Trova tutte le terzine in una matrice ordinata che forma la progressione geometrica
Data una serie ordinata di interi positivi distinti, stampa tutte le terzine che formano la progressione geometrica con rapporto comune integrale.
Una progressione geometrica è una sequenza di numeri in cui ogni termine successivo al primo si trova moltiplicando il precedente per un numero fisso diverso da zero chiamato rapporto comune. Ad esempio la sequenza 2 6 18 54... è una progressione geometrica con rapporto comune 3.
Esempi:
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
L'idea è di partire dal secondo elemento e fissare ogni elemento come elemento centrale e cercare gli altri due elementi in una terzina (uno più piccolo e uno più grande). Affinché un elemento arr[j] sia al centro della progressione geometrica devono esistere elementi arr[i] e arr[k] tali che -
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
Di seguito è riportata l'implementazione dell'idea di cui sopra
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>
Produzione
1 2 4 1 4 16
Complessità temporale della soluzione di cui sopra è O(n 2 ) come per ogni j troviamo i e k nel tempo lineare.
Spazio ausiliario: O(1) poiché non abbiamo utilizzato spazio extra.
