Given an array arr[] consisting of N positive integers, the task is to find the LCM of all unique elements of the given array. If the array does not contain any unique elements, then print “-1“.
Examples:
Input: arr[] = {1, 2, 1, 3, 3, 4}
Output: 4
Explanation:
The unique elements of the given array are: 2 and 4. Therefore, the LCM of (2, 4) is 4.Input: arr[] = {1, 1, 2, 2, 3, 3}
Output: -1
Naive Approach: The simplest approach to solve the given problem is to traverse the given array arr[] for each array element arr[i] and check if arr[i] is unique or not. If found to be true, then store it in another array. After traversing for all the elements, print the LCM of the elements present in the newly created array.
Time Complexity: O(N2 + N * log(M)), where M is the largest element of the array arr[].
Auxiliary Space: O(N)
Efficient Approach: The above approach can also be optimized by using Hashing for finding the unique elements in the array. Follow the steps below to solve the problem:
- Initialize a variable, say ans as 1 to store the LCM of the unique elements of the array.
- Traverse the array arr[] and store the frequency of each element in the Map.
- Now, traverse the map and if the count of any element is equal to 1, then update the value of ans to LCM of ans and the current element.
- After completing the above steps, if the value of ans is 1 then print “-1”. Otherwise, print the value of ans as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find GCD of two numbersint findGCD(int a, int b){ // Base Case if (b == 0) return a; // Recursively find the GCD return findGCD(b, a % b);}// Function to find LCM of two numbersint findLCM(int a, int b){ return (a * b) / findGCD(a, b);}// Function to find LCM of unique elements// present in the arrayint uniqueElementsLCM(int arr[], int N){ // Stores the frequency of each // number of the array unordered_map<int, int> freq; // Store the frequency of each // element of the array for (int i = 0; i < N; i++) { freq[arr[i]]++; } // Store the required result int lcm = 1; // Traverse the map freq for (auto i : freq) { // If the frequency of the // current element is 1, then // update ans if (i.second == 1) { lcm = findLCM(lcm, i.first); } } // If there is no unique element, // set lcm to -1 if (lcm == 1) lcm = -1; // Print the result cout << lcm;}// Driver Codeint main(){ int arr[] = { 1, 2, 1, 3, 3, 4 }; int N = sizeof(arr) / sizeof(arr[0]); // Function Call uniqueElementsLCM(arr, N); return 0;} |
Java
// Java program for the above approachimport java.util.HashMap;import java.util.Map.Entry;class GFG{ // Function to find GCD of two numbersstatic int findGCD(int a, int b){ // Base Case if (b == 0) return a; // Recursively find the GCD return findGCD(b, a % b);}// Function to find LCM of two numbersstatic int findLCM(int a, int b){ return (a * b) / findGCD(a, b);}// Function to find LCM of unique elements// present in the arraystatic void uniqueElementsLCM(int arr[], int N){ // Stores the frequency of each // number of the array HashMap<Integer, Integer> freq = new HashMap<Integer, Integer>(); // Store the frequency of each // element of the array for(int i = 0; i < N; i++) { freq.put(arr[i], freq.getOrDefault(arr[i], 0) + 1); } // Store the required result int lcm = 1; // Traverse the map freq for(Entry<Integer, Integer> i : freq.entrySet()) { // If the frequency of the // current element is 1, then // update ans if (i.getValue() == 1) { lcm = findLCM(lcm, i.getKey()); } } // If there is no unique element, // set lcm to -1 if (lcm == 1) lcm = -1; // Print the result System.out.print(lcm);}// Driver codepublic static void main(String[] args){ int arr[] = { 1, 2, 1, 3, 3, 4 }; int N = arr.length; // Function Call uniqueElementsLCM(arr, N);}}// This code is contributed by abhinavjain194 |
Python3
# Python3 program for the above approachfrom collections import defaultdict# Function to find GCD of two numbersdef findGCD(a, b): # Base Case if (b == 0): return a # Recursively find the GCD return findGCD(b, a % b)# Function to find LCM of two numbersdef findLCM(a, b): return (a * b) // findGCD(a, b)# Function to find LCM of unique elements# present in the arraydef uniqueElementsLCM(arr, N): # Stores the frequency of each # number of the array freq = defaultdict(int) # Store the frequency of each # element of the array for i in range(N): freq[arr[i]] += 1 # Store the required result lcm = 1 # Traverse the map freq for i in freq: # If the frequency of the # current element is 1, then # update ans if (freq[i] == 1): lcm = findLCM(lcm, i) # If there is no unique element, # set lcm to -1 if (lcm == 1): lcm = -1 # Print the result print(lcm)# Driver Codeif __name__ == "__main__": arr = [ 1, 2, 1, 3, 3, 4 ] N = len(arr) # Function Call uniqueElementsLCM(arr, N)# This code is contributed by ukasp |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{ // Function to find GCD of two numbersstatic int findGCD(int a, int b){ // Base Case if (b == 0) return a; // Recursively find the GCD return findGCD(b, a % b);}// Function to find LCM of two numbersstatic int findLCM(int a, int b){ return (a * b) / findGCD(a, b);}// Function to find LCM of unique elements// present in the arraystatic void uniqueElementsLCM(int []arr, int N){ // Stores the frequency of each // number of the array Dictionary<int,int> freq = new Dictionary<int,int>(); // Store the frequency of each // element of the array for (int i = 0; i < N; i++) { if(freq.ContainsKey(arr[i])) freq[arr[i]]++; else freq.Add(arr[i],1); } // Store the required result int lcm = 1; // Traverse the map freq foreach(KeyValuePair<int, int> kvp in freq) { // If the frequency of the // current element is 1, then // update ans if (kvp.Value == 1) { lcm = findLCM(lcm, kvp.Key); } } // If there is no unique element, // set lcm to -1 if (lcm == 1) lcm = -1; // Print the result Console.Write(lcm);}// Driver Codepublic static void Main(){ int []arr = {1, 2, 1, 3, 3, 4 }; int N = arr.Length; // Function Call uniqueElementsLCM(arr, N);}}// This code is contributed by ipg2016107. |
Javascript
<script>// Javascript program for the above approach// Function to find GCD of two numbersfunction findGCD(a, b){ // Base Case if (b == 0) return a; // Recursively find the GCD return findGCD(b, a % b);}// Function to find LCM of two numbersfunction findLCM(a, b){ return (a * b) / findGCD(a, b);}// Function to find LCM of unique elements// present in the arrayfunction uniqueElementsLCM(arr, N){ // Stores the frequency of each // number of the array var freq = new Map(); // Store the frequency of each // element of the array for (var i = 0; i < N; i++) { if(freq.has(arr[i])) { freq.set(arr[i], freq.get(arr[i])+1); } else { freq.set(arr[i], 1); } } // Store the required result var lcm = 1; // Traverse the map freq freq.forEach((value, key) => { // If the frequency of the // current element is 1, then // update ans if (value == 1) { lcm = findLCM(lcm, key); } }); // If there is no unique element, // set lcm to -1 if (lcm == 1) lcm = -1; // Print the result document.write( lcm);}// Driver Codevar arr = [1, 2, 1, 3, 3, 4];var N = arr.length;// Function CalluniqueElementsLCM(arr, N);</script> |
Output:
4
Time Complexity: O(N * log(M)), where M is the largest element of the array arr[]
Auxiliary Space: O(N)
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