Given an array of distinct integers(considering only positive numbers) and a number ‘m’, find the number of triplets with product equal to ‘m’.
Examples:
Input : arr[] = { 1, 4, 6, 2, 3, 8} m = 24 Output : 3 {1, 4, 6} {1, 3, 8} {4, 2, 3} Input : arr[] = { 0, 4, 6, 2, 3, 8} m = 18 Output : 0
Asked in : Microsoft
A Naive approach is to consider each and every triplet one by one and count if their product is equal to m.
C++
// C++ program to count triplets with given // product m #include <iostream> using namespace std; // Function to count such triplets int countTriplets( int arr[], int n, int m) { int count = 0; // Consider all triplets and count if // their product is equal to m for ( int i = 0; i < n - 2; i++) for ( int j = i + 1; j < n - 1; j++) for ( int k = j + 1; k < n; k++) if (arr[i] * arr[j] * arr[k] == m) count++; return count; } // Drivers code int main() { int arr[] = { 1, 4, 6, 2, 3, 8 }; int n = sizeof (arr) / sizeof (arr[0]); int m = 24; cout << countTriplets(arr, n, m); return 0; } |
Java
// Java program to count triplets with given // product m class GFG { // Method to count such triplets static int countTriplets( int arr[], int n, int m) { int count = 0 ; // Consider all triplets and count if // their product is equal to m for ( int i = 0 ; i < n - 2 ; i++) for ( int j = i + 1 ; j < n - 1 ; j++) for ( int k = j + 1 ; k < n; k++) if (arr[i] * arr[j] * arr[k] == m) count++; return count; } // Driver method public static void main(String[] args) { int arr[] = { 1 , 4 , 6 , 2 , 3 , 8 }; int m = 24 ; System.out.println(countTriplets(arr, arr.length, m)); } } |
Python3
# Python3 program to count # triplets with given product m # Method to count such triplets def countTriplets(arr, n, m): count = 0 # Consider all triplets and count if # their product is equal to m for i in range (n - 2 ): for j in range (i + 1 , n - 1 ): for k in range (j + 1 , n): if (arr[i] * arr[j] * arr[k] = = m): count + = 1 return count # Driver code if __name__ = = "__main__" : arr = [ 1 , 4 , 6 , 2 , 3 , 8 ] m = 24 print (countTriplets(arr, len (arr), m)) # This code is contributed by Chitranayal |
C#
// C# program to count triplets // with given product m using System; public class GFG { // Method to count such triplets static int countTriplets( int [] arr, int n, int m) { int count = 0; // Consider all triplets and count if // their product is equal to m for ( int i = 0; i < n - 2; i++) for ( int j = i + 1; j < n - 1; j++) for ( int k = j + 1; k < n; k++) if (arr[i] * arr[j] * arr[k] == m) count++; return count; } // Driver method public static void Main() { int [] arr = { 1, 4, 6, 2, 3, 8 }; int m = 24; Console.WriteLine(countTriplets(arr, arr.Length, m)); } } // This code is contributed by Sam007 |
PHP
<?php // PHP program to count triplets // with given product m // Function to count such triplets function countTriplets( $arr , $n , $m ) { $count = 0; // Consider all triplets and count if // their product is equal to m for ( $i = 0; $i < $n - 2; $i ++) for ( $j = $i + 1; $j < $n - 1; $j ++) for ( $k = $j + 1; $k < $n ; $k ++) if ( $arr [ $i ] * $arr [ $j ] * $arr [ $k ] == $m ) $count ++; return $count ; } // Driver code $arr = array (1, 4, 6, 2, 3, 8); $n = sizeof( $arr ); $m = 24; echo countTriplets( $arr , $n , $m ); // This code is contributed by jit_t. ?> |
Output:
3
Time Complexity: O(n3)
An Efficient Method is to use Hashing.
- Store all the elements in a hash_map with their index.
- Consider all pairs(i, j) and check the following:
- If (arr[i]*arr[j] !=0 && (m % arr[i]*arr[j]) == 0), If yes, then search for ( m / (arr[i]*arr[j]) in the map.
- Also check m / (arr[i]*arr[j]) is not equal to arr[i] and arr[j].
- Also, check that the current triplet is not counted previously by using the index stored in the map.
- If all the above conditions are satisfied, then increment the count.
- Return count.
