Gegeven een gesorteerde reeks van verschillende positieve gehele getallen, worden alle tripletten afgedrukt die een geometrische progressie vormen met een integrale gemeenschappelijke verhouding.
Een geometrische progressie is een reeks getallen waarbij elke term na de eerste wordt gevonden door de vorige te vermenigvuldigen met een vast getal dat niet nul is, de gemeenschappelijke verhouding. De reeks 2 6 18 54... is bijvoorbeeld een geometrische progressie met gemeenschappelijke verhouding 3.
Voorbeelden:
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
Het idee is om vanaf het tweede element te beginnen en elk element als middelste element vast te stellen en de andere twee elementen in een triplet te zoeken (een kleiner en een groter). Wil een element arr[j] zich in het midden van de geometrische progressie bevinden, dan moeten er elementen arr[i] en arr[k] bestaan, zodat -
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
Hieronder ziet u de implementatie van bovenstaand idee
C++// C++ program to find if there exist three elements in // Geometric Progression or not #include
// 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 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# 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.
<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(arrn) { // 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>
Uitvoer
1 2 4 1 4 16
Tijdcomplexiteit van bovenstaande oplossing is O(n2) want voor elke j vinden we i en k in lineaire tijd.
Hulpruimte: O(1) omdat we geen extra ruimte gebruikten.