#practiceLinkDiv { display: ingen !important; }Givet en matrix arr[] af N heltalselementer er opgaven at finde summen af gennemsnittet af alle delmængder af denne matrix.
hvad er mac os
Eksempel:
Input : arr[] = [2 3 5]Recommended Practice Summen af gennemsnittet af alle delmængder Prøv det!
Output : 23.33
Explanation : Subsets with their average are
[2] average = 2/1 = 2
[3] average = 3/1 = 3
[5] average = 5/1 = 5
[2 3] average = (2+3)/2 = 2.5
[2 5] average = (2+5)/2 = 3.5
[3 5] average = (3+5)/2 = 4
[2 3 5] average = (2+3+5)/3 = 3.33
Sum of average of all subset is
2 + 3 + 5 + 2.5 + 3.5 + 4 + 3.33 = 23.33
Naiv tilgang: En naiv løsning er at iterere gennem alle mulige delmængder få en gennemsnit af dem alle og derefter tilføje dem én efter én, men dette vil tage eksponentiel tid og vil være umuligt for større arrays.
Vi kan få et mønster ved at tage et eksempel
arr = [a0 a1 a2 a3]
sum of average =
a0/1 + a1/1 + a2/2 + a3/1 +
(a0+a1)/2 + (a0+a2)/2 + (a0+a3)/2 + (a1+a2)/2 +
(a1+a3)/2 + (a2+a3)/2 +
(a0+a1+a2)/3 + (a0+a2+a3)/3 + (a0+a1+a3)/3 +
(a1+a2+a3)/3 +
(a0+a1+a2+a3)/4
If S = (a0+a1+a2+a3) then above expression
can be rearranged as below
sum of average = (S)/1 + (3*S)/2 + (3*S)/3 + (S)/4
Koefficienten med tællere kan forklares som følger, hvis vi itererer over delmængder med K elementer, så vil nævneren være K, og tælleren vil være r*S, hvor 'r' angiver antallet af gange, et bestemt array-element vil blive tilføjet, mens det itererer over delmængder af samme størrelse. Ved inspektion kan vi se, at r vil være nCr(N - 1 n - 1), fordi efter at have placeret et element i summeringen, skal vi vælge (n – 1) elementer fra (N - 1) elementer, så hvert element vil have en frekvens på nCr(N - 1 n - 1), mens vi overvejer delmængder af samme størrelse som alle gange, og frekvensen er lig med antallet af summering, og frekvensen er lige mange af disse elementer. vil være tælleren i det endelige udtryk.
I nedenstående kode nCr er implementeret ved hjælp af dynamisk programmeringsmetode det kan du læse mere om her
C++// C++ program to get sum of average of all subsets #include
// java program to get sum of // average of all subsets import java.io.*; class GFG { // Returns value of Binomial // Coefficient C(n k) static int nCr(int n int k) { int C[][] = new int[n + 1][k + 1]; int i j; // Calculate value of Binomial // Coefficient in bottom up manner for (i = 0; i <= n; i++) { for (j = 0; j <= Math.min(i k); j++) { // Base Cases if (j == 0 || j == i) C[i][j] = 1; // Calculate value using // previously stored values else C[i][j] = C[i - 1][j - 1] + C[i - 1][j]; } } return C[n][k]; } // method returns sum of average of all subsets static double resultOfAllSubsets(int arr[] int N) { // Initialize result double result = 0.0; // Find sum of elements int sum = 0; for (int i = 0; i < N; i++) sum += arr[i]; // looping once for all subset of same size for (int n = 1; n <= N; n++) /* each element occurs nCr(N-1 n-1) times while considering subset of size n */ result += (double)(sum * (nCr(N - 1 n - 1))) / n; return result; } // Driver code to test above methods public static void main(String[] args) { int arr[] = { 2 3 5 7 }; int N = arr.length; System.out.println(resultOfAllSubsets(arr N)); } } // This code is contributed by vt_m
