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Comproveu la parella amb el producte donat

Proveu -ho a la pràctica GFG ' title=

Donat una matriu arr [] de n nombres enters diferents i a objectiu Valor La tasca és comprovar si hi ha un parell d’elements a la matriu el producte del qual és igual a l’objectiu.

cercador i exemples

Exemples:  



Entrada: arr [] = [1 5 7 -1 5] objectiu = 35
Sortida: lleial
Explicació: Com a 5* 7 = 35 La resposta és certa.

Entrada: arr [] = [-10 20 9 -40] objectiu = 30
Sortida: fals
Explicació: No existeix cap parella amb el producte 30

Taula de contingut



[Enfocament ingenu] generant totes les parelles possibles - O (n 2 ) Temps i O (1) Espai

L’enfocament molt bàsic és generar totes les parelles possibles i comprovar si hi ha algun parell que el producte sigui igual al valor objectiu donat, després tornar lleial . Si no existeix aquest parell, torneu fals .

bucle for a l'script de shell
C++ 
#include    using namespace std; // Function to check if any pair exists whose product // equals the target bool isProduct(vector<int> &arr long long target) {  int n = arr.size();  for (int i = 0; i < n - 1; i++) {  for (int j = i + 1; j < n; j++) {  if (1LL * arr[i] * arr[j] == target) {  return true;  }  }  }  return false; } int main() {  vector<int> arr = {1 5 7 -1 5};  long long target = 35;  cout << isProduct(arr target) << endl;  return 0; }
C 
#include  #include  // Function to check if any pair exists whose product // equals the target bool isProduct(int arr[] int n long long target) {  for (int i = 0; i < n - 1; i++) {  for (int j = i + 1; j < n; j++) {  if (1LL * arr[i] * arr[j] == target) {  return true;  }  }  }  return false; } int main() {  int arr[] = {1 5 7 -1 5};  long long target = 35;   int n = sizeof(arr) / sizeof(arr[0]);  printf('%dn' isProduct(arr n target));    return 0; }
Java 
class GfG {  // Function to check if any pair exists whose product  // equals the target  static boolean isProduct(int[] arr long target) {  int n = arr.length;  for (int i = 0; i < n - 1; i++) {  for (int j = i + 1; j < n; j++) {  if ((long) arr[i] * arr[j] == target) {  return true;  }  }  }  return false;  }  public static void main(String[] args) {  int[] arr = {1 5 7 -1 5};  long target = 35;   System.out.println(isProduct(arr target));  } }
Python 
# Function to check if any pair exists whose product # equals the target def is_product(arr target): n = len(arr) for i in range(n - 1): for j in range(i + 1 n): if arr[i] * arr[j] == target: return True return False arr = [1 5 7 -1 5] target = 35 print(is_product(arr target))
C# 
using System; class GfG {  // Function to check if any pair exists whose product  // equals the target  static bool IsProduct(int[] arr long target) {  int n = arr.Length;  for (int i = 0; i < n - 1; i++) {  for (int j = i + 1; j < n; j++) {  if ((long)arr[i] * arr[j] == target) {  return true;  }  }  }  return false;  }  static void Main() {  int[] arr = { 1 5 7 -1 5 };  long target = 35;   Console.WriteLine(IsProduct(arr target));  } }
JavaScript 
// Function to check if any pair exists whose product // equals the target function isProduct(arr target) {  let n = arr.length;  for (let i = 0; i < n - 1; i++) {  for (let j = i + 1; j < n; j++) {  if (arr[i] * arr[j] === target) {  return true;  }  }  }  return false; } let arr = [1 5 7 -1 5]; let target = 35; console.log(isProduct(arr target));

Producció 
1

Complexitat del temps: O (n²) per utilitzar dos bucles imbricats
Espai auxiliar: O (1)

css per ajustar el text

[Millor enfocament] Utilitzant el temps de dos punter

També podem utilitzar la tècnica de dos punts per a aquest problema, però només és aplicable a dades ordenades. Així que primer ordeneu la matriu i mantingueu dos indicadors un punter al principi ( esquerre ) i un altre al final ( dret ) de la matriu. A continuació, comproveu el producte dels elements en aquests dos indicadors:



