Zählen Sie Paare, deren Produkte im Array vorhanden sind
Bei einem gegebenen Array werden die Paare gezählt, deren Produktwert im Array vorhanden ist.
Beispiele:
Input : arr[] = {6 2 4 12 5 3}
Output : 3
All pairs whose product exist in array
(6 2) (2 3) (4 3)
Input : arr[] = {3 5 2 4 15 8}
Output : 2
C++
A Einfache Lösung besteht darin, alle Paare eines bestimmten Arrays zu generieren und zu prüfen, ob das Produkt im Array vorhanden ist. Wenn vorhanden, wird die Anzahl erhöht. Zum Schluss zählen wir zurück.
Nachfolgend finden Sie die Umsetzung der obigen Idee
Java// C++ program to count pairs whose product exist in array #includeusing namespace std ; // Returns count of pairs whose product exists in arr[] int countPairs ( int arr [] int n ) { int result = 0 ; for ( int i = 0 ; i < n ; i ++ ) { for ( int j = i + 1 ; j < n ; j ++ ) { int product = arr [ i ] * arr [ j ] ; // find product in an array for ( int k = 0 ; k < n ; k ++ ) { // if product found increment counter if ( arr [ k ] == product ) { result ++ ; break ; } } } } // return Count of all pair whose product exist in array return result ; } //Driver program int main () { int arr [] = { 6 2 4 12 5 3 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ]); cout < < countPairs ( arr n ); return 0 ; } Python 3// Java program to count pairs // whose product exist in array import java.io.* ; class GFG { // Returns count of pairs // whose product exists in arr[] static int countPairs ( int arr [] int n ) { int result = 0 ; for ( int i = 0 ; i < n ; i ++ ) { for ( int j = i + 1 ; j < n ; j ++ ) { int product = arr [ i ] * arr [ j ] ; // find product // in an array for ( int k = 0 ; k < n ; k ++ ) { // if product found // increment counter if ( arr [ k ] == product ) { result ++ ; break ; } } } } // return Count of all pair // whose product exist in array return result ; } // Driver Code public static void main ( String [] args ) { int arr [] = { 6 2 4 12 5 3 } ; int n = arr . length ; System . out . println ( countPairs ( arr n )); } } // This code is contributed by anuj_67.C## Python program to count pairs whose # product exist in array # Returns count of pairs whose # product exists in arr[] def countPairs ( arr n ): result = 0 ; for i in range ( 0 n ): for j in range ( i + 1 n ): product = arr [ i ] * arr [ j ] ; # find product in an array for k in range ( 0 n ): # if product found increment counter if ( arr [ k ] == product ): result = result + 1 ; break ; # return Count of all pair whose # product exist in array return result ; # Driver program arr = [ 6 2 4 12 5 3 ] ; n = len ( arr ); print ( countPairs ( arr n )); # This code is contributed # by Shivi_AggarwalJavaScript// C# program to count pairs // whose product exist in array using System ; class GFG { // Returns count of pairs // whose product exists in arr[] public static int countPairs ( int [] arr int n ) { int result = 0 ; for ( int i = 0 ; i < n ; i ++ ) { for ( int j = i + 1 ; j < n ; j ++ ) { int product = arr [ i ] * arr [ j ]; // find product in an array for ( int k = 0 ; k < n ; k ++ ) { // if product found // increment counter if ( arr [ k ] == product ) { result ++ ; break ; } } } } // return Count of all pair // whose product exist in array return result ; } // Driver Code public static void Main ( string [] args ) { int [] arr = new int [] { 6 2 4 12 5 3 }; int n = arr . Length ; Console . WriteLine ( countPairs ( arr n )); } } // This code is contributed by Shrikant13PHP< script > // javascript program to count pairs // whose product exist in array // Returns count of pairs // whose product exists in arr function countPairs ( arr n ) { var result = 0 ; for ( i = 0 ; i < n ; i ++ ) { for ( j = i + 1 ; j < n ; j ++ ) { var product = arr [ i ] * arr [ j ]; // find product // in an array for ( k = 0 ; k < n ; k ++ ) { // if product found // increment counter if ( arr [ k ] == product ) { result ++ ; break ; } } } } // return Count of all pair // whose product exist in array return result ; } // Driver Code var arr = [ 6 2 4 12 5 3 ]; var n = arr . length ; document . write ( countPairs ( arr n )); // This code is contributed by Rajput-Ji < /script>Ausgabe:
3
Zeitkomplexität: An 3 )Hilfsraum: O(1)
C++
Ein Effiziente Lösung besteht darin, „Hash“ zu verwenden, der alle Array-Elemente speichert. Generieren Sie alle möglichen Paare des gegebenen Arrays „arr“ und prüfen Sie, ob das Produkt jedes Paares im „Hash“ enthalten ist. Wenn vorhanden, wird die Anzahl erhöht. Zum Schluss zählen wir zurück.
