# Maximize length of longest increasing prime subsequence from the given array

Given an array, **arr[] **of size **N**, the task is to find the length of the longest increasing prime subsequence possible by performing the following operations.

- If
**arr[i]**is already a prime number, no need to update**arr[i]**. - Update non-prime
**arr[i]**to the closest prime number less than**arr[i]**. - Update non-prime
**arr[i]**to the closest prime number greater than**arr[i]**.

**Examples:**

Input:arr[] = {8, 6, 9, 2, 5}Output:2Explanation:

Possible rearrangements of the array are: {{7, 5, 2, 5}, {7, 7, 2, 5}, {11, 5, 2, 5}, {1, 7, 2, 5}}.

Therefore, the length of the longest increasing prime subsequence = 2.

Input:arr[] = {27, 38, 43, 68, 83, 12, 69, 12}Output :5

**Naive Approach:** The simplest approach is to update all the elements of the given array to either it’s closest smaller prime number or it’s closest greater prime number and then 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(2^{N})**Auxiliary Space:** O(N)

**Efficient Approach:** The idea is to use Dynamic programming approach to optimize the above approach. This problem is a basic variation of the Longest Increasing Prime Subsequence (LIPS) problem. Follow the steps below to solve the problem.

- Initialize a 2-dimensional array, say
**dp[][]**of size**N * 2**, where**dp[i][0]**stores the length of the longest increasing prime subsequence by choosing the closest prime number smaller than**arr[i]**at**i**index and^{th}**dp[i][1]**stores the length of the longest increasing prime subsequence by choosing the closest prime number greater than or equal to arr[i] at**i**index. Below are the recurrence relation:^{th}- If closest smaller prime number to arr[j] < closest smaller prime number to arr[i]: dp[i][0] = 1 + dp[j][0]
- If closest prime number greater than or equal to arr[j] < closest smaller prime number to arr[i]: dp[i][0] = max(dp[i][0], 1 + dp[j][1])
- If closest smaller prime number to arr[j] < closest smaller prime number to arr[i]: dp[i][1] = 1 + dp[j][0]
- If closest greater or equal prime number to arr[j] < closest prime number greater than or equal to arr[i]: dp[i][1] = max(dp[i][1], 1 + dp[j][1])

Here the value of j = 0, 1, â€¦, (i-1)

- Use sieve of Eratosthenes to efficiently compute the prime numbers.
- Traverse the array
**arr[]**and for each index,**i**, update**arr[i]**to the closest prime number of**arr[i]**. - For each index
**i**, find the length of the longest increasing prime subsequence ending at**i**, optimally. - Finally, return the length of the longest increasing prime subsequence.

Below is the implementation of the above approach:

## Java

`// Java Program to implement` `// the above approach` ` ` `import` `java.util.*;` ` ` `public` `class` `Main {` ` ` ` ` `// Stores the closest prime` ` ` `// number for each array element` ` ` `static` `TreeSet<Integer> set` ` ` `= ` `new` `TreeSet<>();` ` ` ` ` `// Function to find the length of longest` ` ` `// increasing prime subsequence` ` ` `public` `static` `int` `LIPS(` `int` `arr[], ` `int` `N)` ` ` `{` ` ` `// Base case` ` ` `if` `(arr.length == ` `0` `)` ` ` `return` `0` `;` ` ` ` ` `int` `dp[][] = ` `new` `int` `[N + ` `1` `][` `2` `];` ` ` ` ` `// Store the length of the longest` ` ` `// increasing prime subsequence` ` ` `int` `max_subsequence = ` `0` `;` ` ` `for` `(` `int` `i = ` `0` `; i < arr.length;` ` ` `i++) {` ` ` `// Store the length of LIPS` ` ` `// by choosing the closest prime` ` ` `// number smaller than arr[i]` ` ` `dp[i][` `0` `] = (arr[i] >= ` `2` `) ? ` `1` `: ` `0` `;` ` ` ` ` `// Store the length of longest LIPS` ` ` `// by choosing the closest prime` ` ` `// number greater than arr[i]` ` ` `dp[i][` `1` `] = ` `1` `;` ` ` `for` `(` `int` `j = ` `0` `; j < i; j++) {` ` ` ` ` `// Store closest smaller` ` ` `// prime number` ` ` `Integer option1 = set.floor(arr[j]);` ` ` ` ` `// Store closest prime number` ` ` `// greater or equal` ` ` `Integer option2 = set.ceiling(arr[j]);` ` ` ` ` `// Recurrence relation` ` ` ` ` `// Fill the value of dp[i][0]` ` ` `if` `(option1 != ` `null` ` ` `&& option1 < set.floor(arr[i]))` ` ` `dp[i][` `0` `]` ` ` `= Math.max(dp[i][` `0` `], dp[j][` `0` `] + ` `1` `);` ` ` ` ` `if` `(option2 != ` `null` ` ` `&& option2 < set.floor(arr[i]))` ` ` `dp[i][` `0` `]` ` ` `= Math.max(dp[i][` `0` `], dp[j][` `1` `] + ` `1` `);` ` ` ` ` `// Fill the value of dp[i][1]` ` ` `if` `(option1 != ` `null` ` ` `&& option1 < set.ceiling(arr[i]))` ` ` `dp[i][` `1` `]` ` ` `= Math.max(dp[i][` `0` `], dp[j][` `0` `] + ` `1` `);` ` ` ` ` `if` `(option2 != ` `null` ` ` `&& option2 < set.ceiling(arr[i]))` ` ` `dp[i][` `1` `]` ` ` `= Math.max(dp[i][` `1` `], dp[j][` `1` `] + ` `1` `);` ` ` `}` ` ` ` ` `// Store the length of the longest` ` ` `// increasing prime subsequence` ` ` `max_subsequence` ` ` `= Math.max(max_subsequence, dp[i][` `0` `]);` ` ` ` ` `max_subsequence` ` ` `= Math.max(max_subsequence, dp[i][` `1` `]);` ` ` `}` ` ` ` ` `return` `max_subsequence;` ` ` `}` ` ` ` ` `// Function to generate all prime numbers` ` ` `public` `static` `void` `prime_sieve()` ` ` `{` ` ` `// Store all prime numbers` ` ` `boolean` `primes[]` ` ` `= ` `new` `boolean` `[` `1000000` `+ ` `5` `];` ` ` ` ` `// Consider all prime numbers` ` ` `// to be true initially` ` ` `Arrays.fill(primes, ` `true` `);` ` ` ` ` `// Mark 0 and 1 non-prime` ` ` `primes[` `0` `] = primes[` `1` `] = ` `false` `;` ` ` ` ` `// Set all even numbers to` ` ` `// non-prime` ` ` `for` `(` `int` `i = ` `4` `; i <= ` `1000000` `;` ` ` `i += ` `2` `)` ` ` `primes[i] = ` `false` `;` ` ` ` ` `for` `(` `int` `i = ` `3` `; i <= ` `1000000` `;` ` ` `i += ` `2` `) {` ` ` ` ` `// If current element is prime` ` ` `if` `(primes[i]) {` ` ` ` ` `// Update all its multiples` ` ` `// as non-prime` ` ` `for` `(` `int` `j = ` `2` `* i; j <= ` `1000000` `;` ` ` `j += i)` ` ` `primes[j] = ` `false` `;` ` ` `}` ` ` `}` ` ` ` ` `// Mark 2 as prime` ` ` `set.add(` `2` `);` ` ` ` ` `// Add all primes to the set` ` ` `for` `(` `int` `i = ` `3` `; i <= ` `1000000` `;` ` ` `i += ` `2` `)` ` ` `if` `(primes[i])` ` ` `set.add(i);` ` ` `}` ` ` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `int` `N = ` `6` `;` ` ` `int` `arr[] = { ` `6` `, ` `7` `, ` `8` `, ` `9` `, ` `10` `, ` `11` `};` ` ` ` ` `prime_sieve();` ` ` ` ` `System.out.println(LIPS(arr, N));` ` ` `}` `}` |

**Output:**

3

**Time Complexity:** O(N^{2}logN)**Auxiliary Space:** O(N)