Dada una array arr[] que consta de N enteros positivos, la tarea es encontrar el número mínimo de pares (arr[i], arr[j]) de la array dada que se necesita reemplazar con su MCM de modo que la array se reduzca a un solo elemento igual al LCM de la array inicial.
Ejemplos:
Entrada: arr[] = {1, 2, 3, 4}
Salida: 3
Explicación:
MCM de la array = 12
Paso 1: MCM(3, 4) = 12. Por lo tanto, la array se modifica a {1, 2, 12}
Paso 2: LCM(1, 12) = 12. Por lo tanto, el arreglo se modifica a {2, 12}
Paso 3: LCM(2, 12) = 12. Por lo tanto, el arreglo se modifica a {12}
Entrada: arr[] = {7 , 9, 3}
Salida: 2
Explicación:
LCM del arreglo = 63
Paso 1: LCM(7, 9) = 63. Por lo tanto el arreglo se modifica a {63, 3}
Paso 2: LCM(63, 3) = 63. Por lo tanto, la array se modifica a {63}
Enfoque ingenuo: la idea es generar todos los pares posibles y, para cada par, reemplazarlos por su LCM y calcular la cantidad de pasos necesarios para reducirlos a un solo elemento de array igual a su LCM. Imprime el número mínimo de operaciones requeridas.
Complejidad de tiempo: O((N!)*log N)
Espacio auxiliar: O(N)
Enfoque eficiente: el enfoque anterior se puede optimizar en función de las siguientes observaciones:
- El MCM de un arreglo es igual al producto de todos los números primos del arreglo.
- En (X – 1) pasos, el MCM de todos los X números primos se puede obtener usando dos números como pares.
- En los siguientes (N – 2) pasos, convierta el resto (N – 2) elementos iguales al LCM de la array.
- Por lo tanto, el número total de pasos viene dado por:
(N – 2) + (X – 1) para N > 2
- Para N = 1 , el número de operaciones es simplemente 0 y para N = 2 , el número de operaciones es 1 .
Pasos:
- Si N = 1, entonces el conteo de pasos es 0 .
- Si N = 2, entonces el conteo de pasos es 1 .
- Genere todos los primos hasta N usando Sieve Of Eratosthenes .
- Almacene el conteo de números primos en una variable, digamos X .
- El conteo total de operaciones viene dado por:
(N – 2) + (X – 1) para N > 2
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; const int maxm = 10001; // Boolean array to set or unset // prime non-prime indices bool prime[maxm]; // Stores the prefix sum of the count // of prime numbers int prime_number[maxm]; // Function to check if a number // is prime or not from 0 to N void SieveOfEratosthenes() { memset(prime, true, sizeof(prime)); for (int p = 2; p * p < maxm; p++) { // If p is a prime if (prime[p] == true) { // Set its multiples as // non-prime for (int i = p * p; i < maxm; i += p) prime[i] = false; } } prime[0] = false; prime[1] = false; } // Function to store the count of // prime numbers void num_prime() { prime_number[0] = 0; for (int i = 1; i <= maxm; i++) prime_number[i] = prime_number[i - 1] + prime[i]; } // Function to count the operations // to reduce the array to one element // by replacing each pair with its LCM void min_steps(int arr[], int n) { // Generating Prime Number SieveOfEratosthenes(); num_prime(); // Corner Case if (n == 1) cout << "0\n"; else if (n == 2) cout << "1\n"; else cout << prime_number[n] - 1 + (n - 2); } // Driver Code int main() { // Given array arr[] int arr[] = { 5, 4, 3, 2, 1 }; int N = sizeof(arr) / sizeof(arr[0]); // Function Call min_steps(arr, N); return 0; }
Java
// Java program for the above approach import java.util.*; class GFG{ static final int maxm = 10001; // Boolean array to set or unset // prime non-prime indices static boolean prime[]; // Stores the prefix sum of the count // of prime numbers static int prime_number[]; // Function to check if a number // is prime or not from 0 to N static void SieveOfEratosthenes() { Arrays.fill(prime,true); for(int p = 2; p * p < maxm; p++) { // If p is a prime if (prime[p] == true) { // Set its multiples as // non-prime for(int i = p * p; i < maxm; i += p) prime[i] = false; } } prime[0] = false; prime[1] = false; } // Function to store the count of // prime numbers static void num_prime() { prime_number[0] = 0; for(int i = 1; i <= maxm; i++) { int tmp; if(prime[i] == true) { tmp = 1; } else { tmp = 0; } prime_number[i] = prime_number[i - 1] + tmp; } } // Function to count the operations // to reduce the array to one element // by replacing each pair with its LCM static void min_steps(int arr[], int n) { // Generating Prime Number SieveOfEratosthenes(); num_prime(); // Corner Case if (n == 1) { System.out.println("0"); } else if (n == 2) { System.out.println("1"); } else { System.out.println(prime_number[n] - 1 + (n - 2)); } } // Driver code public static void main(String[] args) { prime = new boolean[maxm + 1]; // Stores the prefix sum of the count // of prime numbers prime_number = new int[maxm + 1]; // Given array arr[] int arr[] = { 5, 4, 3, 2, 1 }; int N = arr.length; // Function call min_steps(arr, N); } } // This code is contributed by rutvik_56
