Finden Sie alle Drillinge in einem sortierten Array, das die geometrische Progression bildet
Geben Sie bei einem sortierten Array unterschiedlicher positiver Ganzzahlen alle Tripel aus, die eine geometrische Progression mit ganzzahligem gemeinsamem Verhältnis bilden.
Eine geometrische Folge ist eine Folge von Zahlen, bei der jeder Term nach dem ersten durch Multiplikation des vorherigen mit einer festen Zahl ungleich Null, dem sogenannten gemeinsamen Verhältnis, ermittelt wird. Beispielsweise ist die Folge 2 6 18 54... eine geometrische Folge mit dem gemeinsamen Verhältnis 3.
Beispiele:
Input: arr = [1 2 6 10 18 54] Output: 2 6 18 6 18 54 Input: arr = [2 8 10 15 16 30 32 64] Output: 2 8 32 8 16 32 16 32 64 Input: arr = [ 1 2 6 18 36 54] Output: 2 6 18 1 6 36 6 18 54
Die Idee besteht darin, vom zweiten Element auszugehen und jedes Element als mittleres Element festzulegen und nach den anderen beiden Elementen in einem Triplett zu suchen (ein kleineres und ein größeres). Damit ein Element arr[j] in der Mitte der geometrischen Progression liegt, müssen die Elemente arr[i] und arr[k] vorhanden sein, sodass -
arr[j] / arr[i] = r and arr[k] / arr[j] = r where r is an positive integer and 0 <= i < j and j < k <= n - 1
Nachfolgend finden Sie die Umsetzung der obigen Idee
C++ // C++ program to find if there exist three elements in // Geometric Progression or not #include using namespace std ; // The function prints three elements in GP if exists // Assumption: arr[0..n-1] is sorted. void findGeometricTriplets ( int arr [] int n ) { // One by fix every element as middle element for ( int j = 1 ; j < n - 1 ; j ++ ) { // Initialize i and k for the current j int i = j - 1 k = j + 1 ; // Find all i and k such that (i j k) // forms a triplet of GP while ( i >= 0 && k <= n - 1 ) { // if arr[j]/arr[i] = r and arr[k]/arr[j] = r // and r is an integer (i j k) forms Geometric // Progression while ( arr [ j ] % arr [ i ] == 0 && arr [ k ] % arr [ j ] == 0 && arr [ j ] / arr [ i ] == arr [ k ] / arr [ j ]) { // print the triplet cout < < arr [ i ] < < ' ' < < arr [ j ] < < ' ' < < arr [ k ] < < endl ; // Since the array is sorted and elements // are distinct. k ++ i -- ; } // if arr[j] is multiple of arr[i] and arr[k] is // multiple of arr[j] then arr[j] / arr[i] != // arr[k] / arr[j]. We compare their values to // move to next k or previous i. if ( arr [ j ] % arr [ i ] == 0 && arr [ k ] % arr [ j ] == 0 ) { if ( arr [ j ] / arr [ i ] < arr [ k ] / arr [ j ]) i -- ; else k ++ ; } // else if arr[j] is multiple of arr[i] then // try next k. Else try previous i. else if ( arr [ j ] % arr [ i ] == 0 ) k ++ ; else i -- ; } } } // Driver code int main () { // int arr[] = {1 2 6 10 18 54}; // int arr[] = {2 8 10 15 16 30 32 64}; // int arr[] = {1 2 6 18 36 54}; int arr [] = { 1 2 4 16 }; // int arr[] = {1 2 3 6 18 22}; int n = sizeof ( arr ) / sizeof ( arr [ 0 ]); findGeometricTriplets ( arr n ); return 0 ; }