// C# program to get sum of // average of all subsets using System; class GFG { // Returns value of Binomial // Coefficient C(n k) static int nCr(int n int k) { int[ ] C = new int[n + 1 k + 1]; int i j; // Calculate value of Binomial // Coefficient in bottom up manner for (i = 0; i <= n; i++) { for (j = 0; j <= Math.Min(i k); j++) { // Base Cases if (j == 0 || j == i) C[i j] = 1; // Calculate value using // previously stored values else C[i j] = C[i - 1 j - 1] + C[i - 1 j]; } } return C[n k]; } // method returns sum of average // of all subsets static double resultOfAllSubsets(int[] arr int N) { // Initialize result double result = 0.0; // Find sum of elements int sum = 0; for (int i = 0; i < N; i++) sum += arr[i]; // looping once for all subset // of same size for (int n = 1; n <= N; n++) /* each element occurs nCr(N-1 n-1) times while considering subset of size n */ result += (double)(sum * (nCr(N - 1 n - 1))) / n; return result; } // Driver code to test above methods public static void Main() { int[] arr = { 2 3 5 7 }; int N = arr.Length; Console.WriteLine(resultOfAllSubsets(arr N)); } } // This code is contributed by Sam007
<script> // javascript program to get sum of // average of all subsets // Returns value of Binomial // Coefficient C(n k) function nCr(n k) { let C = new Array(n + 1); for (let i = 0; i <= n; i++) { C[i] = new Array(k + 1); for (let j = 0; j <= k; j++) { C[i][j] = 0; } } let i j; // Calculate value of Binomial // Coefficient in bottom up manner for (i = 0; i <= n; i++) { for (j = 0; j <= Math.min(i k); j++) { // Base Cases if (j == 0 || j == i) C[i][j] = 1; // Calculate value using // previously stored values else C[i][j] = C[i - 1][j - 1] + C[i - 1][j]; } } return C[n][k]; } // method returns sum of average of all subsets function resultOfAllSubsets(arr N) { // Initialize result let result = 0.0; // Find sum of elements let sum = 0; for (let i = 0; i < N; i++) sum += arr[i]; // looping once for all subset of same size for (let n = 1; n <= N; n++) /* each element occurs nCr(N-1 n-1) times while considering subset of size n */ result += (sum * (nCr(N - 1 n - 1))) / n; return result; } let arr = [ 2 3 5 7 ]; let N = arr.length; document.write(resultOfAllSubsets(arr N)); </script>
// PHP program to get sum // of average of all subsets // Returns value of Binomial // Coefficient C(n k) function nCr($n $k) { $C[$n + 1][$k + 1] = 0; $i; $j; // Calculate value of Binomial // Coefficient in bottom up manner for ($i = 0; $i <= $n; $i++) { for ($j = 0; $j <= min($i $k); $j++) { // Base Cases if ($j == 0 || $j == $i) $C[$i][$j] = 1; // Calculate value using // previously stored values else $C[$i][$j] = $C[$i - 1][$j - 1] + $C[$i - 1][$j]; } } return $C[$n][$k]; } // method returns sum of // average of all subsets function resultOfAllSubsets($arr $N) { // Initialize result $result = 0.0; // Find sum of elements $sum = 0; for ($i = 0; $i < $N; $i++) $sum += $arr[$i]; // looping once for all // subset of same size for ($n = 1; $n <= $N; $n++) /* each element occurs nCr(N-1 n-1) times while considering subset of size n */ $result += (($sum * (nCr($N - 1 $n - 1))) / $n); return $result; } // Driver Code $arr = array( 2 3 5 7 ); $N = sizeof($arr) / sizeof($arr[0]); echo resultOfAllSubsets($arr $N) ; // This code is contributed by nitin mittal. ?> # Python3 program to get sum # of average of all subsets # Returns value of Binomial # Coefficient C(n k) def nCr(n k): C = [[0 for i in range(k + 1)] for j in range(n + 1)] # Calculate value of Binomial # Coefficient in bottom up manner for i in range(n + 1): for j in range(min(i k) + 1): # Base Cases if (j == 0 or j == i): C[i][j] = 1 # Calculate value using # previously stored values else: C[i][j] = C[i-1][j-1] + C[i-1][j] return C[n][k] # Method returns sum of # average of all subsets def resultOfAllSubsets(arr N): result = 0.0 # Initialize result # Find sum of elements sum = 0 for i in range(N): sum += arr[i] # looping once for all subset of same size for n in range(1 N + 1): # each element occurs nCr(N-1 n-1) times while # considering subset of size n */ result += (sum * (nCr(N - 1 n - 1))) / n return result # Driver code arr = [2 3 5 7] N = len(arr) print(resultOfAllSubsets(arr N)) # This code is contributed by Anant Agarwal.