Nachfolgend finden Sie die Umsetzung der obigen Idee
Java// A hashing based C++ program to count pairs whose product // exists in arr[] #includeusing namespace std ; // Returns count of pairs whose product exists in arr[] int countPairs ( int arr [] int n ) { int result = 0 ; // Create an empty hash-set that store all array element set < int > Hash ; // Insert all array element into set for ( int i = 0 ; i < n ; i ++ ) Hash . insert ( arr [ i ]); // Generate all pairs and check is exist in 'Hash' or not for ( int i = 0 ; i < n ; i ++ ) { for ( int j = i + 1 ; j < n ; j ++ ) { int product = arr [ i ] * arr [ j ]; // if product exists in set then we increment // count by 1 if ( Hash . find ( product ) != Hash . end ()) result ++ ; } } // return count of pairs whose product exist in array return result ; } // Driver program int main () { int arr [] = { 6 2 4 12 5 3 }; int n = sizeof ( arr ) / sizeof ( arr [ 0 ]); cout < < countPairs ( arr n ) ; return 0 ; } Python3// A hashing based Java program to count pairs whose product // exists in arr[] import java.util.* ; class GFG { // Returns count of pairs whose product exists in arr[] static int countPairs ( int arr [] int n ) { int result = 0 ; // Create an empty hash-set that store all array element HashSet < Integer > Hash = new HashSet <> (); // Insert all array element into set for ( int i = 0 ; i < n ; i ++ ) { Hash . add ( arr [ i ] ); } // Generate all pairs and check is exist in 'Hash' or not for ( int i = 0 ; i < n ; i ++ ) { for ( int j = i + 1 ; j < n ; j ++ ) { int product = arr [ i ] * arr [ j ] ; // if product exists in set then we increment // count by 1 if ( Hash . contains ( product )) { result ++ ; } } } // return count of pairs whose product exist in array return result ; } // Driver program public static void main ( String [] args ) { int arr [] = { 6 2 4 12 5 3 }; int n = arr . length ; System . out . println ( countPairs ( arr n )); } } // This code has been contributed by 29AjayKumarC## A hashing based C++ program to count # pairs whose product exists in arr[] # Returns count of pairs whose product # exists in arr[] def countPairs ( arr n ): result = 0 # Create an empty hash-set that # store all array element Hash = set () # Insert all array element into set for i in range ( n ): Hash . add ( arr [ i ]) # Generate all pairs and check is # exist in 'Hash' or not for i in range ( n ): for j in range ( i + 1 n ): product = arr [ i ] * arr [ j ] # if product exists in set then # we increment count by 1 if product in ( Hash ): result += 1 # return count of pairs whose # product exist in array return result # Driver Code if __name__ == '__main__' : arr = [ 6 2 4 12 5 3 ] n = len ( arr ) print ( countPairs ( arr n )) # This code is contributed by # Sanjit_PrasadJavaScript// A hashing based C# program to count pairs whose product // exists in arr[] using System ; using System.Collections.Generic ; class GFG { // Returns count of pairs whose product exists in arr[] static int countPairs ( int [] arr int n ) { int result = 0 ; // Create an empty hash-set that store all array element HashSet < int > Hash = new HashSet < int > (); // Insert all array element into set for ( int i = 0 ; i < n ; i ++ ) { Hash . Add ( arr [ i ]); } // Generate all pairs and check is exist in 'Hash' or not for ( int i = 0 ; i < n ; i ++ ) { for ( int j = i + 1 ; j < n ; j ++ ) { int product = arr [ i ] * arr [ j ]; // if product exists in set then we increment // count by 1 if ( Hash . Contains ( product )) { result ++ ; } } } // return count of pairs whose product exist in array return result ; } // Driver code public static void Main ( String [] args ) { int [] arr = { 6 2 4 12 5 3 }; int n = arr . Length ; Console . WriteLine ( countPairs ( arr n )); } } /* This code contributed by PrinciRaj1992 */Ausgabe:
3
Zeitkomplexität: An 2 ) „Unter der Annahme, dass die Einfüge- und Suchoperation O(1) Zeit in Anspruch nimmt“Hilfsraum: An)
Methode 3: Verwendung einer ungeordneten Karte
Ansatz:
1.Erstellen Sie eine leere Karte, um die Elemente des Arrays und ihre Häufigkeiten zu speichern.
2. Durchlaufen Sie das Array und fügen Sie jedes Element zusammen mit seiner Häufigkeit in die Karte ein.
3. Initialisieren Sie eine Zählvariable auf 0, um die Anzahl der Paare zu verfolgen.
4. Durchlaufen Sie das Array erneut und prüfen Sie für jedes Element, ob es einen Faktor (außer sich selbst) hat, der in der Karte vorhanden ist.
