Given a positive integer N, the task is to construct the longest sorted sequence of unique elements whose LCM of the is equal to N.
Examples:
Input: N = 12
Output: 1 2 3 4 6 12
Explanation:
LCM of {1, 2, 3, 4, 6, 12 } is N( = 12).
Therefore, the longest possible sequence is {1, 2, 3, 4, 6, 12 }.Input: N = 9
Output: 1 3 9
Explanation:
LCM of { 1, 2, 9 } is N( = 9).
Therefore, the longest possible sequence is {1, 3, 9 }.
Approach: The problem can be solved based on the following observation:
If an array element is not a factor of N then the LCM of the array elements never be N. Therefore, the array elements must be the factor of N.
Follow the below steps to solve this problem:
- Initialize an array, say newArr[], to store all unique array elements whose LCM is equal to N.
- Find all the factors of N and store it in newArr[].
- Sort the array newArr[].
- Print all array elements of newArr[].
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to construct an array of// unique elements whose LCM is Nvoid constructArrayWithGivenLCM(int N){ // Stores array elements // whose LCM is N vector<int> newArr; // Iterate over the range // [1, sqrt(N)] for (int i = 1; i * i <= N; i++) { // If N is divisible // by i if (N % i == 0) { // Insert i into newArr[] newArr.push_back(i); // If N is not perfect square if (N / i != i) { newArr.push_back(N / i); } } } // Sort the array newArr[] sort(newArr.begin(), newArr.end()); // Print array elements for (auto i : newArr) { cout << i << " "; }}// Driver Codeint main(){ // Given N int N = 12; // Function Call constructArrayWithGivenLCM(N); return 0;} |
Java
// Java program to implement// the above approachimport java.util.Arrays;class GFG{// Function to construct an array of// unique elements whose LCM is Nstatic void constructArrayWithGivenLCM(int N){ // Stores array elements // whose LCM is N int newArr[] = new int[N]; int j = 0; // Iterate over the range // [1, sqrt(N)] for(int i = 1; i * i <= N; i++) { // If N is divisible // by i if (N % i == 0) { // Insert i into newArr[] newArr[j] = i; j++; // If N is not perfect square if (N / i != i) { newArr[j] = N / i; j++; } } } // Sort the array newArr[] Arrays.sort(newArr); // Print array elements for(int i = j; i < N; i++) { System.out.print(newArr[i] + " "); }}// Driver Codepublic static void main (String[] args){ // Given N int N = 12; // Function Call constructArrayWithGivenLCM(N);}}// This code is contributed by AnkThon |
Python3
# Python3 program to implement# the above approachfrom math import sqrt,ceil,floor# Function to construct an array of# unique elements whose LCM is Ndef constructArrayWithGivenLCM(N): # Stores array elements # whose LCM is N newArr = [] # Iterate over the range # [1, sqrt(N)] for i in range(1, ceil(sqrt(N + 1))): # If N is divisible # by i if (N % i == 0): # Insert i into newArr[] newArr.append(i) # If N is not perfect square if (N // i != i): newArr.append(N // i) # Sort the array newArr[] newArr = sorted(newArr) # Print array elements for i in newArr: print(i, end = " ")# Driver Codeif __name__ == '__main__': # Given N N = 12 # Function Call constructArrayWithGivenLCM(N)# This code is contributed by mohit kumar 29 |
C#
// C# program to implement// the above approachusing System;class GFG{ // Function to construct an array of // unique elements whose LCM is N static void constructArrayWithGivenLCM(int N) { // Stores array elements // whose LCM is N int []newArr = new int[N]; int j = 0; // Iterate over the range // [1, sqrt(N)] for(int i = 1; i * i <= N; i++) { // If N is divisible // by i if (N % i == 0) { // Insert i into newArr[] newArr[j] = i; j++; // If N is not perfect square if (N / i != i) { newArr[j] = N / i; j++; } } } // Sort the array newArr[] Array.Sort(newArr); // Print array elements for(int i = j; i < N; i++) { Console.Write(newArr[i] + " "); } } // Driver Code public static void Main(String[] args) { // Given N int N = 12; // Function Call constructArrayWithGivenLCM(N); }}// This code is contributed by Princi Singh |
Javascript
<script> // JavaScript program to implement the above approach // Function to construct an array of // unique elements whose LCM is N function constructArrayWithGivenLCM(N) { // Stores array elements // whose LCM is N let newArr = new Array(N); newArr.fill(0); let j = 0; // Iterate over the range // [1, sqrt(N)] for(let i = 1; i * i <= N; i++) { // If N is divisible // by i if (N % i == 0) { // Insert i into newArr[] newArr[j] = i; j++; // If N is not perfect square if (parseInt(N / i, 10) != i) { newArr[j] = parseInt(N / i, 10); j++; } } } // Sort the array newArr[] newArr.sort(function(a, b){return a - b}); // Print array elements for(let i = j; i < N; i++) { document.write(newArr[i] + " "); } } // Given N let N = 12; // Function Call constructArrayWithGivenLCM(N);</script> |
Output:
1 2 3 4 6 12
Time Complexity: O(? N)
Auxiliary Space: O(1)
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