Given an integer K, the task is to construct an array of maximum length with product of all array elements equal to K, such that each array element except the first one is divisible by its previous adjacent element.
Note: Every array element in the generated array must be greater than 1.
Examples:
Input: K = 4
Output: {2, 2}
Explanation:
The second element, i.e. arr[1] (= 2) is divisible by the first element, i.e. arr[0] (= 2).
Product of the array elements = 2 * 2 = 4(= K).
Therefore, the array satisfies the required condition.Input: K = 23
Output: {23}
Approach: The idea to solve this problem is to find all the prime factors of K with their respective powers such that:
prime_factor[1]x * prime_factor[2]y … * primefactor[i]z = K
Follow the steps below to solve this problem:
- Find the prime factor, say X, with the greatest power, say Y.
- Then, Y will be the length of the required array.
- Therefore, construct an array of length Y with all elements in it equal to X.
- To make the product of array equal to K, multiply the last element by K / X y.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to construct longest array// with product K such that each element// is divisible by its previous elementvoid findLongestArray(int K){ // Stores the prime factors of K vector<pair<int, int> > primefactors; int K_temp = K; for (int i = 2; i * i <= K; i++) { // Stores the power to which // primefactor i is raised int count = 0; while (K_temp % i == 0) { K_temp /= i; count++; } if (count > 0) primefactors.push_back({ count, i }); } if (K_temp != 1) primefactors.push_back( { 1, K_temp }); // Sort prime factors in descending order sort(primefactors.rbegin(), primefactors.rend()); // Stores the final array vector<int> answer( primefactors[0].first, primefactors[0].second); // Multiply the last element by K answer.back() *= K; for (int i = 0; i < primefactors[0].first; i++) { answer.back() /= primefactors[0].second; } // Print the constructed array cout << "{"; for (int i = 0; i < (int)answer.size(); i++) { if (i == answer.size() - 1) cout << answer[i] << "}"; else cout << answer[i] << ", "; }}// Driver Codeint main(){ int K = 4; findLongestArray(K);} |
Java
// java program for the above approachimport java.io.*;import java.lang.*;import java.util.*;class GFG { // Function to construct longest array // with product K such that each element // is divisible by its previous element static void findLongestArray(int K) { // Stores the prime factors of K ArrayList<int[]> primefactors = new ArrayList<>(); int K_temp = K; for (int i = 2; i * i <= K; i++) { // Stores the power to which // primefactor i is raised int count = 0; while (K_temp % i == 0) { K_temp /= i; count++; } if (count > 0) primefactors.add(new int[] { count, i }); } if (K_temp != 1) primefactors.add(new int[] { 1, K_temp }); // Sort prime factors in descending order Collections.sort(primefactors, (x, y) -> { if (x[0] != y[0]) return y[0] - x[0]; return y[1] - x[1]; }); // Stores the final array int n = primefactors.get(0)[0]; int val = primefactors.get(0)[1]; int answer[] = new int[n]; Arrays.fill(answer, val); // Multiply the last element by K answer[n - 1] *= K; for (int i = 0; i < n; i++) { answer[n - 1] /= val; } // Print the constructed array System.out.print("{"); for (int i = 0; i < answer.length; i++) { if (i == answer.length - 1) System.out.print(answer[i] + "}"); else System.out.print(answer[i] + ", "); } } // Driver Code public static void main(String[] args) { int K = 4; findLongestArray(K); }}// This code is contributed by Kingash. |
Python3
# Python 3 program for the above approach# Function to construct longest array# with product K such that each element# is divisible by its previous elementdef findLongestArray(K): # Stores the prime factors of K primefactors = [] K_temp = K i = 2 while i * i <= K: # Stores the power to which # primefactor i is raised count = 0 while (K_temp % i == 0): K_temp //= i count += 1 if (count > 0): primefactors.append([count, i]) i += 1 if (K_temp != 1): primefactors.append( [1, K_temp]) # Sort prime factors in descending order primefactors.sort() # Stores the final array answer = [primefactors[0][0], primefactors[0][1]] # Multiply the last element by K answer[-1] *= K for i in range(primefactors[0][0]): answer[-1] //= primefactors[0][1] # Print the constructed array print("{", end = "") for i in range(len(answer)): if (i == len(answer) - 1): print(answer[i], end = "}") else: print(answer[i], end = ", ")# Driver Codeif __name__ == "__main__": K = 4 findLongestArray(K) # This code is contributed by ukasp. |
C#
using System;using System.Collections.Generic;using System.Linq;namespace ConsoleApp{ class Program { // Function to construct longest array // with product K such that each element // is divisible by its previous element static void FindLongestArray(int K) { // Stores the prime factors of K List<int[]> primefactors = new List<int[]>(); int K_temp = K; for (int i = 2; i * i <= K; i++) { // Stores the power to which // primefactor i is raised int count = 0; while (K_temp % i == 0) { K_temp /= i; count++; } if (count > 0) primefactors.Add(new int[] { count, i }); } if (K_temp != 1) primefactors.Add(new int[] { 1, K_temp }); // Sort prime factors in descending order primefactors = primefactors.OrderByDescending(x => x[0]).ThenByDescending(y => y[1]).ToList(); // Stores the final array int n = primefactors[0][0]; int val = primefactors[0][1]; int[] answer = new int[n]; for (int i = 0; i < answer.Length; i++) { answer[i] = val; } // Multiply the last element by K answer[n - 1] *= K; for (int i = 0; i < n; i++) { answer[n - 1] /= val; } // Print the constructed array Console.Write("{"); for (int i = 0; i < answer.Length; i++) { if (i == answer.Length - 1) Console.Write(answer[i] + "}"); else Console.Write(answer[i] + ", "); } } // Driver Code static void Main(string[] args) { int K = 4; FindLongestArray(K); } }}// This code is contributed by phasing17. |
Javascript
<script>// JavaScript program for the above approach// Function to construct longest array // with product K such that each element // is divisible by its previous elementfunction findLongestArray(K){ // Stores the prime factors of K let primefactors = []; let K_temp = K; for (let i = 2; i * i <= K; i++) { // Stores the power to which // primefactor i is raised let count = 0; while (K_temp % i == 0) { K_temp = Math.floor(K_temp/i); count++; } if (count > 0) primefactors.push([ count, i ]); } if (K_temp != 1) primefactors.push([ 1, K_temp ]); // Sort prime factors in descending order primefactors.sort(function(x, y) { if (x[0] != y[0]) return y[0] - x[0]; return y[1] - x[1]; }); // Stores the final array let n = primefactors[0][0]; let val = primefactors[0][1]; let answer = new Array(n); for(let i=0;i<n;i++) { answer[i]=val; } // Multiply the last element by K answer[n - 1] *= K; for (let i = 0; i < n; i++) { answer[n - 1] = Math.floor(answer[n - 1]/val); } // Print the constructed array document.write("{"); for (let i = 0; i < answer.length; i++) { if (i == answer.length - 1) document.write(answer[i] + "}"); else document.write(answer[i] + ", "); }}// Driver Codelet K = 4;findLongestArray(K);// This code is contributed by avanitrachhadiya2155</script> |
Output:
{2, 2}
Time Complexity: O(√(K) * log(√(K)))
Auxiliary Space: O(√(K))