5. Wenn beide Faktoren in der Karte vorhanden sind, erhöhen Sie die Anzahl der Paare.
6.Gibt die Anzahl der Paare zurück.Durchführung:
C++Java#includeusing namespace std ; // Function to count pairs whose product value is present in array int count_Pairs ( int arr [] int n ) { map < int int > mp ; // Create a map to store the elements of the array and their frequencies // Initialize the map with the frequencies of the elements in the array for ( int i = 0 ; i < n ; i ++ ) { mp [ arr [ i ]] ++ ; } int count = 0 ; // Initialize the count of pairs to zero // Traverse the array and check if arr[i] has a factor in the map for ( int i = 0 ; i < n ; i ++ ) { for ( int j = 1 ; j * j <= arr [ i ]; j ++ ) { if ( arr [ i ] % j == 0 ) { int factor1 = j ; int factor2 = arr [ i ] / j ; // If both factors are present in the map then increment the count of pairs if ( mp . count ( factor1 ) && mp . count ( factor2 )) { if ( factor1 == factor2 && mp [ factor1 ] < 2 ) { continue ; } count ++ ; } } } } // Return the count of pairs return count ; } // Driver code int main () { // Example input int arr [] = { 6 2 4 12 5 3 }; int n = sizeof ( arr ) / sizeof ( arr [ 0 ]); // Count pairs whose product value is present in array int count = count_Pairs ( arr n ); // Print the count cout < < count < < endl ; return 0 ; } Pythonimport java.util.HashMap ; import java.util.Map ; public class Main { // Function to count pairs whose product value is // present in the array static int countPairs ( int [] arr ) { Map < Integer Integer > frequencyMap = new HashMap <> (); // Initialize the map with the frequencies of the // elements in the array for ( int num : arr ) { frequencyMap . put ( num frequencyMap . getOrDefault ( num 0 ) + 1 ); } int count = 0 ; // Initialize the count of pairs to zero // Traverse the array and check if arr[i] has a // factor in the map for ( int num : arr ) { for ( int j = 1 ; j * j <= num ; j ++ ) { if ( num % j == 0 ) { int factor1 = j ; int factor2 = num / j ; // If both factors are present in the // map then increment the count of // pairs if ( frequencyMap . containsKey ( factor1 ) && frequencyMap . containsKey ( factor2 )) { if ( factor1 == factor2 && frequencyMap . get ( factor1 ) < 2 ) { continue ; } count ++ ; } } } } // Return the count of pairs return count ; } public static void main ( String [] args ) { // Example input int [] arr = { 6 2 4 12 5 3 }; // Count pairs whose product value is present in the // array int count = countPairs ( arr ); // Print the count System . out . println ( count ); } }C## Function to count pairs whose product value is present in the array def count_pairs ( arr ): # Create a dictionary to store the elements of the array and their frequencies mp = {} # Initialize the dictionary with the frequencies of the elements in the array for num in arr : if num in mp : mp [ num ] += 1 else : mp [ num ] = 1 count = 0 # Initialize the count of pairs to zero # Traverse the array and check if arr[i] has a factor in the dictionary for num in arr : for j in range ( 1 int ( num ** 0.5 ) + 1 ): if num % j == 0 : factor1 = j factor2 = num // j # If both factors are present in the dictionary # then increment the count of pairs if factor1 in mp and factor2 in mp : if factor1 == factor2 and mp [ factor1 ] < 2 : continue count += 1 return count # Driver code if __name__ == '__main__' : # Example input arr = [ 6 2 4 12 5 3 ] # Count pairs whose product value is present in the array count = count_pairs ( arr ) # Print the count print ( count )JavaScriptusing System ; using System.Collections.Generic ; class GFG { // Function to count pairs whose product value is // present in array static int CountPairs ( int [] arr int n ) { Dictionary < int int > mp = new Dictionary < int int > (); // Create a dictionary to store the // elements of the array and their // frequencies // Initialize the dictionary with the frequencies of // the elements in the array for ( int i = 0 ; i < n ; i ++ ) { if ( ! mp . ContainsKey ( arr [ i ])) mp [ arr [ i ]] = 1 ; else mp [ arr [ i ]] ++ ; } int count = 0 ; // Initialize the count of pairs to zero // Traverse the array and check if arr[i] has a // factor in the dictionary for ( int i = 0 ; i < n ; i ++ ) { for ( int j = 1 ; j * j <= arr [ i ]; j ++ ) { if ( arr [ i ] % j == 0 ) { int factor1 = j ; int factor2 = arr [ i ] / j ; // If both factors are present in the // dictionary then increment the count // of pairs if ( mp . ContainsKey ( factor1 ) && mp . ContainsKey ( factor2 )) { if ( factor1 == factor2 && mp [ factor1 ] < 2 ) { continue ; } count ++ ; } } } } // Return the count of pairs return count ; } // Driver code static void Main ( string [] args ) { // Example input int [] arr = { 6 2 4 12 5 3 }; int n = arr . Length ; // Count pairs whose product value is present in // array int count = CountPairs ( arr n ); // Print the count Console . WriteLine ( count ); } }Ausgabe:
3
Zeitkomplexität: O(n log n)
Hilfsraum: An)
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