C++
// C++ program to count triplets with given // product m #include <bits/stdc++.h> using namespace std; // Function to count such triplets int countTriplets( int arr[], int n, int m) { // Store all the elements in a set unordered_map< int , int > occ; for ( int i = 0; i < n; i++) occ[arr[i]] = i; int count = 0; // Consider all pairs and check for a // third number so their product is equal to m for ( int i = 0; i < n - 1; i++) { for ( int j = i + 1; j < n; j++) { // Check if current pair divides m or not // If yes, then search for (m / arr[i]*arr[j]) if ((arr[i] * arr[j] <= m) && (arr[i] * arr[j] != 0) && (m % (arr[i] * arr[j]) == 0)) { int check = m / (arr[i] * arr[j]); auto it = occ.find(check); // Check if the third number is present // in the map and it is not equal to any // other two elements and also check if // this triplet is not counted already // using their indexes if (check != arr[i] && check != arr[j] && it != occ.end() && it->second > i && it->second > j) count++; } } } // Return number of triplets return count; } // Drivers code int main() { int arr[] = { 1, 4, 6, 2, 3, 8 }; int n = sizeof (arr) / sizeof (arr[0]); int m = 24; cout << countTriplets(arr, n, m); return 0; } |
Java
// Java program to count triplets with given // product m import java.util.HashMap; class GFG { // Method to count such triplets static int countTriplets( int arr[], int n, int m) { // Store all the elements in a set HashMap<Integer, Integer> occ = new HashMap<Integer, Integer>(n); for ( int i = 0 ; i < n; i++) occ.put(arr[i], i); int count = 0 ; // Consider all pairs and check for a // third number so their product is equal to m for ( int i = 0 ; i < n - 1 ; i++) { for ( int j = i + 1 ; j < n; j++) { // Check if current pair divides m or not // If yes, then search for (m / arr[i]*arr[j]) if ((arr[i] * arr[j] <= m) && (arr[i] * arr[j] != 0 ) && (m % (arr[i] * arr[j]) == 0 )) { int check = m / (arr[i] * arr[j]); occ.containsKey(check); // Check if the third number is present // in the map and it is not equal to any // other two elements and also check if // this triplet is not counted already // using their indexes if (check != arr[i] && check != arr[j] && occ.containsKey(check) && occ.get(check) > i && occ.get(check) > j) count++; } } } // Return number of triplets return count; } // Driver method public static void main(String[] args) { int arr[] = { 1 , 4 , 6 , 2 , 3 , 8 }; int m = 24 ; System.out.println(countTriplets(arr, arr.length, m)); } } |
Python3
# Python3 program for the above approach # Function to find the triplet def countTriplets(li,product): flag = 0 count = 0 # Consider all pairs and check # for a third number so their # product is equal to product for i in range ( len (li)): # Check if current pair # divides product or not # If yes, then search for # (product / li[i]*li[j]) if li[i]! = 0 and product % li[i] = = 0 : for j in range (i + 1 , len (li)): # Check if the third number is present # in the map and it is not equal to any # other two elements and also check if # this triplet is not counted already # using their indexes if li[j]! = 0 and product % (li[j] * li[i]) = = 0 : if product / / (li[j] * li[i]) in li: n = li.index(product / / (li[j] * li[i])) if n > i and n > j: flag = 1 count + = 1 print (count) # Driver code li = [ 1 , 4 , 6 , 2 , 3 , 8 ] product = 24 # Function call countTriplets(li,product) |
C#
// C# implementation of the above // approach using System; using System.Collections.Generic; class GFG{ // Method to count such triplets static int countTriplets( int [] arr, int n, int m) { // Store all the elements // in a set Dictionary< int , int > occ = new Dictionary< int , int >(n); for ( int i = 0; i < n; i++) occ.Add(arr[i], i); int count = 0; // Consider all pairs and // check for a third number // so their product is equal to m for ( int i = 0; i < n - 1; i++) { for ( int j = i + 1; j < n; j++) { // Check if current pair divides // m or not If yes, then search // for (m / arr[i]*arr[j]) if ((arr[i] * arr[j] <= m) && (arr[i] * arr[j] != 0) && (m % (arr[i] * arr[j]) == 0)) { int check = m / (arr[i] * arr[j]); //occ.containsKey(check); // Check if the third number // is present in the map and // it is not equal to any // other two elements and also // check if this triplet is not // counted already using their indexes if (check != arr[i] && check != arr[j] && occ.ContainsKey(check) && occ[check] > i && occ[check] > j) count++; } } } // Return number of triplets return count; } // Driver code static void Main() { int [] arr = {1, 4, 6, 2, 3, 8}; int m = 24; Console.WriteLine(countTriplets(arr, arr.Length, m)); } } // This code is contributed by divyeshrabadiya07 |
Output:
3
Time Complexity : O(n2)
Auxiliary Space : O(n)
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