Java // Java program to find if there exist three elements in // Geometric Progression or not import java.util.* ; class GFG { // The function prints three elements in GP if exists // Assumption: arr[0..n-1] is sorted. static void findGeometricTriplets ( int arr [] int n ) { // One by fix every element as middle element for ( int j = 1 ; j < n - 1 ; j ++ ) { // Initialize i and k for the current j int i = j - 1 k = j + 1 ; // Find all i and k such that (i j k) // forms a triplet of GP while ( i >= 0 && k <= n - 1 ) { // if arr[j]/arr[i] = r and arr[k]/arr[j] = r // and r is an integer (i j k) forms Geometric // Progression while ( i >= 0 && arr [ j ] % arr [ i ] == 0 && arr [ k ] % arr [ j ] == 0 && arr [ j ] / arr [ i ] == arr [ k ] / arr [ j ] ) { // print the triplet System . out . println ( arr [ i ] + ' ' + arr [ j ] + ' ' + arr [ k ] ); // Since the array is sorted and elements // are distinct. k ++ ; i -- ; } // if arr[j] is multiple of arr[i] and arr[k] is // multiple of arr[j] then arr[j] / arr[i] != // arr[k] / arr[j]. We compare their values to // move to next k or previous i. if ( i >= 0 && arr [ j ] % arr [ i ] == 0 && arr [ k ] % arr [ j ] == 0 ) { if ( i >= 0 && arr [ j ] / arr [ i ] < arr [ k ] / arr [ j ] ) i -- ; else k ++ ; } // else if arr[j] is multiple of arr[i] then // try next k. Else try previous i. else if ( i >= 0 && arr [ j ] % arr [ i ] == 0 ) k ++ ; else i -- ; } } } // Driver code public static void main ( String [] args ) { // int arr[] = {1 2 6 10 18 54}; // int arr[] = {2 8 10 15 16 30 32 64}; // int arr[] = {1 2 6 18 36 54}; int arr [] = { 1 2 4 16 }; // int arr[] = {1 2 3 6 18 22}; int n = arr . length ; findGeometricTriplets ( arr n ); } } // This code is contributed by Rajput-Ji
Python 3 # Python 3 program to find if # there exist three elements in # Geometric Progression or not # The function prints three elements # in GP if exists. # Assumption: arr[0..n-1] is sorted. def findGeometricTriplets ( arr n ): # One by fix every element # as middle element for j in range ( 1 n - 1 ): # Initialize i and k for # the current j i = j - 1 k = j + 1 # Find all i and k such that # (i j k) forms a triplet of GP while ( i >= 0 and k <= n - 1 ): # if arr[j]/arr[i] = r and # arr[k]/arr[j] = r and r # is an integer (i j k) forms # Geometric Progression while ( arr [ j ] % arr [ i ] == 0 and arr [ k ] % arr [ j ] == 0 and arr [ j ] // arr [ i ] == arr [ k ] // arr [ j ]): # print the triplet print ( arr [ i ] ' ' arr [ j ] ' ' arr [ k ]) # Since the array is sorted and # elements are distinct. k += 1 i -= 1 # if arr[j] is multiple of arr[i] # and arr[k] is multiple of arr[j] # then arr[j] / arr[i] != arr[k] / arr[j]. # We compare their values to # move to next k or previous i. if ( arr [ j ] % arr [ i ] == 0 and arr [ k ] % arr [ j ] == 0 ): if ( arr [ j ] // arr [ i ] < arr [ k ] // arr [ j ]): i -= 1 else : k += 1 # else if arr[j] is multiple of # arr[i] then try next k. Else # try previous i. elif ( arr [ j ] % arr [ i ] == 0 ): k += 1 else : i -= 1 # Driver code if __name__ == '__main__' : arr = [ 1 2 4 16 ] n = len ( arr ) findGeometricTriplets ( arr n ) # This code is contributed # by ChitraNayal
