# Longest sub-sequence that satisfies the given conditions

Given an array arr[] of N integers, the task is to find the longest sub-sequence in the given array such that for all pairs from the sub-sequence (arr[i], arr[j]) where i != j either arr[i] divides arr[j] or vice versa. If no such sub-sequence exists then print -1.

Examples:

Input: arr[] = {2, 4, 6, 1, 3, 11}
Output: 3
Longest valid sub-sequences are {1, 2, 6} and {1, 3, 6}.

Input: arr[] = {21, 22, 6, 4, 13, 7, 332}
Output: 2

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: This problem is a simple variation of the longest increasing sub-sequence problem. What changes is the base condition and the trick to reduce the number of computations by sorting the given array.

• First sort the given array so that we only need to check values where arr[i] > arr[j] for i > j.
• Then we move forward using two loops, outer loop runs from 1 to N and the inner loop runs from 0 to i.
• Now in the inner loop we have to find the number of arr[j] where j is from 0 to i – 1 which divides the element arr[i].
• And the recurrence relation will be dp[i] = max(dp[i], 1 + dp[j]).
• We will update the max dp[i] value in a variable named res which will be the final answer.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the length of the ` `// longest required sub-sequence ` `int` `find(``int` `n, ``int` `a[]) ` `{ ` `    ``// Sort the array ` `    ``sort(a, a + n); ` ` `  `    ``// To store the resultant length ` `    ``int` `res = 1; ` `    ``int` `dp[n]; ` ` `  `    ``// If array contains only one element ` `    ``// then it divides itself ` `    ``dp = 1; ` ` `  `    ``for` `(``int` `i = 1; i < n; i++) { ` ` `  `        ``// Every elemnet divides itself ` `        ``dp[i] = 1; ` ` `  `        ``// Count for all numbers which ` `        ``// are leser than a[i] ` `        ``for` `(``int` `j = 0; j < i; j++) { ` ` `  `            ``if` `(a[i] % a[j] == 0) { ` ` `  `                ``// If a[i] % a[j] then update the maximum ` `                ``// subsequence length, ` `                ``// dp[i] = max(dp[i], 1 + dp[j]) ` `                ``// where j is in the range [0, i-1] ` `                ``dp[i] = std::max(dp[i], 1 + dp[j]); ` `            ``} ` `        ``} ` ` `  `        ``res = std::max(res, dp[i]); ` `    ``} ` ` `  `    ``// If array contains only one element ` `    ``// then i = j which doesn't satisfy the condition ` `    ``return` `(res == 1) ? -1 : res; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `a[] = { 2, 4, 6, 1, 3, 11 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(``int``); ` ` `  `    ``cout << find(n, a); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.Arrays;  ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` `     `  `// Function to return the length of the ` `// longest required sub-sequence ` `static` `int` `find(``int` `n, ``int` `a[]) ` `{ ` `    ``// Sort the array ` `    ``Arrays.sort(a); ` ` `  `    ``// To store the resultant length ` `    ``int` `res = ``1``; ` `    ``int` `dp[] = ``new` `int``[n]; ` ` `  `    ``// If array contains only one element ` `    ``// then it divides itself ` `    ``dp[``0``] = ``1``; ` ` `  `    ``for` `(``int` `i = ``1``; i < n; i++)  ` `    ``{ ` ` `  `        ``// Every elemnet divides itself ` `        ``dp[i] = ``1``; ` ` `  `        ``// Count for all numbers which ` `        ``// are leser than a[i] ` `        ``for` `(``int` `j = ``0``; j < i; j++)  ` `        ``{ ` ` `  `            ``if` `(a[i] % a[j] == ``0``) ` `            ``{ ` ` `  `                ``// If a[i] % a[j] then update the maximum ` `                ``// subsequence length, ` `                ``// dp[i] = Math.max(dp[i], 1 + dp[j]) ` `                ``// where j is in the range [0, i-1] ` `                ``dp[i] = Math.max(dp[i], ``1` `+ dp[j]); ` `            ``} ` `        ``} ` ` `  `        ``res = Math.max(res, dp[i]); ` `    ``} ` ` `  `    ``// If array contains only one element ` `    ``// then i = j which doesn't satisfy the condition ` `    ``return` `(res == ``1``) ? -``1` `: res; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main (String[] args)  ` `{ ` `    ``int` `a[] = { ``2``, ``4``, ``6``, ``1``, ``3``, ``11` `}; ` `    ``int` `n = a.length; ` ` `  `    ``System.out.println (find(n, a)); ` `} ` `} ` ` `  `// This code is contributed by jit_t `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the length of the  ` `# longest required sub-sequence  ` `def` `find(n, a) : ` ` `  `    ``# Sort the array  ` `    ``a.sort();  ` ` `  `    ``# To store the resultant length  ` `    ``res ``=` `1``;  ` `    ``dp ``=` `[``0``]``*``n;  ` ` `  `    ``# If array contains only one element  ` `    ``# then it divides itself  ` `    ``dp[``0``] ``=` `1``;  ` ` `  `    ``for` `i ``in` `range``(``1``, n) : ` `         `  `        ``# Every elemnet divides itself  ` `        ``dp[i] ``=` `1``;  ` ` `  `        ``# Count for all numbers which  ` `        ``# are leser than a[i]  ` `        ``for` `j ``in` `range``(i) : ` ` `  `            ``if` `(a[i] ``%` `a[j] ``=``=` `0``) : ` ` `  `                ``# If a[i] % a[j] then update the maximum  ` `                ``# subsequence length,  ` `                ``# dp[i] = max(dp[i], 1 + dp[j])  ` `                ``# where j is in the range [0, i-1]  ` `                ``dp[i] ``=` `max``(dp[i], ``1` `+` `dp[j]);  ` `     `  `        ``res ``=` `max``(res, dp[i]);  ` ` `  `    ``# If array contains only one element  ` `    ``# then i = j which doesn't satisfy the condition  ` `    ``if` `(res ``=``=` `1``):  ` `        ``return` `-``1` `    ``else` `: ` `        ``return` `res;  ` ` `  ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``a ``=` `[ ``2``, ``4``, ``6``, ``1``, ``3``, ``11` `]; ` `    ``n ``=` `len``(a);  ` ` `  `    ``print``(find(n, a));  ` `     `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `     `  `// Function to return the length of the ` `// longest required sub-sequence ` `static` `int` `find(``int` `n, ``int` `[]a) ` `{ ` `    ``// Sort the array ` `    ``Array.Sort(a); ` ` `  `    ``// To store the resultant length ` `    ``int` `res = 1; ` `    ``int` `[]dp = ``new` `int``[n]; ` ` `  `    ``// If array contains only one element ` `    ``// then it divides itself ` `    ``dp = 1; ` ` `  `    ``for` `(``int` `i = 1; i < n; i++)  ` `    ``{ ` ` `  `        ``// Every elemnet divides itself ` `        ``dp[i] = 1; ` ` `  `        ``// Count for all numbers which ` `        ``// are leser than a[i] ` `        ``for` `(``int` `j = 0; j < i; j++)  ` `        ``{ ` ` `  `            ``if` `(a[i] % a[j] == 0) ` `            ``{ ` ` `  `                ``// If a[i] % a[j] then update the maximum ` `                ``// subsequence length, ` `                ``// dp[i] = Math.max(dp[i], 1 + dp[j]) ` `                ``// where j is in the range [0, i-1] ` `                ``dp[i] = Math.Max(dp[i], 1 + dp[j]); ` `            ``} ` `        ``} ` ` `  `        ``res = Math.Max(res, dp[i]); ` `    ``} ` ` `  `    ``// If array contains only one element ` `    ``// then i = j which doesn't satisfy the condition ` `    ``return` `(res == 1) ? -1 : res; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main ()  ` `{ ` `    ``int` `[]a = { 2, 4, 6, 1, 3, 11 }; ` `    ``int` `n = a.Length; ` ` `  `    ``Console.WriteLine(find(n, a)); ` `} ` `} ` ` `  `// This code is contributed by anuj_67.. `

Output:

```3
```

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Improved By : jit_t, vt_m, AnkitRai01