  • Si el producte és igual al objectiu Hem trobat la parella.
  • Si el producte és inferior al objectiu moure el esquerre punter al dret per augmentar el producte.
  • Si el producte és més gran que el objectiu moure el dret punter al esquerre per disminuir el producte.
C++ 
#include    using namespace std; // Function to check if any pair exists whose product equals the target. bool isProduct(vector<int> &arr long long target) {    // Sort the array  sort(arr.begin() arr.end());  int left = 0 right = arr.size() - 1;  while (left < right) {  // Calculate the current product  long long currProd = 1LL*arr[left]*arr[right];  // If the product matches the target return true.  if (currProd == target) return true;  // Move the pointers based on comparison with target.  if (currProd > target) right--;  else left++;  }  return false; } int main() {  vector<int> arr = {1 5 7 -1 5};  long long target = 35;  cout << isProduct(arr target) << endl;  return 0; }
C 
#include  #include  #include  // Function to compare two integers (used in qsort) int compare(const void *a const void *b) {  return (*(int *)a - *(int *)b); } // Function to check if any pair exists whose product // equals the target. bool isProduct(int arr[] int n long long target) {  // Sort the array  qsort(arr n sizeof(int) compare);  int left = 0 right = n - 1;  while (left < right)  {  // Calculate the current product  long long currProd = (long long)arr[left] * arr[right];  // If the product matches the target return true.  if (currProd == target)  return true;  // Move the pointers based on comparison with target.  if (currProd > target)  right--;  else  left++;  }  return false; } int main() {  int arr[] = {1 5 7 -1 5};  long long target = 35;  int n = sizeof(arr) / sizeof(arr[0]);  printf('%dn' isProduct(arr n target));  return 0; }
Java 
import java.util.Arrays; class GfG {  // Function to check if any pair exists whose product equals the target.  static boolean isProduct(int[] arr long target) {  // Sort the array  Arrays.sort(arr);  int left = 0 right = arr.length - 1;  while (left < right) {    // Calculate the current product  long currProd = (long) arr[left] * arr[right];  // If the product matches the target return true.  if (currProd == target) return true;  // Move the pointers based on comparison with target.  if (currProd > target) right--;  else left++;  }  return false;  }  public static void main(String[] args) {  int[] arr = {1 5 7 -1 5};  long target = 35;   System.out.println(isProduct(arr target));  } }
Python 
# Function to check if any pair exists whose product equals the target. def isProduct(arr target): # Sort the array arr.sort() left right = 0 len(arr) - 1 while left < right: # Calculate the current product currProd = arr[left] * arr[right] # If the product matches the target return True. if currProd == target: return True # Move the pointers based on comparison with target. if currProd > target: right -= 1 else: left += 1 return False if __name__ == '__main__': arr = [1 5 7 -1 5] target = 35 print(isProduct(arr target))
C# 
using System; using System.Linq; class GfG {  // Function to check if any pair exists whose product  // equals the target.  static bool isProduct(int[] arr long target) {    // Sort the array  Array.Sort(arr);  int left = 0 right = arr.Length - 1;  while (left < right) {  // Calculate the current product  long currProd = (long) arr[left] * arr[right];  // If the product matches the target return true.  if (currProd == target) return true;  // Move the pointers based on comparison with target.  if (currProd > target) right--;  else left++;  }  return false;  }  static void Main(string[] args) {  int[] arr = { 1 5 7 -1 5 };  long target = 35;   Console.WriteLine(isProduct(arr target));  } }
JavaScript 
// Function to check if any pair exists whose product // equals the target. function isProduct(arr target) {  // Sort the array  arr.sort((a b) => a - b);  let left = 0 right = arr.length - 1;  while (left < right) {  // Calculate the current product  let currProd = arr[left] * arr[right];  // If the product matches the target return true.  if (currProd === target) return true;  // Move the pointers based on comparison with target.  if (currProd > target) right--;  else left++;  }  return false; } let arr = [1 5 7 -1 5]; let target = 35; console.log(isProduct(arr target));

Producció 
1

Complexitat del temps: O (n log (n)) per ordenar la matriu
Espai auxiliar: O (1)

[Enfocament esperat] Utilitzant Hashset - O (N) Temps i O (N) Espai

Podem utilitzar un Conjunt de hash per buscar de manera eficient. A mesura que iterem a través de la matriu comprovem si cada número és un factor de l'objectiu. Si ho és, veiem si el seu factor corresponent ja està al conjunt. Si és així, tornem lleial ; En cas contrari, afegim el número actual al conjunt i continuem.

