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Generation of n numbers with given set of factors

  • Difficulty Level : Expert
  • Last Updated : 28 Apr, 2021
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Given an array of k numbers factor[], the task is to print first n numbers (in ascending order) whose factors are from the given array.

Examples: 

Input  : factor[] = {2, 3, 4, 7} 
         n = 8
Output : 2 3 4 6 7 8 9 10

Input  :  factor[] = {3, 5, 7}
          n = 10
Output :  3 5 6 7 9 10 12 14 15 18

This problem is mainly an extension of Ugly Number Problem. We create an auxiliary array next[] of size same as factor[] to Keep track of next multiples of factors. In Each iteration, we output the minimum element from next, then increment it by the respective factor. If the minimum value is equal to the previous output value, increment it (to avoid duplicates). Else we print the minimum value.

C++




// C++ program to generate n numbers with
// given factors
#include<bits/stdc++.h>
using namespace std;
 
// Generate n numbers with factors in factor[]
void generateNumbers(int factor[], int n, int k)
{
    // array of k to store next multiples of
    // given factors
    int next[k] = {0};
 
    // Prints n numbers 
    int output = 0;  // Next number to print as output
    for (int i=0; i<n; )
    {
        // Find the next smallest multiple
        int toincrement = 0;
        for (int j=0; j<k; j++)
            if (next[j] < next[toincrement])
                toincrement = j;
 
        // Printing minimum in each iteration
        // print the value if output is not equal to
        // current value(to avoid the duplicates)
        if (output != next[toincrement])
        {
            output = next[toincrement];
            printf("%d ",  next[toincrement]);
            i++;
        }
 
        // incrementing the current value by the
        // respective factor
        next[toincrement] += factor[toincrement];
    }
}
 
// Driver code
int main()
{
    int factor[] = {3, 5, 7};
    int n = 10;
    int k = sizeof(factor)/sizeof(factor[0]);
    generateNumbers(factor, n, k);
    return 0;
}

Java




// Java program to generate
// n numbers with given factors
import java.io.*;
 
class GFG
{
     
// Generate n numbers with
// factors in factor[]
static void generateNumbers(int factor[],
                            int n, int k)
{
    // array of k to store
    // next multiples of
    // given factors
    int next[] = new int[k];
 
    // Prints n numbers
    int output = 0; // Next number to
                    // print as output
    for (int i = 0; i < n;)
    {
        // Find the next
        // smallest multiple
        int toincrement = 0;
        for (int j = 0; j < k; j++)
            if (next[j] < next[toincrement])
                toincrement = j;
 
        // Printing minimum in
        // each iteration print
        // the value if output
        // is not equal to current
        // value(to avoid the
        // duplicates)
        if (output != next[toincrement])
        {
            output = next[toincrement];
            System.out.print(
                   next[toincrement] + " ");
            i++;
        }
 
        // incrementing the
        // current value by the
        // respective factor
        next[toincrement] +=
               factor[toincrement];
    }
}
 
// Driver code
public static void main (String[] args)
{
    int factor[] = {3, 5, 7};
    int n = 10;
    int k = factor.length;
    generateNumbers(factor, n, k);
}
}
 
// This code is contributed
// by ajit

Python3




# Python3 program to generate n
# numbers with given factors
 
# Generate n numbers with
# factors in factor[]
def generateNumbers(factor, n, k):
     
    # array of k to store next
    # multiples of given factors
    next = [0] * k;
 
    # Prints n numbers
    output = 0; # Next number to
                # print as output
    i = 0;
    while(i < n):
         
        # Find the next smallest
        # multiple
        toincrement = 0;
        for j in range(k):
            if(next[j] < next[toincrement]):
                toincrement = j;
 
        # Printing minimum in each
        # iteration print the value
        # if output is not equal to
        # current value(to avoid the
        # duplicates)
        if(output != next[toincrement]):
            output = next[toincrement];
            print(next[toincrement], end = " ");
            i += 1;
 