C# // C# program to find if there exist three elements // in Geometric Progression or not using System ; class GFG { // The function prints three elements in GP if exists // Assumption: arr[0..n-1] is sorted. static void findGeometricTriplets ( int [] arr int n ) { // One by fix every element as middle element for ( int j = 1 ; j < n - 1 ; j ++ ) { // Initialize i and k for the current j int i = j - 1 k = j + 1 ; // Find all i and k such that (i j k) // forms a triplet of GP while ( i >= 0 && k <= n - 1 ) { // if arr[j]/arr[i] = r and arr[k]/arr[j] = r // and r is an integer (i j k) forms Geometric // Progression while ( i >= 0 && arr [ j ] % arr [ i ] == 0 && arr [ k ] % arr [ j ] == 0 && arr [ j ] / arr [ i ] == arr [ k ] / arr [ j ]) { // print the triplet Console . WriteLine ( arr [ i ] + ' ' + arr [ j ] + ' ' + arr [ k ]); // Since the array is sorted and elements // are distinct. k ++ ; i -- ; } // if arr[j] is multiple of arr[i] and arr[k] is // multiple of arr[j] then arr[j] / arr[i] != // arr[k] / arr[j]. We compare their values to // move to next k or previous i. if ( i >= 0 && arr [ j ] % arr [ i ] == 0 && arr [ k ] % arr [ j ] == 0 ) { if ( i >= 0 && arr [ j ] / arr [ i ] < arr [ k ] / arr [ j ]) i -- ; else k ++ ; } // else if arr[j] is multiple of arr[i] then // try next k. Else try previous i. else if ( i >= 0 && arr [ j ] % arr [ i ] == 0 ) k ++ ; else i -- ; } } } // Driver code static public void Main () { // int arr[] = {1 2 6 10 18 54}; // int arr[] = {2 8 10 15 16 30 32 64}; // int arr[] = {1 2 6 18 36 54}; int [] arr = { 1 2 4 16 }; // int arr[] = {1 2 3 6 18 22}; int n = arr . Length ; findGeometricTriplets ( arr n ); } } // This code is contributed by ajit.
JavaScript < script > // Javascript program to find if there exist three elements in // Geometric Progression or not // The function prints three elements in GP if exists // Assumption: arr[0..n-1] is sorted. function findGeometricTriplets ( arr n ) { // One by fix every element as middle element for ( let j = 1 ; j < n - 1 ; j ++ ) { // Initialize i and k for the current j let i = j - 1 k = j + 1 ; // Find all i and k such that (i j k) // forms a triplet of GP while ( i >= 0 && k <= n - 1 ) { // if arr[j]/arr[i] = r and arr[k]/arr[j] = r // and r is an integer (i j k) forms Geometric // Progression while ( i >= 0 && arr [ j ] % arr [ i ] == 0 && arr [ k ] % arr [ j ] == 0 && arr [ j ] / arr [ i ] == arr [ k ] / arr [ j ]) { // print the triplet document . write ( arr [ i ] + ' ' + arr [ j ] + ' ' + arr [ k ] + '
' ); // Since the array is sorted and elements // are distinct. k ++ ; i -- ; } // if arr[j] is multiple of arr[i] and arr[k] is // multiple of arr[j] then arr[j] / arr[i] != // arr[k] / arr[j]. We compare their values to // move to next k or previous i. if ( i >= 0 && arr [ j ] % arr [ i ] == 0 && arr [ k ] % arr [ j ] == 0 ) { if ( i >= 0 && arr [ j ] / arr [ i ] < arr [ k ] / arr [ j ]) i -- ; else k ++ ; } // else if arr[j] is multiple of arr[i] then // try next k. Else try previous i. else if ( i >= 0 && arr [ j ] % arr [ i ] == 0 ) k ++ ; else i -- ; } } } // Driver code // int arr[] = {1 2 6 10 18 54}; // int arr[] = {2 8 10 15 16 30 32 64}; // int arr[] = {1 2 6 18 36 54}; let arr = [ 1 2 4 16 ]; // int arr[] = {1 2 3 6 18 22}; let n = arr . length ; findGeometricTriplets ( arr n ); // This code is contributed by avanitrachhadiya2155 < /script>
Ausgabe
1 2 4 1 4 16
Zeitkomplexität der obigen Lösung ist O(n 2 ), da wir für jedes j i und k in linearer Zeit finden.
Hilfsraum: O(1) da wir keinen zusätzlichen Platz beansprucht haben.
