Find integers that divides maximum number of elements of the array
Given an array arr[] of integers, the task is to find the element (other than 1) which is the factor of maximum number of elements from the array. If multiple such factors exist, print all the factors in ascending order.
Examples:
Input: arr[] = {10, 20}
Output: 2 5 10
The factors of 10 are 1, 2, 5, 10.
The factors of 20 are 1, 2, 4, 5, 10, 20.
The factors other than 1 which occur most number of times (twice) are 2, 5, 10.Input: arr[] = {120, 15, 24, 63, 18}
Output: 3
Approach:
- Initialize two lists one to store the rank (the number of elements that the integer is a factor of) of the factor and another to store the factor.
- Start from 2 till the maximum element of the array.
- Count the number of elements from the array the current integer is a factor of.
- Add the count to the rank list and the integer to factor list.
- Find the integer with the maximum rank.
- Print all the elements with the same rank.
Below is the implementation of the above approach:
C++
// CPP implementation of the approach #include<bits/stdc++.h> using namespace std; // Function to print the integers that divide // the maximum number of elements from the array void maximumFactor(vector<int>arr) { // Initialize two lists // to store rank and factors int n = arr.size(); vector<int> rank; vector<int> factors; int max = *max_element(arr.begin(), arr.end()); // Start from 2 till the maximum element in arr for (int i = 2; i <= max; i++) { // Initialize a variable // to count the number of elements // it is a factor of int count = 0; for (int j = 0; j < n; j++) { if (arr[j] % i == 0) count+= 1; rank.push_back(count); factors.push_back(i); } } // Maximum rank in the rank list int m = *max_element(rank.begin(),rank.end()); for (int i = 0; i < rank.size(); i++) { // Print all the elements with rank m if (rank[i] == m) cout << factors[i] <<" "; } } // Driver code int main() { vector<int>arr = {120, 15, 24, 63, 18}; maximumFactor(arr); } // This code is contributed by // Surendra_Gangwar |
Java
// Java implementation of the approach import java.util.*; class GFG { // Function to print the integers that // divide the maximum number of // elements from the array static void maximumFactor(int []arr) { // Initialize two lists to store // rank and factors int[] rank = new int[Arrays.stream(arr).max().getAsInt() + 1]; int[] factors = new int[Arrays.stream(arr).max().getAsInt() + 1]; int g = 0; // Start from 2 till the maximum // element in arr for (int i = 2; i <= Arrays.stream(arr).max().getAsInt(); i++) { // Initialize a variable to count // the number of elements it is a // factor of int count = 0; for (int j = 0; j < arr.length; j++) if (arr[j] % i == 0) count += 1; rank[g] = count; factors[g] = i; g++; } // Maximum rank in the rank list int m = Arrays.stream(rank).max().getAsInt(); for (int i = 0; i < rank.length; i++) { // Print all the elements with rank m if (rank[i] == m) System.out.print(factors[i] + " "); } } // Driver code public static void main (String[] args) { int []arr = {120, 15, 24, 63, 18}; maximumFactor(arr); } } // This code is contributed by // chandan_jnu |
Python
# Python3 implementation of the approach # Function to print the integers that divide # the maximum number of elements from the array def maximumFactor(arr): # Initialize two lists # to store rank and factors rank, factors = [], [] # Start from 2 till the maximum element in arr for i in range(2, max(arr)+1): # Initialize a variable # to count the number of elements # it is a factor of count = 0 for j in arr: if j % i == 0:count+= 1 rank.append(count) factors.append(i) # Maximum rank in the rank list m = max(rank) for i in range(len(rank)): # Print all the elements with rank m if rank[i]== m: print(factors[i], end =" ") # Driver code arr = [120, 15, 24, 63, 18] maximumFactor(arr) |
C#
// C# implementation of the approach using System; using System.Collections; using System.Linq; class GFG { // Function to print the integers that // divide the maximum number of // elements from the array static void maximumFactor(int []arr) { // Initialize two lists to store // rank and factors int[] rank = new int[arr.Max() + 1]; int[] factors = new int[arr.Max() + 1]; int g = 0; // Start from 2 till the maximum // element in arr for (int i = 2; i <= arr.Max(); i++) { // Initialize a variable to count // the number of elements it is a // factor of int count = 0 ; for (int j = 0; j < arr.Length; j++) if (arr[j] % i == 0) count += 1; rank[g]=count; factors[g]=i; g++; } // Maximum rank in the rank list int m = rank.Max(); for (int i = 0; i < rank.Length; i++) { // Print all the elements with rank m if ((int)rank[i] == m) Console.Write(factors[i]+" "); } } // Driver code static void Main() { int []arr = {120, 15, 24, 63, 18}; maximumFactor(arr); } } // This code is contributed by chandan_jnu |
PHP
<?php // PHP implementation of the approach // Function to print the integers that // divide the maximum number of // elements from the array function maximumFactor($arr) { // Initialize two lists to store // rank and factors $rank = array(); $factors = array(); // Start from 2 till the maximum // element in arr for ($i = 2; $i <= max($arr); $i++) { // Initialize a variable to count // the number of elements it is a // factor of $count = 0 ; for ($j = 0; $j < sizeof($arr); $j++) if ($arr[$j] % $i == 0) $count += 1; array_push($rank, $count); array_push($factors, $i); } // Maximum rank in the rank list $m = max($rank); for ($i = 0; $i < sizeof($rank); $i++) { // Print all the elements with rank m if ($rank[$i] == $m) echo $factors[$i], " "; } } // Driver code $arr = array(120, 15, 24, 63, 18); maximumFactor($arr) // This code is contributed by Ryuga ?> |
Output:
3
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