        # incrementing the current
        # value by the respective factor
        next[toincrement] += factor[toincrement];
 
# Driver code
factor = [3, 5, 7];
n = 10;
k = len(factor);
generateNumbers(factor, n, k);
     
# This code is contributed by mits

C#




// C# program to generate
// n numbers with given factors
using System;
 
class GFG
{
     
// Generate n numbers with
// factors in factor[]
static void generateNumbers(int []factor,
                            int n, int k)
{
    // array of k to store
    // next multiples of
    // given factors
    int []next = new int[k];
 
    // Prints n numbers
    int output = 0; // Next number to
                    // print as output
    for (int i = 0; i < n;)
    {
        // Find the next
        // smallest multiple
        int toincrement = 0;
        for (int j = 0; j < k; j++)
            if (next[j] < next[toincrement])
                toincrement = j;
 
        // Printing minimum in
        // each iteration print
        // the value if output
        // is not equal to current
        // value(to avoid the
        // duplicates)
        if (output != next[toincrement])
        {
            output = next[toincrement];
            Console.Write(
                next[toincrement] + " ");
            i++;
        }
 
        // incrementing the
        // current value by the
        // respective factor
        next[toincrement] +=
            factor[toincrement];
    }
}
 
// Driver code
static public void Main ()
{
    int []factor = {3, 5, 7};
    int n = 10;
    int k = factor.Length;
    generateNumbers(factor, n, k);
}
}
 
// This code is contributed
// by akt_mit

PHP




<?php
// PHP program to generate n numbers
// with given factors
 
// Generate n numbers with factors
// in factor[]
function generateNumbers($factor, $n, $k)
{
    // array of k to store next
    // multiples of given factors
    $next = array_fill(0, $k, 0);
 
    // Prints n numbers
    $output = 0; // Next number to print
                 // as output
    for ($i = 0; $i < $n; )
    {
        // Find the next smallest multiple
        $toincrement = 0;
        for ($j = 0; $j < $k; $j++)
            if ($next[$j] < $next[$toincrement])
                $toincrement = $j;
 
        // Printing minimum in each iteration
        // print the value if output is not
        // equal to current value(to avoid
        // the duplicates)
        if ($output != $next[$toincrement])
        {
            $output = $next[$toincrement];
            echo $next[$toincrement] . " ";
            $i++;
        }
 
        // incrementing the current value
        // by the respective factor
        $next[$toincrement] += $factor[$toincrement];
    }
}
 
// Driver code
$factor = array(3, 5, 7);
$n = 10;
$k = count($factor);
generateNumbers($factor, $n, $k);
 
// This code is contributed by mits
?>

Javascript




<script>
 
// Javascript program to generate
// n numbers with given factors
 
// Generate n numbers with
// factors in factor[]
function generateNumbers(factor, n, k)
{
     
    // Array of k to store
    // next multiples of
    // given factors
    let next = new Array(k);
    next.fill(0);
 
    // Prints n numbers
    let output = 0; // Next number to
                    // print as output
                     
    for(let i = 0; i < n;)
    {
         
        // Find the next
        // smallest multiple
        let toincrement = 0;
        for(let j = 0; j < k; j++)
            if (next[j] < next[toincrement])
                toincrement = j;
 
        // Printing minimum in
        // each iteration print
        // the value if output
        // is not equal to current
        // value(to avoid the
        // duplicates)
        if (output != next[toincrement])
        {
            output = next[toincrement];
            document.write(next[toincrement] + " ");
            i++;
        }
 
        // Incrementing the
        // current value by the
        // respective factor
        next[toincrement] += factor[toincrement];
    }
}
 
// Driver code
let factor = [ 3, 5, 7 ];
let n = 10;
let k = factor.length;
 
generateNumbers(factor, n, k);
 
// This code is contributed by divyesh072019
 
</script>

Output : 

3 5 6 7 9 10 12 14 15 18 

Time complexity – O(n * k) 
Auxiliary Space – O(k)

This article is contributed by Niteesh kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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