# 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)

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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 = 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 > 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); ` `    ``int` `m = ``sizeof``(arr2) / ``sizeof``(arr2); ` ` `  `    ``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 s1 = ``new` `HashSet(); ` `    ``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 s2 = ``new` `HashSet(); ` ` `  `    ``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 = 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 s2 = ``new` `HashSet(); ` ` `  `    ``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 `

Output:

```8
```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

1

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.