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 numberunsigned 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 pairsint 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 codeint 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 approachimport 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 numberstatic 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 pairsstatic 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 codepublic 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 approachusing 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 numberstatic 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 pairsstatic 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 codepublic 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 numberfunction 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 pairsfunction 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 codelet 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)
METHOD 2:Using brute force method
APPROACH:
This program takes two arrays of integers as input, and then finds the total number of distinct pairs from these two arrays, where both numbers in the pair are “ugly” numbers. An ugly number is defined as a number whose only prime factors are 2, 3 or 5.
ALGORITHM:
1.Define a function is_ugly that checks whether a number is an ugly number or not.
2.Take two arrays as input and define a variable count as 0 to keep track of the count of pairs.
3.Loop through each element of arr1 and arr2 using nested for loops.
4.Multiply each element of arr1 with each element of arr2 and check whether their product is an ugly number or not using the is_ugly function.
5.If the product is an ugly number, increment the count variable.
6.Finally, print the count variable, which represents the total number of distinct pairs where both numbers in the pair are ugly numbers.
C++
#include <iostream>using namespace std;
// Function to check if a number is uglybool isUgly(int n)
{ while (n % 2 == 0) {
n /= 2;
}
while (n % 3 == 0) {
n /= 3;
}
while (n % 5 == 0) {
n /= 5;
}
return n == 1;
}int main()
{ int arr1[] = { 3, 7, 1 };
int arr2[] = { 5, 1, 10, 4 };
int count = 0;
// Iterate over the elements of both arrays
for (int i = 0; i < sizeof(arr1) / sizeof(arr1[0]);
i++) {
for (int j = 0; j < sizeof(arr2) / sizeof(arr2[0]);
j++) {
// Check if the product of the elements is an ugly number
if (isUgly(arr1[i] * arr2[j])) {
count++;
}
}
}
cout << count << endl;
return 0;
} |
Java
import java.util.Arrays;
// Function to check if a number is uglyclass Main {
public static boolean isUgly(int n)
{
while (n % 2 == 0) {
n /= 2;
}
while (n % 3 == 0) {
n /= 3;
}
while (n % 5 == 0) {
n /= 5;
}
return n == 1;
}
public static void main(String[] args)
{
int[] arr1 = { 3, 7, 1 };
int[] arr2 = { 5, 1, 10, 4 };
int count = 0;
// Iterate over the elements of both arrays
for (int i = 0; i < arr1.length; i++) {
for (int j = 0; j < arr2.length; j++) {
// Check if the product of the elements is
// an ugly number
if (isUgly(arr1[i] * arr2[j])) {
count++;
}
}
}
System.out.println(count);
}
} |
Python3
def is_ugly(n):
while n % 2 == 0:
n //= 2
while n % 3 == 0:
n //= 3
while n % 5 == 0:
n //= 5
return n == 1
arr1 = [3, 7, 1]
arr2 = [5, 1, 10, 4]
count = 0
for i in range(len(arr1)):
for j in range(len(arr2)):
if is_ugly(arr1[i] * arr2[j]):
count += 1
print(count)
|
C#
using System;
class Program {
// Function to check if a number is ugly
static bool IsUgly(int n)
{
while (n % 2 == 0) {
n /= 2;
}
while (n % 3 == 0) {
n /= 3;
}
while (n % 5 == 0) {
n /= 5;
}
return n == 1;
}
static void Main(string[] args)
{
int[] arr1 = { 3, 7, 1 };
int[] arr2 = { 5, 1, 10, 4 };
int count = 0;
// Iterate over the elements of both arrays
for (int i = 0; i < arr1.Length; i++) {
for (int j = 0; j < arr2.Length; j++) {
// Check if the product of the elements is
// an ugly number
if (IsUgly(arr1[i] * arr2[j])) {
count++;
}
}
}
Console.WriteLine(count);
}
} |
Javascript
// Function to check if a number is an ugly numberfunction is_ugly(n) {
while (n % 2 == 0) {
n /= 2;
}
while (n % 3 == 0) {
n /= 3;
}
while (n % 5 == 0) {
n /= 5;
}
return n == 1;
}let arr1 = [3, 7, 1];let arr2 = [5, 1, 10, 4];let count = 0;// Iterate over the elements of both arraysfor (let i = 0; i < arr1.length; i++) {
for (let j = 0; j < arr2.length; j++) {
// Check if the product of the elements is an ugly number
if (is_ugly(arr1[i] * arr2[j])) {
count += 1;
}
}
}console.log(count);// This code is contributed by Kanchan Agarwal |
Output
8
Time complexity: O(N^2), where N is the maximum length of the input arrays.
Auxiliary Space: O(1), as we are not using any extra space in the program apart from the input arrays and the count variable.