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# Length of the longest increasing subsequence such that no two adjacent elements are coprime

Given an array arr[] of size N. The task is to find the length of the longest subsequence from the given array such that the sequence is strictly increasing and no two adjacent elements are coprime.
Note: The elements in the given array are strictly increasing in order (1 <= a[i] <= 105)

Examples:

Input : a[] = { 1, 2, 3, 4, 5, 6}
Output :
Explanation : Possible sub sequences are {1}, {2}, {3}, {4}, {5}, {6}, {2, 4}, {2, 4, 6}, {2, 6}, {4, 6}, {3, 6}.
The subsequence {2, 4, 6} has the longest length.

Input : a[] = { 1, 1, 1, 1}
Output :

Approach: The main idea is to use the concept of Dynamic programming. Let’s define dp[x] as the maximal value of the length of the subsequence whose last element is x, and define d[i] as the (maximal value of dp[x] where x is divisible by i).
We should calculate dp[x] in the increasing order of x. The value of dp[x] is (maximal value of d[i] where i is a divisor of x) + 1. After we calculate dp[x], for each divisor i of x, we should update d[i] too. This algorithm works in O(N*logN) because the sum of the number of the divisors from 1 to N is O(N*logN).

Note: There is a corner case. When the set is {1}, you should output 1.

Below is the implementation of the above approach:

## C++

 `// CPP program to find the length of the``// longest  increasing sub sequence from the given array``// such that no two adjacent elements are co prime` `#include ``using` `namespace` `std;``#define N 100005` `// Function to find the length of the``// longest  increasing sub sequence from the given array``// such that no two adjacent elements are co prime``int` `LIS(``int` `a[], ``int` `n)``{``    ``// To store dp and d value``    ``int` `dp[N], d[N];` `    ``// To store required answer``    ``int` `ans = 0;` `    ``// For all elements in the array``    ``for` `(``int` `i = 0; i < n; i++) {``        ``// Initially answer is one``        ``dp[a[i]] = 1;` `        ``// For all it's divisors``        ``for` `(``int` `j = 2; j * j <= a[i]; j++) {``            ``if` `(a[i] % j == 0) {``                ``// Update the dp value``                ``dp[a[i]] = max(dp[a[i]], dp[d[j]] + 1);``                ``dp[a[i]] = max(dp[a[i]], dp[d[a[i] / j]] + 1);` `                ``// Update the divisor value``                ``d[j] = a[i];``                ``d[a[i] / j] = a[i];``            ``}``        ``}` `        ``// Check for required answer``        ``ans = max(ans, dp[a[i]]);` `        ``// Update divisor of a[i]``        ``d[a[i]] = a[i];``    ``}` `    ``// Return required answer``    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``int` `a[] = { 1, 2, 3, 4, 5, 6 };` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]);` `    ``cout << LIS(a, n);` `    ``return` `0;``}`

## Java

 `// Java program to find the length``// of the longest increasing sub``// sequence from the given array``// such that no two adjacent``// elements are co prime` `class` `GFG``{``    ``static` `int` `N=``100005``;` `// Function to find the length of the``// longest increasing sub sequence``// from the given array such that``// no two adjacent elements are co prime``static` `int` `LIS(``int` `a[], ``int` `n)``{``    ``// To store dp and d value``    ``int` `dp[]=``new` `int``[N], d[]=``new` `int``[N];` `    ``// To store required answer``    ``int` `ans = ``0``;` `    ``// For all elements in the array``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{``        ``// Initially answer is one``        ``dp[a[i]] = ``1``;` `        ``// For all it's divisors``        ``for` `(``int` `j = ``2``; j * j <= a[i]; j++)``        ``{``            ``if` `(a[i] % j == ``0``)``            ``{``                ``// Update the dp value``                ``dp[a[i]] = Math.max(dp[a[i]], dp[d[j]] + ``1``);``                ``dp[a[i]] = Math.max(dp[a[i]], dp[d[a[i] / j]] + ``1``);` `                ``// Update the divisor value``                ``d[j] = a[i];``                ``d[a[i] / j] = a[i];``            ``}``        ``}` `        ``// Check for required answer``        ``ans = Math.max(ans, dp[a[i]]);` `        ``// Update divisor of a[i]``        ``d[a[i]] = a[i];``    ``}` `    ``// Return required answer``    ``return` `ans;``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``int` `a[] = { ``1``, ``2``, ``3``, ``4``, ``5``, ``6` `};` `    ``int` `n = a.length;` `    ``System.out.print( LIS(a, n));``}``}` `// This code is contributed by Arnab Kundu`

