# Find the longest subsequence of an array having LCM at most K

Given an array arr[] of N elements and a positive integer K. The task is to find the longest sub-sequence in the array having LCM (Least Common Multiple) at most K. Print the LCM and the length of the sub-sequence, following the indexes (starting from 0) of the elements of the obtained sub-sequence. Print -1 if it is not possible to do so.

Examples:

Input: arr[] = {2, 3, 4, 5}, K = 14
Output:
LCM = 12, Length = 3
Indexes = 0 1 2

Input: arr[] = {12, 33, 14, 52}, K = 4
Output: -1

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

Approach: Find all the unique elements of the array and their respective frequencies. Now the highest LCM that you are supposed to get is K. Suppose you have a number X such that 1 ≤ X ≤ K, obtain all the unique numbers from the array whom X is a multiple of and add their frequencies to numCount of X. The answer will be the number with highest numCount, let it be your LCM. Now, to obtain the indexes of the numbers of the sub-sequence, start traversing the array from the beginning and print the index i if LCM % arr[i] = 0.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the longest subsequence ` `// having LCM less than or equal to K ` `void` `findSubsequence(``int``* arr, ``int` `n, ``int` `k) ` `{ ` ` `  `    ``// Map to store unique elements ` `    ``// and their frequencies ` `    ``map<``int``, ``int``> M; ` ` `  `    ``// Update the frequencies ` `    ``for` `(``int` `i = 0; i < n; ++i) ` `        ``++M[arr[i]]; ` ` `  `    ``// Array to store the count of numbers whom ` `    ``// 1 <= X <= K is a multiple of ` `    ``int``* numCount = ``new` `int``[k + 1]; ` ` `  `    ``for` `(``int` `i = 0; i <= k; ++i) ` `        ``numCount[i] = 0; ` ` `  `    ``// Check every unique element ` `    ``for` `(``auto` `p : M) { ` `        ``if` `(p.first <= k) { ` ` `  `            ``// Find all its multiples <= K ` `            ``for` `(``int` `i = 1;; ++i) { ` `                ``if` `(p.first * i > k) ` `                    ``break``; ` ` `  `                ``// Store its frequency ` `                ``numCount[p.first * i] += p.second; ` `            ``} ` `        ``} ` `        ``else` `            ``break``; ` `    ``} ` ` `  `    ``int` `lcm = 0, length = 0; ` ` `  `    ``// Obtain the number having maximum count ` `    ``for` `(``int` `i = 1; i <= k; ++i) { ` `        ``if` `(numCount[i] > length) { ` `            ``length = numCount[i]; ` `            ``lcm = i; ` `        ``} ` `    ``} ` ` `  `    ``// Condition to check if answer ` `    ``// doesn't exist ` `    ``if` `(lcm == 0) ` `        ``cout << -1 << endl; ` `    ``else` `{ ` ` `  `        ``// Print the answer ` `        ``cout << ``"LCM = "` `<< lcm ` `             ``<< ``", Length = "` `<< length << endl; ` ` `  `        ``cout << ``"Indexes = "``; ` `        ``for` `(``int` `i = 0; i < n; ++i) ` `            ``if` `(lcm % arr[i] == 0) ` `                ``cout << i << ``" "``; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `k = 14; ` `    ``int` `arr[] = { 2, 3, 4, 5 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``findSubsequence(arr, n, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` `    ``// Function to find the longest subsequence ` `    ``// having LCM less than or equal to K ` `    ``static` `void` `findSubsequence(``int` `[]arr, ``int` `n, ``int` `k) ` `    ``{ ` `     `  `        ``// Map to store unique elements ` `        ``// and their frequencies ` `        ``HashMap M = ``new` `HashMap(); ` `     `  `        ``// Update the frequencies ` `        ``for` `(``int` `i = ``0``; i < n; ++i) ` `        ``{ ` `            ``if``(M.containsKey(arr[i])) ` `                ``M.put(arr[i], M.get(arr[i])+``1``); ` `            ``else` `                ``M.put(arr[i], ``1``); ` `        ``} ` `         `  `        ``// Array to store the count of numbers whom ` `        ``// 1 <= X <= K is a multiple of ` `        ``int` `[] numCount = ``new` `int``[k + ``1``]; ` `     `  `        ``for` `(``int` `i = ``0``; i <= k; ++i) ` `            ``numCount[i] = ``0``; ` `     `  `        ``Iterator> itr = M.entrySet().iterator();  ` `         `  `        ``// Check every unique element ` `        ``while``(itr.hasNext())  ` `        ``{ ` `            ``HashMap.Entry entry = itr.next(); ` `            ``if` `(entry.getKey() <= k)  ` `            ``{ ` `     `  `                ``// Find all its multiples <= K ` `                ``for` `(``int` `i = ``1``;; ++i)  ` `                ``{ ` `                    ``if` `(entry.getKey() * i > k) ` `                        ``break``; ` `     `  `                    ``// Store its frequency ` `                    ``numCount[entry.getKey() * i] += entry.getValue(); ` `                ``} ` `            ``} ` `            ``else` `                ``break``; ` `        ``} ` `     `  `        ``int` `lcm = ``0``, length = ``0``; ` `     `  `        ``// Obtain the number having maximum count ` `        ``for` `(``int` `i = ``1``; i <= k; ++i)  ` `        ``{ ` `            ``if` `(numCount[i] > length)  ` `            ``{ ` `                ``length = numCount[i]; ` `                ``lcm = i; ` `            ``} ` `        ``} ` `     `  `        ``// Condition to check if answer ` `        ``// doesn't exist ` `        ``if` `(lcm == ``0``) ` `            ``System.out.println(-``1``); ` `        ``else` `        ``{ ` `     `  `            ``// Print the answer ` `            ``System.out.println(``"LCM = "` `+ lcm ` `                ``+ ``" Length = "` `+ length ); ` `     `  `            ``System.out.print( ``"Indexes = "``); ` `            ``for` `(``int` `i = ``0``; i < n; ++i) ` `                ``if` `(lcm % arr[i] == ``0``) ` `                    ``System.out.print(i + ``" "``); ` `        ``} ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `     `  `        ``int` `k = ``14``; ` `        ``int` `arr[] = { ``2``, ``3``, ``4``, ``5` `}; ` `        ``int` `n = arr.length; ` `     `  `        ``findSubsequence(arr, n, k); ` `    ``} ` `} ` ` `  `// This code is contributed by ihritik `

