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++
#include <bits/stdc++.h>
using namespace std;
unsigned uglyNumber( int n)
{
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;
}
int totalPairs( int arr1[], int arr2[], int n, int m)
{
unordered_set< int > s1;
int i = 1;
while (1) {
int next_ugly_number = uglyNumber(i);
if (next_ugly_number > 1000)
break ;
s1.insert(next_ugly_number);
i++;
}
set<pair< int , int > > s2;
for ( int i = 0; i < n; i++) {
if (s1.find(arr1[i]) != s1.end()) {
for ( int j = 0; j < m; j++) {
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 s2.size();
}
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
import java.util.*;
class GFG
{
static class pair
{
int first, second;
public pair( int first, int second)
{
this .first = first;
this .second = second;
}
}
static int uglyNumber( int n)
{
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;
}
static int totalPairs( int arr1[], int arr2[],
int n, int m)
{
HashSet<Integer> s1 = new HashSet<Integer>();
int i = 1 ;
while ( true )
{
int next_ugly_number = uglyNumber(i);
if (next_ugly_number > 1000 )
break ;
s1.add(next_ugly_number);
i++;
}
HashSet<pair> s2 = new HashSet<pair>();
for (i = 0 ; i < n; i++)
{
if (s1.contains(arr1[i]))
{
for ( int j = 0 ; j < m; j++)
{
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 s2.size();
}
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));
}
}
|
Python3
def uglyNumber(n):
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
def totalPairs(arr1, arr2, n, m):
s1 = set ()
i = 1
while True :
next_ugly_number = uglyNumber(i)
if (next_ugly_number > 1000 ):
break
s1.add(next_ugly_number)
i + = 1
s2 = set ()
for i in range ( 0 , n):
if arr1[i] in s1:
for j in range ( 0 , m):
if arr2[j] in s1:
if (arr1[i] < arr2[j]):
s2.add((arr1[i], arr2[j]))
else :
s2.add((arr2[j], arr1[i]))
return len (s2)
if __name__ = = "__main__" :
arr1 = [ 3 , 7 , 1 ]
arr2 = [ 5 , 1 , 10 , 4 ]
n = len (arr1)
m = len (arr2)
print (totalPairs(arr1, arr2, n, m))
|
C#
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;
}
}
static int uglyNumber( int n)
{
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;
}
static int totalPairs( int []arr1, int []arr2,
int n, int m)
{
HashSet< int > s1 = new HashSet< int >();
int i = 1;
while ( true )
{
int next_ugly_number = uglyNumber(i);
if (next_ugly_number > 1000)
break ;
s1.Add(next_ugly_number);
i++;
}
HashSet<pair> s2 = new HashSet<pair>();
for (i = 0; i < n; i++)
{
if (s1.Contains(arr1[i]))
{
for ( int j = 0; j < m; j++)
{
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 s2.Count;
}
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));
}
}
|
Javascript
<script>
function uglyNumber(n) {
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 totalPairs(arr1, arr2, n, m) {
let s1 = new Set();
let i = 1;
while (1) {
let next_ugly_number = uglyNumber(i);
if (next_ugly_number > 1000)
break ;
s1.add(next_ugly_number);
i++;
}
let s2 = new Set();
for (let i = 0; i < n; i++) {
if (s1.has(arr1[i])) {
for (let j = 0; j < m; j++) {
if (s1.has(arr2[j])) {
if (arr1[i] < arr2[j])
s2.add([arr1[i], arr2[j]]);
else
s2.add([arr2[j], arr1[i]]);
}
}
}
}
return s2.size;
}
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));
</script>
|
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;
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;
}
int main()
{
int arr1[] = { 3, 7, 1 };
int arr2[] = { 5, 1, 10, 4 };
int count = 0;
for ( int i = 0; i < sizeof (arr1) / sizeof (arr1[0]);
i++) {
for ( int j = 0; j < sizeof (arr2) / sizeof (arr2[0]);
j++) {
if (isUgly(arr1[i] * arr2[j])) {
count++;
}
}
}
cout << count << endl;
return 0;
}
|
Java
import java.util.Arrays;
class 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 ;
for ( int i = 0 ; i < arr1.length; i++) {
for ( int j = 0 ; j < arr2.length; j++) {
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 {
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;
for ( int i = 0; i < arr1.Length; i++) {
for ( int j = 0; j < arr2.Length; j++) {
if (IsUgly(arr1[i] * arr2[j])) {
count++;
}
}
}
Console.WriteLine(count);
}
}
|
Javascript
function 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;
for (let i = 0; i < arr1.length; i++) {
for (let j = 0; j < arr2.length; j++) {
if (is_ugly(arr1[i] * arr2[j])) {
count += 1;
}
}
}
console.log(count);
|
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.
Last Updated :
13 Nov, 2023
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