Total distinct pairs of ugly numbers from two arrays

Given two arrays arr1[] and arr2[] of sizes N and M where 0 ≤ arr1[i], arr2[i] ≤ 1000 for all valid i, the task is to take one element from first array and one element from second array such that both of them are ugly numbers. We call it a pair (a, b). You have to find the count of all such distinct pairs. Note that (a, b) and (b, a) are not distinct. 
Ugly numbers are numbers whose only prime factors are 2, 3 or 5. 
The sequence 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ….. shows first few ugly numbers. By convention, 1 is included.
Examples:

Input: arr1[] = {7, 2, 3, 14}, arr2[] = {2, 11, 10} 
Output: 4 
All distinct pairs are (2, 2), (2, 10), (3, 2) and (3, 10)
Input: arr1[] = {1, 2, 3}, arr2[] = {1, 1} 
Output: 3 
All distinct pairs are (1, 1), (1, 2) and (1, 3) 

Approach:

• First generate all ugly numbers and insert them in a unordered_set s1.
• Take another empty set s2.
• Run two nested loops to generate all possible pairs from the two arrays taking one element from first array(call it a) and one from second array(call it b).
• Check if a is present in s1. If yes then check for each element of arr2[] if it is also present in s1.
• If both a and b are ugly numbers, then insert pair (a, b) in s2 if a is less than b, or (b, a) otherwise. This is done to avoid duplicacy.
• Total pairs is the size of the set s2.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to get the nth ugly number
unsigned uglyNumber(int n)
{

    // To store ugly numbers
    int ugly[n];
    int i2 = 0, i3 = 0, i5 = 0;
    int next_multiple_of_2 = 2;
    int next_multiple_of_3 = 3;
    int next_multiple_of_5 = 5;
    int next_ugly_no = 1;

    ugly[0] = 1;
    for (int i = 1; i < n; i++) {
        next_ugly_no = min(next_multiple_of_2,
                           min(next_multiple_of_3,
                               next_multiple_of_5));
        ugly[i] = next_ugly_no;
        if (next_ugly_no == next_multiple_of_2) {
            i2 = i2 + 1;
            next_multiple_of_2 = ugly[i2] * 2;
        }
        if (next_ugly_no == next_multiple_of_3) {
            i3 = i3 + 1;
            next_multiple_of_3 = ugly[i3] * 3;
        }
        if (next_ugly_no == next_multiple_of_5) {
            i5 = i5 + 1;
            next_multiple_of_5 = ugly[i5] * 5;
        }
    }

    return next_ugly_no;
}

// Function to return the required count of pairs
int totalPairs(int arr1[], int arr2[], int n, int m)
{
    unordered_set<int> s1;
    int i = 1;

    // Insert ugly numbers in set
    // which are less than 1000
    while (1) {
        int next_ugly_number = uglyNumber(i);
        if (next_ugly_number > 1000)
            break;
        s1.insert(next_ugly_number);
        i++;
    }

    // Set is used to avoid duplicate pairs
    set<pair<int, int> > s2;

    for (int i = 0; i < n; i++) {

        // Check if arr1[i] is an ugly number
        if (s1.find(arr1[i]) != s1.end()) {

            for (int j = 0; j < m; j++) {

                // Check if arr2[i] is an ugly number
                if (s1.find(arr2[j]) != s1.end()) {
                    if (arr1[i] < arr2[j])
                        s2.insert(make_pair(arr1[i], arr2[j]));
                    else
                        s2.insert(make_pair(arr2[j], arr1[i]));
                }
            }
        }
    }

    // Return the size of the set s2
    return s2.size();
}

// Driver code
int main()
{
    int arr1[] = { 3, 7, 1 };
    int arr2[] = { 5, 1, 10, 4 };
    int n = sizeof(arr1) / sizeof(arr1[0]);
    int m = sizeof(arr2) / sizeof(arr2[0]);

    cout << totalPairs(arr1, arr2, n, m);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{
static class pair
{
    int first, second;
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}