## Python3

 `# Python3 implementation of the approach ` `from` `collections ``import` `defaultdict ` ` `  `# Function to find the longest subsequence ` `# having LCM less than or equal to K ` `def` `findSubsequence(arr, n, k): ` ` `  `    ``# Map to store unique elements ` `    ``# and their frequencies ` `    ``M ``=` `defaultdict(``lambda``:``0``) ` ` `  `    ``# Update the frequencies ` `    ``for` `i ``in` `range``(``0``, n): ` `        ``M[arr[i]] ``+``=` `1` ` `  `    ``# Array to store the count of numbers ` `    ``# whom 1 <= X <= K is a multiple of ` `    ``numCount ``=` `[``0``] ``*` `(k ``+` `1``) ` ` `  `    ``# Check every unique element ` `    ``for` `p ``in` `M:  ` `        ``if` `p <``=` `k:  ` ` `  `            ``# Find all its multiples <= K ` `            ``i ``=` `1` `            ``while` `p ``*` `i <``=` `k:  ` `                 `  `                ``# Store its frequency ` `                ``numCount[p ``*` `i] ``+``=` `M[p] ` `                ``i ``+``=` `1` `         `  `        ``else``: ` `            ``break` `     `  `    ``lcm, length ``=` `0``, ``0` ` `  `    ``# Obtain the number having maximum count ` `    ``for` `i ``in` `range``(``1``, k ``+` `1``):  ` `        ``if` `numCount[i] > length:  ` `            ``length ``=` `numCount[i] ` `            ``lcm ``=` `i ` ` `  `    ``# Condition to check if answer doesn't exist ` `    ``if` `lcm ``=``=` `0``: ` `        ``print``(``-``1``) ` `    ``else``: ` ` `  `        ``# Print the answer ` `        ``print``(``"LCM = {0}, Length = {1}"``.``format``(lcm, length)) ` ` `  `        ``print``(``"Indexes = "``, end ``=` `"") ` `        ``for` `i ``in` `range``(``0``, n): ` `            ``if` `lcm ``%` `arr[i] ``=``=` `0``: ` `                ``print``(i, end ``=` `" "``) ` `     `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``k ``=` `14` `    ``arr ``=` `[``2``, ``3``, ``4``, ``5``]  ` `    ``n ``=` `len``(arr) ` ` `  `    ``findSubsequence(arr, n, k) ` ` `  `# This code is contributed by Rituraj Jain `

