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
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 <bits/stdc++.h>using namespace std;// Function to find the longest subsequence// having LCM less than or equal to Kvoid 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 codeint main(){ int k = 14; int arr[] = { 2, 3, 4, 5 }; int n = sizeof(arr) / sizeof(arr[0]); findSubsequence(arr, n, k); return 0;} |
Java
// Java implementation of the approachimport 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<Integer, Integer> M = new HashMap<Integer,Integer>(); // 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<HashMap.Entry<Integer, Integer>> itr = M.entrySet().iterator(); // Check every unique element while(itr.hasNext()) { HashMap.Entry<Integer, Integer> 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 approachfrom collections import defaultdict# Function to find the longest subsequence# having LCM less than or equal to Kdef 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 codeif __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 approachusing 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
<?php// PHP implementation of the approach // Function to find the longest subsequence // having LCM less than or equal to K function findSubsequence($arr, $n, $k) { // Map to store unique elements // and their frequencies $M = array(); for($i = 0; $i < $n; $i++) $M[$arr[$i]] = 0 ; // Update the frequencies for ($i = 0; $i < $n; ++$i) ++$M[$arr[$i]]; // Array to store the count of numbers // whom 1 <= X <= K is a multiple of $numCount = array(); for ($i = 0; $i <= $k; ++$i) $numCount[$i] = 0; // Check every unique element foreach($M as $key => $value) { if ($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?> |
Javascript
<script>// JavaScript implementation of the approach// Function to find the longest subsequence// having LCM less than or equal to Kfunction findSubsequence(arr, n, k) { // Map to store unique elements // and their frequencies let M = new Map(); // Update the frequencies for (let i = 0; i < n; ++i) { if (M.has(arr[i])) { M.set(arr[i], M.get(arr[i]) + 1) } else { M.set(arr[i], 1) } } // Array to store the count of numbers whom // 1 <= X <= K is a multiple of let numCount = new Array(k + 1); for (let i = 0; i <= k; ++i) numCount[i] = 0; // Check every unique element for (let p of M) { if (p[0] <= k) { // Find all its multiples <= K for (let i = 1; ; ++i) { if (p[0] * i > k) break; // Store its frequency numCount[p[0] * i] += p[1]; } } else break; } let lcm = 0, length = 0; // Obtain the number having maximum count for (let 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) document.write(-1 + "<br>"); else { // Print the answer document.write( "LCM = " + lcm + ", Length = " + length + "<br>" ); document.write("Indexes = "); for (let i = 0; i < n; ++i) if (lcm % arr[i] == 0) document.write(i + " "); }}// Driver codelet k = 14;let arr = [2, 3, 4, 5];let n = arr.length;findSubsequence(arr, n, k);// This code is contributed by _saurabh_jaiswal</script> |
Output:
LCM = 12, Length = 3 Indexes = 0 1 2
Time Complexity: O(nlogn)
Auxiliary Space: O(n)
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