// Function to get the nth ugly number
static int uglyNumber(int n)
{

    // To store ugly numbers
    int []ugly = new int[n];
    int i2 = 0, i3 = 0, i5 = 0;
    int next_multiple_of_2 = 2;
    int next_multiple_of_3 = 3;
    int next_multiple_of_5 = 5;
    int next_ugly_no = 1;

    ugly[0] = 1;
    for (int i = 1; i < n; i++)
    {
        next_ugly_no = Math.min(next_multiple_of_2,
                       Math.min(next_multiple_of_3,
                                next_multiple_of_5));
        ugly[i] = next_ugly_no;
        if (next_ugly_no == next_multiple_of_2)
        {
            i2 = i2 + 1;
            next_multiple_of_2 = ugly[i2] * 2;
        }
        if (next_ugly_no == next_multiple_of_3)
        {
            i3 = i3 + 1;
            next_multiple_of_3 = ugly[i3] * 3;
        }
        if (next_ugly_no == next_multiple_of_5)
        {
            i5 = i5 + 1;
            next_multiple_of_5 = ugly[i5] * 5;
        }
    }

    return next_ugly_no;
}

// Function to return the required count of pairs
static int totalPairs(int arr1[], int arr2[],
                      int n, int m)
{
    HashSet<Integer> s1 = new HashSet<Integer>();
    int i = 1;

    // Insert ugly numbers in set
    // which are less than 1000
    while (true)
    {
        int next_ugly_number = uglyNumber(i);
        if (next_ugly_number > 1000)
            break;
        s1.add(next_ugly_number);
        i++;
    }

    // Set is used to avoid duplicate pairs
    HashSet<pair> s2 = new HashSet<pair>();

    for (i = 0; i < n; i++)
    {

        // Check if arr1[i] is an ugly number
        if (s1.contains(arr1[i]))
        {
            for (int j = 0; j < m; j++)
            {

                // Check if arr2[i] is an ugly number
                if (s1.contains(arr2[j]))
                {
                    if (arr1[i] < arr2[j])
                        s2.add(new pair(arr1[i], arr2[j]));
                    else
                        s2.add(new pair(arr2[j], arr1[i]));
                }
            }
        }
    }

    // Return the size of the set s2
    return s2.size();
}

// Driver code
public static void main(String[] args)
{
    int arr1[] = { 3, 7, 1 };
    int arr2[] = { 5, 1, 10, 4 };
    int n = arr1.length;
    int m = arr2.length;

    System.out.println(totalPairs(arr1, arr2, n, m));
}
}

// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach

# Function to get the nth ugly number
def uglyNumber(n):

    # To store ugly numbers
    ugly = [None] * n
    i2 = i3 = i5 = 0
    next_multiple_of_2 = 2
    next_multiple_of_3 = 3
    next_multiple_of_5 = 5
    next_ugly_no = 1

    ugly[0] = 1
    for i in range(1, n):
        next_ugly_no = min(next_multiple_of_2,
                        min(next_multiple_of_3,
                            next_multiple_of_5))
        ugly[i] = next_ugly_no
        if (next_ugly_no == next_multiple_of_2):
            i2 = i2 + 1
            next_multiple_of_2 = ugly[i2] * 2

        if (next_ugly_no == next_multiple_of_3):
            i3 = i3 + 1
            next_multiple_of_3 = ugly[i3] * 3

        if (next_ugly_no == next_multiple_of_5):
            i5 = i5 + 1
            next_multiple_of_5 = ugly[i5] * 5

    return next_ugly_no

# Function to return the required count of pairs
def totalPairs(arr1, arr2, n, m):

    s1 = set()
    i = 1

    # Insert ugly numbers in set
    # which are less than 1000
    while True:
        next_ugly_number = uglyNumber(i)
        if (next_ugly_number > 1000):
            break
        s1.add(next_ugly_number)
        i += 1

    # Set is used to avoid duplicate pairs
    s2 = set()

    for i in range(0, n):

        # Check if arr1[i] is an ugly number
        if arr1[i] in s1:

            for j in range(0, m):

                # Check if arr2[i] is an ugly number
                if arr2[j] in s1:
                    if (arr1[i] < arr2[j]):
                        s2.add((arr1[i], arr2[j]))
                    else:
                        s2.add((arr2[j], arr1[i]))

    # Return the size of the set s2
    return len(s2)

# Driver code
if __name__ == "__main__":

    arr1 = [3, 7, 1]
    arr2 = [5, 1, 10, 4]
    n = len(arr1)
    m = len(arr2)

    print(totalPairs(arr1, arr2, n, m))

# This code is contributed by Rituraj Jain

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;

class GFG
{
public class pair
{
    public int first, second;
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}

