function which computes the length
// of the LIS of Prime Numbers
static int LISPrime(int arr[], int n)
{
      
    // Create an array of size n
    int []lisp = new int[n];
  
    // Create boolean array to
    // mark prime numbers
    boolean []prime = new boolean[N + 1];
  
    // Initialize all values to true
    for(int i = 0; i < prime.length; i++)
        prime[i] = true;
  
    // Precompute N primes
    SieveOfEratosthenes(prime, N);
  
    lisp[0] = prime[arr[0]] ? 1 : 0;
  
    // Compute optimized LIS having
    // prime numbers in bottom up manner
    for(int i = 1; i < n; i++)
    {
        if (!prime[arr[i]]) 
        {
            lisp[i] = 0;
            continue;
        }
  
        lisp[i] = 1;
        for(int j = 0; j < i; j++) 
        {
              
            // Check for LIS and prime
            if (prime[arr[j]] && 
                arr[i] > arr[j] && 
               lisp[i] < lisp[j] + 1
            {
                lisp[i] = lisp[j] + 1;
            }
        }
    }
  
    // Return maximum value in lis[]
    return Arrays.stream(lisp).max().getAsInt();
}
  
// Driver Code
public static void main(String[] args)
{
      
    // Given array
    int arr[] = { 1, 2, 5, 3, 2, 5, 1, 7 };
  
    // Size of array
    int M = arr.length;
  
    // Function call
    System.out.print(LISPrime(arr, M));
}
}
  
// This code is contributed by Amit Katiyar

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Python3

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# Python3 program for 
# the above approach 
N = 100005
   
# Function to find the prime numbers
# till 10^5 using Sieve of Eratosthenes
def SieveOfEratosthenes(prime, p_size):
  
  # False here indicates
  # that it is not prime
  prime[0] = False
  prime[1] = False
  
  p = 2
  while p * p <= p_size:
  
    # If prime[p] is not changed,
    # then it is a prime
    if (prime[p]):
  
      # Update all multiples of p,
      # set them to non-prime
      for i in range (p * 2,
                      p_size + 1, p):
        prime[i] = False
  
        p += 1
  
# Function which computes the length
# of the LIS of Prime Numbers
def LISPrime(arr, n):
  
  # Create an array of size n
  lisp = [0] * n
  
  # Create boolean array to
  # mark prime numbers
  prime = [True] * (N + 1)
  
  # Precompute N primes
  SieveOfEratosthenes(prime, N)
  
  if prime[arr[0]]:
    lisp[0] = 1
    else:
      lisp[0] = 0
  
      # Compute optimized LIS having
      # prime numbers in bottom up manner
      for i in range (1, n):
        if (not prime[arr[i]]):
          lisp[i] = 0
          continue
  
          lisp[i] = 1
          for j in range (i):
            # check for LIS and prime
            if (prime[arr[j]] and 
                arr[i] > arr[j] and 
                lisp[i] < lisp[j] + 1):
              lisp[i] = lisp[j] + 1
  
              # Return maximum value in lis[]
              return max(lisp)
  
# Driver Code
if __name__ == "__main__":
  
  # Given array
  arr = [1, 2, 5, 3
         2, 5, 1, 7]
  
  # Size of array
  M = len(arr)
  
  # Function Call
  print (LISPrime(arr, M))
  
# This code is contributed by Chitranayal

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C#

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// C# program for the above approach
using System;
using System.Linq;
  
class GFG{
      
static readonly int N = 100005;
  
// Function to find the prime numbers
// till 10^5 using Sieve of Eratosthenes
static void SieveOfEratosthenes(bool []prime,
                                int p_size)
{
      
    // False here indicates
    // that it is not prime
    prime[0] = false;
    prime[1] = false;
  
    for(int p = 2; p * p <= p_size; p++) 
    {
  
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p]) 
        {
  
            // Update all multiples of p,
            // set them to non-prime
            for(int i = p * 2;
                    i <= p_size;
                    i += p)
                prime[i] = false;
        }
    }
}
  
// Function which computes the length
// of the LIS of Prime Numbers
static int LISPrime(int []arr, int n)
{
      
    // Create an array of size n
    int []lisp = new int[n];
  
    // Create bool array to
    // mark prime numbers
    bool []prime = new bool[N + 1];
  
    // Initialize all values to true
    for(int i = 0; i < prime.Length; i++)
        prime[i] = true;
  
    // Precompute N primes
    SieveOfEratosthenes(prime, N);
  
    lisp[0] = prime[arr[0]] ? 1 : 0;
  
    // Compute optimized LIS having
    // prime numbers in bottom up manner
    for(int i = 1; i < n; i++)
    {
        if (!prime[arr[i]]) 
        {
            lisp[i] = 0;
            continue;
        }
  
        lisp[i] = 1;
        for(int j = 0; j < i; j++) 
        {
              
            // Check for LIS and prime
            if (prime[arr[j]] && 
                arr[i] > arr[j] && 
               lisp[i] < lisp[j] + 1) 
            {
                lisp[i] = lisp[j] + 1;
            }
        }
    }
  
    // Return maximum value in lis[]
    return lisp.Max();
}
  
// Driver Code
public static void Main(String[] args)
{
      
    // Given array
    int []arr = { 1, 2, 5, 3, 2, 5, 1, 7 };
  
    // Size of array
    int M = arr.Length;
  
    // Function call
    Console.Write(LISPrime(arr, M));
}
}
  
// This code is contributed by Amit Katiyar 

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Output: 

4

Time Complexity: O(N2)
Auxiliary Space: O(N)

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