C++ 
#include    #include  #include  using namespace std; // Function to check if any pair exists whose product // equals the target. bool isProduct(vector<int> &arr long long target) {    // Use an unordered set to store previously seen numbers.  unordered_set<int> st;  for (int num : arr) {  // If target is 0 and current number is 0 return true.  if (target == 0 && num == 0) return true;  // Check if current number can be a factor of the target.  if (target % num == 0) {  int secondNum = target / num;    // If the secondNum has been seen before return true.  if (st.find(secondNum) != st.end()) {  return true;  }    // Mark the current number as seen.  st.insert(num);  }  }  return false; } int main() {  vector<int> arr = {1 5 7 -1 5};  long long target = 35;  cout << isProduct(arr target) << endl;  return 0; }
Java 
import java.util.HashSet; class GfG {  // Function to check if any pair exists whose product  // equals the target.  static boolean isProduct(int[] arr long target)  {  // Use a hash set to store previously seen numbers.  HashSet<Integer> set = new HashSet<>();  for (int num : arr) {  // If target is 0 and current number is 0  // return true.  if (target == 0 && num == 0)  return true;  // Check if current number can be a factor of  // the target.  if (target % num == 0) {  int secondNum = (int)(target / num);  // If the secondNum has been seen before  // return true.  if (set.contains(secondNum))  return true;  // Mark the current number as seen.  set.add(num);  }  }  return false;  }  public static void main(String[] args)  {  int[] arr = { 1 5 7 -1 5 };  long target = 35;  System.out.println(isProduct(arr target));  } }
Python 
# Function to check if any pair exists whose product equals the target. def isProduct(arr target): # Use a set to store previously seen numbers. st = set() for num in arr: # If target is 0 and current number is 0 return True. if target == 0 and num == 0: return True # Check if current number can be a factor of the target. if target % num == 0: secondNum = target // num # If the secondNum has been seen before return True. if secondNum in st: return True # Mark the current number as seen. st.add(num) return False if __name__ == '__main__': arr = [1 5 7 -1 5] target = 35 print(isProduct(arr target))
C# 
using System; using System.Collections.Generic; class GfG {  // Function to check if any pair exists whose product  // equals the target.  static bool isProduct(int[] arr long target)  {  // Use a hash set to store previously seen numbers.  HashSet<int> set = new HashSet<int>();  foreach(int num in arr)  {  // If target is 0 and current number is 0  // return true.  if (target == 0 && num == 0)  return true;  // Check if current number can be a factor of  // the target.  if (target % num == 0) {  int secondNum = (int)(target / num);  // If the secondNum has been seen before  // return true.  if (set.Contains(secondNum))  return true;  // Mark the current number as seen.  set.Add(num);  }  }  return false;  }  static void Main(string[] args)  {  int[] arr = { 1 5 7 -1 5 };  long target = 35;  Console.WriteLine(isProduct(arr target));  } }
JavaScript 
// Function to check if any pair exists whose product equals // the target. function isProduct(arr target) {  // Use a set to store previously seen numbers.  let seen = new Set();  for (let num of arr) {  // If target is 0 and current number is 0 return  // true.  if (target === 0 && num === 0)  return true;  // Check if current number can be a factor of the  // target.  if (target % num === 0) {  let secondNum = target / num;  // If the secondNum has been seen before return  // true.  if (seen.has(secondNum))  return true;  // Mark the current number as seen.  seen.add(num);  }  }  return false; } let arr = [ 1 5 7 -1 5 ]; let target = 35; console.log(isProduct(arr target));

Producció 
1

Complexitat del temps: O (n) per a una iteració única
Espai auxiliar: O (n) per emmagatzemar elements al conjunt de hash