// Function to get the nth ugly number
static int uglyNumber(int n)
{

    // To store ugly numbers
    int []ugly = new int[n];
    int i2 = 0, i3 = 0, i5 = 0;
    int next_multiple_of_2 = 2;
    int next_multiple_of_3 = 3;
    int next_multiple_of_5 = 5;
    int next_ugly_no = 1;

    ugly[0] = 1;
    for (int i = 1; i < n; i++)
    {
        next_ugly_no = Math.Min(next_multiple_of_2,
                       Math.Min(next_multiple_of_3,
                                next_multiple_of_5));
        ugly[i] = next_ugly_no;
        if (next_ugly_no == next_multiple_of_2)
        {
            i2 = i2 + 1;
            next_multiple_of_2 = ugly[i2] * 2;
        }
        if (next_ugly_no == next_multiple_of_3)
        {
            i3 = i3 + 1;
            next_multiple_of_3 = ugly[i3] * 3;
        }
        if (next_ugly_no == next_multiple_of_5)
        {
            i5 = i5 + 1;
            next_multiple_of_5 = ugly[i5] * 5;
        }
    }

    return next_ugly_no;
}

// Function to return the required count of pairs
static int totalPairs(int []arr1, int []arr2,
                      int n, int m)
{
    HashSet<int> s1 = new HashSet<int>();
    int i = 1;

    // Insert ugly numbers in set
    // which are less than 1000
    while (true)
    {
        int next_ugly_number = uglyNumber(i);
        if (next_ugly_number > 1000)
            break;
        s1.Add(next_ugly_number);
        i++;
    }

    // Set is used to avoid duplicate pairs
    HashSet<pair> s2 = new HashSet<pair>();

    for (i = 0; i < n; i++)
    {

        // Check if arr1[i] is an ugly number
        if (s1.Contains(arr1[i]))
        {
            for (int j = 0; j < m; j++)
            {

                // Check if arr2[i] is an ugly number
                if (s1.Contains(arr2[j]))
                {
                    if (arr1[i] < arr2[j])
                        s2.Add(new pair(arr1[i],
                                        arr2[j]));
                    else
                        s2.Add(new pair(arr2[j],
                                        arr1[i]));
                }
            }
        }
    }

    // Return the size of the set s2
    return s2.Count;
}

// Driver code
public static void Main(String[] args)
{
    int []arr1 = { 3, 7, 1 };
    int []arr2 = { 5, 1, 10, 4 };
    int n = arr1.Length;
    int m = arr2.Length;

    Console.WriteLine(totalPairs(arr1, arr2, n, m));
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// JavaScript implementation of the approach


// Function to get the nth ugly number
function uglyNumber(n) {

    // To store ugly numbers
    let ugly = new Array(n);
    let i2 = 0, i3 = 0, i5 = 0;
    let next_multiple_of_2 = 2;
    let next_multiple_of_3 = 3;
    let next_multiple_of_5 = 5;
    let next_ugly_no = 1;

    ugly[0] = 1;
    for (let i = 1; i < n; i++) {
        next_ugly_no = Math.min(next_multiple_of_2,
            Math.min(next_multiple_of_3,
                next_multiple_of_5));
        ugly[i] = next_ugly_no;
        if (next_ugly_no == next_multiple_of_2) {
            i2 = i2 + 1;
            next_multiple_of_2 = ugly[i2] * 2;
        }
        if (next_ugly_no == next_multiple_of_3) {
            i3 = i3 + 1;
            next_multiple_of_3 = ugly[i3] * 3;
        }
        if (next_ugly_no == next_multiple_of_5) {
            i5 = i5 + 1;
            next_multiple_of_5 = ugly[i5] * 5;
        }
    }

    return next_ugly_no;
}

// Function to return the required count of pairs
function totalPairs(arr1, arr2, n, m) {
    let s1 = new Set();
    let i = 1;

    // Insert ugly numbers in set
    // which are less than 1000
    while (1) {
        let next_ugly_number = uglyNumber(i);
        if (next_ugly_number > 1000)
            break;
        s1.add(next_ugly_number);
        i++;
    }

    // Set is used to avoid duplicate pairs
    let s2 = new Set();

    for (let i = 0; i < n; i++) {

        // Check if arr1[i] is an ugly number
        if (s1.has(arr1[i])) {

            for (let j = 0; j < m; j++) {

                // Check if arr2[i] is an ugly number
                if (s1.has(arr2[j])) {
                    if (arr1[i] < arr2[j])
                        s2.add([arr1[i], arr2[j]]);
                    else
                        s2.add([arr2[j], arr1[i]]);
                }
            }
        }
    }

    // Return the size of the set s2
    return s2.size;
}

// Driver code

let arr1 = [3, 7, 1];
let arr2 = [5, 1, 10, 4];
let n = arr1.length;
let m = arr2.length;

document.write(totalPairs(arr1, arr2, n, m));

// This code is contributed by _saurabh_jaiswal

</script>

Output:

8

Time Complexity: O(k²+n*m*log(n+m)) where k = 1000 and n & m are the sizes of the array
Auxiliary Space: O(k+n+m)