Produktion
63.75
Tidskompleksitet: På3)
Hjælpeplads: På2)
Effektiv tilgang: Rumoptimering O(1)
For at optimere pladskompleksiteten af ovenstående tilgang kan vi bruge en mere effektiv tilgang, der undgår behovet for hele matrixen C[][] at gemme binomiale koefficienter. I stedet kan vi bruge en kombinationsformel til at beregne binomialkoefficienten direkte, når det er nødvendigt.
Implementeringstrin:
- Iterér over elementerne i arrayet og beregn summen af alle elementer.
- Iterér over hver delmængdestørrelse fra 1 til N.
- Beregn inde i løkken gennemsnit af summen af elementer ganget med den binomiale koefficient for delmængdens størrelse. Tilføj det beregnede gennemsnit til resultatet.
- Returner det endelige resultat.
Implementering:
C++#include
import java.util.Arrays; public class Main { // Method to calculate binomial coefficient C(n k) static int binomialCoeff(int n int k) { int res = 1; // Since C(n k) = C(n n-k) if (k > n - k) k = n - k; // Calculate value of [n * (n-1) * ... * (n-k+1)] / [k * (k-1) * ... * 1] for (int i = 0; i < k; i++) { res *= (n - i); res /= (i + 1); } return res; } // Method to calculate the sum of the average of all subsets static double resultOfAllSubsets(int arr[] int N) { double result = 0.0; int sum = 0; // Calculate the sum of elements for (int i = 0; i < N; i++) sum += arr[i]; // Loop for each subset size for (int n = 1; n <= N; n++) result += (double) (sum * binomialCoeff(N - 1 n - 1)) / n; return result; } // Driver code to test the above methods public static void main(String[] args) { int arr[] = {2 3 5 7}; int N = arr.length; System.out.println(resultOfAllSubsets(arr N)); } }
using System; public class MainClass { // Method to calculate binomial coefficient C(n k) static int BinomialCoeff(int n int k) { int res = 1; // Since C(n k) = C(n n-k) if (k > n - k) k = n - k; // Calculate value of [n * (n-1) * ... * (n-k+1)] / [k * (k-1) * ... * 1] for (int i = 0; i < k; i++) { res *= (n - i); res /= (i + 1); } return res; } // Method to calculate the sum of the average of all subsets static double ResultOfAllSubsets(int[] arr int N) { double result = 0.0; int sumVal = 0; // Calculate the sum of elements for (int i = 0; i < N; i++) sumVal += arr[i]; // Loop for each subset size for (int n = 1; n <= N; n++) result += (double)(sumVal * BinomialCoeff(N - 1 n - 1)) / n; return result; } // Driver code to test the above methods public static void Main() { int[] arr = { 2 3 5 7 }; int N = arr.Length; Console.WriteLine(ResultOfAllSubsets(arr N)); } }
// Function to calculate binomial coefficient C(n k) function binomialCoeff(n k) { let res = 1; // Since C(n k) = C(n n-k) if (k > n - k) k = n - k; // Calculate value of [n * (n-1) * ... * (n-k+1)] / [k * (k-1) * ... * 1] for (let i = 0; i < k; i++) { res *= (n - i); res /= (i + 1); } return res; } // Function to calculate the sum of the average of all subsets function resultOfAllSubsets(arr) { let result = 0.0; let sum = arr.reduce((acc val) => acc + val 0); // Loop for each subset size for (let n = 1; n <= arr.length; n++) { result += (sum * binomialCoeff(arr.length - 1 n - 1)) / n; } return result; } const arr = [2 3 5 7]; console.log(resultOfAllSubsets(arr));
# Method to calculate binomial coefficient C(n k) def binomialCoeff(n k): res = 1 # Since C(n k) = C(n n-k) if k > n - k: k = n - k # Calculate value of [n * (n-1) * ... * (n-k+1)] / [k * (k-1) * ... * 1] for i in range(k): res *= (n - i) res //= (i + 1) return res # Method to calculate the sum of the average of all subsets def resultOfAllSubsets(arr N): result = 0.0 sum_val = 0 # Calculate the sum of elements for i in range(N): sum_val += arr[i] # Loop for each subset size for n in range(1 N + 1): result += (sum_val * binomialCoeff(N - 1 n - 1)) / n return result # Driver code to test the above methods arr = [2 3 5 7] N = len(arr) print(resultOfAllSubsets(arr N))
Produktion
63.75 Tidskompleksitet: O(n^2)
Hjælpeplads: O(1)