Find the element having maximum premutiples in the array

• Last Updated : 01 Jun, 2021

Given an array arr[], the task is to find the element which has the maximum number of pre-multiples present in the set. For any index i, pre-multiple is the number which is multiple of i and is present before the ith index of the array. Also, print the count of maximum multiples of that element in that array.
Examples:

Input: arr[] = {8, 1, 28, 4, 2, 6, 7}
Output: Element = 2 , Count of Premultiples = 3
Explanation: For the array, arr[] = {8, 1, 28, 4, 2, 6, 7} the number 2 has maximum
number of premultiples i.e. {8, 28, 4}. Therefore count is 3.
Input: arr[] = {8, 12, 5, 8, 17, 5, 28, 4, 3, 8}
Output: Element = 4, 3, Count of Premultiples = 3
for the array, a[] = {8, 12, 5, 8, 17, 5, 6, 15, 4, 3, 8} the number 4 and 3 has maximum
number of premultiples i.e. {8, 12, 8} and {12, 6, 15}. Therefore count is 3.

Approach: The idea is to use another array to store the count of multiples of i before the index. The following steps can be followed to compute the result:

1. Iterate over every element of the array, and for each valid i, the count is equal to the number of valid indexes j < i, such that, the element at index j is divisible by the element at index i.
2. Store the value of the count of the element at index i of temp_count array.
3. Find the maximum element in array temp_count[] and store its value in max.
4. Iterate over every element of array temp_count, such that, if the element at index i of temp_count is equal to max than print the corresponding ith element of original array arr.
5. Finally, print the maximum value stored in max.

Below is the implementation of the above approach:

C++

 // C++ program to find the element which has maximum // number of premultiples and also print its count. #include using namespace std; #define MAX 1000   // Function to find the elements having // maximum number of premultiples. void printMaxMultiple(int arr[], int n) {       int i, j, count, max;       // Initialize of temp_count array with zero     int temp_count[n] = { 0 };       for (i = 1; i < n; i++) {         // Initialize count with zero for         // every ith element of arr[]         count = 0;           // Loop to calculate the count of multiples         // for every ith element of arr[] before it         for (j = 0; j < i; j++) {             // Condition to check whether the element             // at a[i] divides element at a[j]             if (arr[j] % arr[i] == 0)                 count = count + 1;         }         temp_count[i] = count;     }       cout<<"Element = ";     // To get the maximum value in temp_count[]     max = *max_element(temp_count, temp_count + n);       // To print all the elements having maximum     // number of multiples before them.     for (i = 0; i < n; i++) {         if (temp_count[i] == max)             cout << arr[i] << ", ";     }     cout << "Count of Premultiples = ";     // To print the count of maximum number     // of multiples     cout << max << "\n"; }   // Driver function int main() {     int arr[] = { 8, 6, 2, 5, 8, 6, 3, 4 };     int n = sizeof(arr) / sizeof(arr);     printMaxMultiple(arr, n);     return 0; }

Java

 // Java program to find the element which has maximum // number of premultiples and also print its count. import java.io.*; import java.util.Arrays;   class GFG{       public static int MAX = 1000;       // Function to find the elements having // maximum number of premultiples. public static void printMaxMultiple(int[] arr, int n) {     int i, j, count, max;           // Initialize of temp_count array with zero     int[] temp_count = new int[n];     for(i = 0; i < temp_count.length; i++)     {         temp_count[i] = 0;     }           for(i = 1; i < n; i++)     {        // Initialize count with zero for        // every ith element of arr[]        count = 0;              // Loop to calculate the count of multiples        // for every ith element of arr[] before it        for(j = 0; j < i; j++)        {           // Condition to check whether the element           // at a[i] divides element at a[j]           if (arr[j] % arr[i] == 0)               count = count + 1;        }        temp_count[i] = count;     }     System.out.print("Element = ");           // To get the maximum value in temp_count[]     max = Arrays.stream(temp_count).max().getAsInt();           // To print all the elements having maximum     // number of multiples before them.     for(i = 0; i < n; i++)     {        if (temp_count[i] == max)            System.out.print(arr[i] + ", ");     }     System.out.print("Count of Premultiples = ");           // To print the count of maximum number     // of multiples     System.out.println(max); }       // Driver Code public static void main(String[] args) {     int[] arr = { 8, 6, 2, 5, 8, 6, 3, 4 };     int n = arr.length;     printMaxMultiple(arr, n); } }   // This code is contributed by shubhamsingh10

Python3

 # Python3 program to find the element which has maximum # number of premultiples and also print count.   MAX = 1000   # Function to find the elements having # maximum number of premultiples. def printMaxMultiple(arr, n):           # Initialize of temp_count array with zero     temp_count= *n           for i in range(1, n):                   # Initialize count with zero for         # every ith element of arr[]         count = 0                   # Loop to calculate the count of multiples         # for every ith element of arr[] before it         for j in range(i):                           # Condition to check whether the element             # at a[i] divides element at a[j]             if (arr[j] % arr[i] == 0):                 count = count + 1                   temp_count[i] = count               print("Element = ",end="")     # To get the maximum value in temp_count[]     maxx = max(temp_count)           # To prall the elements having maximum     # number of multiples before them.     for i in range(n):         if (temp_count[i] == maxx):             print(arr[i],end=", ",sep="")           print("Count of Premultiples = ",end="")     # To print the count of the maximum number     # of multiples           print(maxx)   # Driver function   arr = [8, 6, 2, 5, 8, 6, 3, 4 ] n = len(arr) printMaxMultiple(arr, n)   # This code is contributed by shubhamsingh10

C#

 // C# program to find the element which has maximum // number of premultiples and also print its count. using System; using System.Linq;   class GFG {           // Function to find the elements having     // maximum number of premultiples.     public static void printMaxMultiple(int[] arr, int n)     {               int i, j, count, max;               // Initialize of temp_count array with zero         int[] temp_count = new int[n];         for(i = 0; i < temp_count.Length; i++)             temp_count[i] = 0;               for (i = 1; i < n; i++) {                           // Initialize count with zero for             // every ith element of arr[]             count = 0;                   // Loop to calculate the count of multiples             // for every ith element of arr[] before it             for (j = 0; j < i; j++) {                                   // Condition to check whether the element                 // at a[i] divides element at a[j]                 if (arr[j] % arr[i] == 0)                     count = count + 1;             }             temp_count[i] = count;         }               Console.Write("Element = ");                   // To get the maximum value in temp_count[]         max = temp_count.Max();;               // To print all the elements having maximum         // number of multiples before them.         for (i = 0; i < n; i++) {             if (temp_count[i] == max)                 Console.Write(arr[i]+ ", ");         }         Console.Write("Count of Premultiples = ");                   // To print the count of maximum         // number of multiples         Console.WriteLine(max);     }           // Driver function     public static void Main()     {         int[] arr = { 8, 6, 2, 5, 8, 6, 3, 4 };         int n = arr.Length;         printMaxMultiple(arr, n);     } }   // This code is contributed by Shubhamsingh10

Javascript



Output:

Element = 2, 3, 4, Count of Premultiples = 2

Time Complexity: O(N2)

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