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 that 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:
- 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.
- Store the value of the count of the element at index i of temp_count array.
- Find the maximum element in array temp_count[] and store its value in max.
- Iterate over every element of array temp_count, such that, if the element at index i of temp_count is equal to max then print the corresponding ith element of original array arr.
- Finally, print the maximum value stored in max.
Below is the implementation of the above approach:
// C++ program to find the element which has maximum // number of premultiples and also print its count. #include <bits/stdc++.h> 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[0]);
printMaxMultiple(arr, n);
return 0;
} |
// 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 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 = [ 0 ] * 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 print all 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# 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 |
<script> // JavaScript program to find
// the element which has maximum
// number of premultiples and also print its count.
// Function to find the elements having
// maximum number of premultiples.
function printMaxMultiple(arr,n)
{
let i, j, count, max;
// Initialize of temp_count array with zero
let temp_count = new Array(n);
temp_count.fill(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;
}
document.write( "Element = " );
// To get the maximum value in temp_count[]
max = Number.MIN_VALUE;
for (i = 0; i < n; i++)
{
max = Math.max(max, temp_count[i]);
}
// To print all the elements having maximum
// number of multiples before them.
for (i = 0; i < n; i++) {
if (temp_count[i] == max)
document.write(arr[i] + ", " );
}
document.write( "Count of Premultiples = " );
// To print the count of maximum number
// of multiples
document.write(max + "</br>" );
}
let arr = [ 8, 6, 2, 5, 8, 6, 3, 4 ];
let n = arr.length;
printMaxMultiple(arr, n);
</script> |
Element = 2, 3, 4, Count of Premultiples = 2
Time Complexity: O(N2)
Auxiliary Space: O(N), where N is the size of the given array.