## Python3

 `# Python3 program to find the length of the``# longest increasing sub sequence from the``# given array such that no two adjacent``# elements are co prime``N ``=` `100005` `# Function to find the length of the``# longest increasing sub sequence from``# the given array such that no two``# adjacent elements are co prime``def` `LIS(a, n):``    ` `    ``# To store dp and d value``    ``dp ``=` `[``0` `for` `i ``in` `range``(N)]``    ``d ``=` `[``0` `for` `i ``in` `range``(N)]` `    ``# To store required answer``    ``ans ``=` `0` `    ``# For all elements in the array``    ``for` `i ``in` `range``(n):``        ` `        ``# Initially answer is one``        ``dp[a[i]] ``=` `1` `        ``# For all it's divisors``        ``for` `j ``in` `range``(``2``, a[i]):``            ``if` `j ``*` `j > a[i]:``                ``break``            ``if` `(a[i] ``%` `j ``=``=` `0``):``                ` `                ``# Update the dp value``                ``dp[a[i]] ``=` `max``(dp[a[i]],``                               ``dp[d[j]] ``+` `1``)``                ``dp[a[i]] ``=` `max``(dp[a[i]],``                               ``dp[d[a[i] ``/``/` `j]] ``+` `1``)` `                ``# Update the divisor value``                ``d[j] ``=` `a[i]``                ``d[a[i] ``/``/` `j] ``=` `a[i]` `        ``# Check for required answer``        ``ans ``=` `max``(ans, dp[a[i]])` `        ``# Update divisor of a[i]``        ``d[a[i]] ``=` `a[i]` `    ``# Return required answer``    ``return` `ans` `# Driver code``a ``=` `[``1``, ``2``, ``3``, ``4``, ``5``, ``6``]` `n ``=` `len``(a)` `print``(LIS(a, n))` `# This code is contributed by mohit kumar`

## C#

 `// C# program to find the length``// of the longest increasing sub``// sequence from the given array``// such that no two adjacent``// elements are co prime``using` `System;` `class` `GFG``{``    ``static` `int` `N = 100005;` `    ``// Function to find the length of the``    ``// longest increasing sub sequence``    ``// from the given array such that``    ``// no two adjacent elements are co prime``    ``static` `int` `LIS(``int` `[]a, ``int` `n)``    ``{``        ``// To store dp and d value``        ``int` `[]dp = ``new` `int``[N];``        ``int` `[]d = ``new` `int``[N];``    ` `        ``// To store required answer``        ``int` `ans = 0;``    ` `        ``// For all elements in the array``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``// Initially answer is one``            ``dp[a[i]] = 1;``    ` `            ``// For all it's divisors``            ``for` `(``int` `j = 2; j * j <= a[i]; j++)``            ``{``                ``if` `(a[i] % j == 0)``                ``{``                    ``// Update the dp value``                    ``dp[a[i]] = Math.Max(dp[a[i]], dp[d[j]] + 1);``                    ``dp[a[i]] = Math.Max(dp[a[i]], dp[d[a[i] / j]] + 1);``    ` `                    ``// Update the divisor value``                    ``d[j] = a[i];``                    ``d[a[i] / j] = a[i];``                ``}``            ``}``    ` `            ``// Check for required answer``            ``ans = Math.Max(ans, dp[a[i]]);``    ` `            ``// Update divisor of a[i]``            ``d[a[i]] = a[i];``        ``}``    ` `        ``// Return required answer``        ``return` `ans;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `[]a = { 1, 2, 3, 4, 5, 6 };``    ` `        ``int` `n = a.Length;``    ` `        ``Console.WriteLine(LIS(a, n));``    ``}``}` `// This code is contributed by Ryuga`

## PHP

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## Javascript

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Output:

`3`

Time Complexity: O(N* log(N))

Auxiliary Space: O(N), since N extra space has been taken.