Python3
# Python3 program for # the above approach maxm = 10001; # Boolean array to set or unset # prime non-prime indices prime = [True] * (maxm + 1); # Stores the prefix sum of the count # of prime numbers prime_number = [0] * (maxm + 1); # Function to check if a number # is prime or not from 0 to N def SieveOfEratosthenes(): for p in range(2, (int(maxm ** 1 / 2))): # If p is a prime if (prime[p] == True): # Set its multiples as # non-prime for i in range(p * p, maxm, p): prime[i] = False; prime[0] = False; prime[1] = False; # Function to store the count of # prime numbers def num_prime(): prime_number[0] = 0; for i in range(1, maxm + 1): tmp = -1; if (prime[i] == True): tmp = 1; else: tmp = 0; prime_number[i] = prime_number[i - 1] + tmp; # Function to count the operations # to reduce the array to one element # by replacing each pair with its LCM def min_steps(arr, n): # Generating Prime Number SieveOfEratosthenes(); num_prime(); # Corner Case if (n == 1): print("0"); elif (n == 2): print("1"); else: print(prime_number[n] - 1 + (n - 2)); # Driver code if __name__ == '__main__': # Given array arr arr = [5, 4, 3, 2, 1]; N = len(arr); # Function call min_steps(arr, N); # This code is contributed by Rajput-Ji
C#
// C# program for the above approach using System; class GFG{ static readonly int maxm = 10001; // Boolean array to set or unset // prime non-prime indices static bool []prime; // Stores the prefix sum of the count // of prime numbers static int []prime_number; // Function to check if a number // is prime or not from 0 to N static void SieveOfEratosthenes() { for(int i = 0; i < prime.Length; i++) prime[i] = true; for(int p = 2; p * p < maxm; p++) { // If p is a prime if (prime[p] == true) { // Set its multiples as // non-prime for(int i = p * p; i < maxm; i += p) prime[i] = false; } } prime[0] = false; prime[1] = false; } // Function to store the count of // prime numbers static void num_prime() { prime_number[0] = 0; for(int i = 1; i <= maxm; i++) { int tmp; if(prime[i] == true) { tmp = 1; } else { tmp = 0; } prime_number[i] = prime_number[i - 1] + tmp; } } // Function to count the operations // to reduce the array to one element // by replacing each pair with its LCM static void min_steps(int []arr, int n) { // Generating Prime Number SieveOfEratosthenes(); num_prime(); // Corner Case if (n == 1) { Console.WriteLine("0"); } else if (n == 2) { Console.WriteLine("1"); } else { Console.WriteLine(prime_number[n] - 1 + (n - 2)); } } // Driver code public static void Main(String[] args) { prime = new bool[maxm + 1]; // Stores the prefix sum of the count // of prime numbers prime_number = new int[maxm + 1]; // Given array []arr int []arr = {5, 4, 3, 2, 1}; int N = arr.Length; // Function call min_steps(arr, N); } } // This code is contributed by Rajput-Ji
Javascript
<script> // javascript program for the above approach const maxm = 10001; // Boolean array to set or unset // prime non-prime indices var prime = Array(); // Stores the prefix sum of the count // of prime numbers var prime_number = Array(); // Function to check if a number // is prime or not from 0 to N function SieveOfEratosthenes() { for(i = 0; i < maxm; i++) prime[i] = true; for (p = 2; p * p < maxm; p++) { // If p is a prime if (prime[p] == true) { // Set its multiples as // non-prime for (i = p * p; i < maxm; i += p) prime[i] = false; } } prime[0] = false; prime[1] = false; } // Function to store the count of // prime numbers function num_prime() { prime_number[0] = 0; for (i = 1; i <= maxm; i++) { var tmp; if (prime[i] == true) { tmp = 1; } else { tmp = 0; } prime_number[i] = prime_number[i - 1] + tmp; } } // Function to count the operations // to reduce the array to one element // by replacing each pair with its LCM function min_steps(arr , n) { // Generating Prime Number SieveOfEratosthenes(); num_prime(); // Corner Case if (n == 1) { document.write("0"); } else if (n == 2) { document.write("1"); } else { document.write(prime_number[n] - 1 + (n - 2)); } } // Driver code // Stores the prefix sum of the count // of prime numbers prime_number.fill(0); // Given array arr var arr = [ 5, 4, 3, 2, 1 ]; var N = arr.length; // Function call min_steps(arr, N); // This code is contributed by aashish1995 </script>
5
Complejidad de tiempo: O(N + log(log(maxm))
Espacio auxiliar: O(maxm)