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG ` `{ ` `    ``// Function to find the longest subsequence ` `    ``// having LCM less than or equal to K ` `    ``static` `void` `findSubsequence(``int` `[]arr, ``int` `n, ``int` `k) ` `    ``{ ` `     `  `        ``// Map to store unique elements ` `        ``// and their frequencies ` `        ``Dictionary<``int``, ``int``> M = ``new` `Dictionary<``int``, ``int``>(); ` `     `  `        ``// Update the frequencies ` `        ``for` `(``int` `i = 0; i < n; ++i) ` `        ``{ ` `            ``if``(M.ContainsKey(arr[i])) ` `                ``M[arr[i]]++; ` `            ``else` `                ``M[arr[i]] = 1; ` `        ``} ` `         `  `        ``// Array to store the count of numbers whom ` `        ``// 1 <= X <= K is a multiple of ` `        ``int` `[] numCount = ``new` `int``[k + 1]; ` `     `  `        ``for` `(``int` `i = 0; i <= k; ++i) ` `            ``numCount[i] = 0; ` `     `  `        ``Dictionary<``int``, ``int``>.KeyCollection keyColl = M.Keys; ` `         `  `        ``// Check every unique element ` `        ``foreach``(``int` `key ``in` `keyColl) ` `        ``{ ` `            ``if` `( key <= k)  ` `            ``{ ` `     `  `                ``// Find all its multiples <= K ` `                ``for` `(``int` `i = 1;; ++i)  ` `                ``{ ` `                    ``if` `(key * i > k) ` `                        ``break``; ` `     `  `                    ``// Store its frequency ` `                    ``numCount[key * i] += M[key]; ` `                ``} ` `            ``} ` `            ``else` `                ``break``; ` `        ``} ` `     `  `        ``int` `lcm = 0, length = 0; ` `     `  `        ``// Obtain the number having maximum count ` `        ``for` `(``int` `i = 1; i <= k; ++i) ` `        ``{ ` `            ``if` `(numCount[i] > length)  ` `            ``{ ` `                ``length = numCount[i]; ` `                ``lcm = i; ` `            ``} ` `        ``} ` `     `  `        ``// Condition to check if answer ` `        ``// doesn't exist ` `        ``if` `(lcm == 0) ` `            ``Console.WriteLine(-1); ` `        ``else`  `        ``{ ` `     `  `            ``// Print the answer ` `            ``Console.WriteLine(``"LCM = "` `+ lcm ` `                ``+ ``" Length = "` `+ length ); ` `     `  `            ``Console.Write( ``"Indexes = "``); ` `            ``for` `(``int` `i = 0; i < n; ++i) ` `                ``if` `(lcm % arr[i] == 0) ` `                    ``Console.Write(i + ``" "``); ` `        ``} ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main ()  ` `    ``{ ` `     `  `        ``int` `k = 14; ` `        ``int` `[]arr = { 2, 3, 4, 5 }; ` `        ``int` `n = arr.Length; ` `     `  `        ``findSubsequence(arr, n, k); ` `    ``} ` `} ` ` `  `// This code is contributed by ihritik `

## PHP

 ` ``\$value``) ` `    ``{ ` `        ``if` `(``\$key` `<= ``\$k``) ` `        ``{  ` ` `  `            ``// Find all its multiples <= K  ` `            ``for` `(``\$i` `= 1;; ++``\$i``)  ` `            ``{  ` `                ``if` `(``\$key` `* ``\$i` `> ``\$k``)  ` `                    ``break``;  ` ` `  `                ``// Store its frequency  ` `                ``\$numCount``[``\$key` `* ``\$i``] += ``\$value``;  ` `            ``}  ` `        ``}  ` `        ``else` `            ``break``;  ` `    ``}  ` ` `  `    ``\$lcm` `= 0; ``\$length` `= 0;  ` ` `  `    ``// Obtain the number having ` `    ``// maximum count  ` `    ``for` `(``\$i` `= 1; ``\$i` `<= ``\$k``; ++``\$i``)  ` `    ``{  ` `        ``if` `(``\$numCount``[``\$i``] > ``\$length``) ` `        ``{  ` `            ``\$length` `= ``\$numCount``[``\$i``];  ` `            ``\$lcm` `= ``\$i``;  ` `        ``}  ` `    ``}  ` ` `  `    ``// Condition to check if answer  ` `    ``// doesn't exist  ` `    ``if` `(``\$lcm` `== 0)  ` `        ``echo` `-1 << ``"\n"``;  ` `    ``else`  `    ``{  ` ` `  `        ``// Print the answer  ` `        ``echo` `"LCM = "``, ``\$lcm``,  ` `             ``", Length = "``, ``\$length``, ``"\n"``;  ` ` `  `        ``echo` `"Indexes = "``;  ` `        ``for` `(``\$i` `= 0; ``\$i` `< ``\$n``; ++``\$i``)  ` `            ``if` `(``\$lcm` `% ``\$arr``[``\$i``] == 0)  ` `                ``echo` `\$i``, ``" "``;  ` `    ``}  ` `}  ` ` `  `// Driver code  ` `\$k` `= 14;  ` `\$arr` `= ``array``( 2, 3, 4, 5 );  ` `\$n` `= ``count``(``\$arr``); ` ` `  `findSubsequence(``\$arr``, ``\$n``, ``\$k``);  ` ` `  `// This code is contributed by Ryuga ` `?> `

Output:

```LCM = 12, Length = 3
Indexes = 0 1 2
```

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