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

  • Difficulty Level : Medium
  • Last Updated : 13 Jul, 2021

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 <bits/stdc++.h>
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




<?php
// PHP 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
function LIS($a, $n)
{
     
    // To store dp and d value
    $dp = array();
    $d = array();
 
    // To store required answer
    $ans = 0;
 
    // For all elements in the array
    for ($i = 0; $i < $n; $i++)
    {
        // Initially answer is one
        $dp[$a[$i]] = 1;
 
        // For all it's divisors
        for ($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
$a = array(1, 2, 3, 4, 5, 6);
 
$n = sizeof($a);
 
echo LIS($a, $n);
 
// This code is contributed
// by Akanksha Rai
?>

Javascript




<script>
// Javascript program to find the length of the
// longest increasing sub sequence from
// the given array such that no two adjacent
// elements are co prime
let 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
function LIS(a, n)
{
     
    // To store dp and d value
    let dp = new Array();
    let d = new Array();
 
    // To store required answer
    let ans = 0;
 
    // For all elements in the array
    for (let i = 0; i < n; i++)
    {
        // Initially answer is one
        dp[a[i]] = 1;
 
        // For all it's divisors
        for (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
let a = [1, 2, 3, 4, 5, 6];
 
let n = a.length;
 
document.write(LIS(a, n));
 
// This code is contributed
// by _saurabh_jaiswal
</script>
Output: 
3

 

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

Auxiliary Space: O(N)

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