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Imprimer les n premiers nombres avec exactement deux bits définis

Étant donné un nombre n, imprimez d'abord n ​​entiers positifs avec exactement deux bits définis dans leur représentation binaire.
Exemples : 

Input: n = 3  
Output: 3 5 6  
The first 3 numbers with two set bits are 3 (0011)  
5 (0101) and 6 (0110)  
Input: n = 5  
Output: 3 5 6 9 10 12

UN Solution simple consiste à considérer tous les entiers positifs un par un en commençant par 1. Pour chaque nombre, vérifiez s'il a exactement deux ensembles de bits. Si un nombre a exactement deux bits définis, imprimez-le et incrémentez le nombre de ces nombres.
Un Solution efficace est de générer directement de tels nombres. Si nous observons clairement les nombres, nous pouvons les réécrire comme indiqué ci-dessous pow(21)+pow(20) pow(22)+pow(20) pow(22)+pow(21) pow(23)+pow(20) pow(23)+pow(21) pow(23)+pow(22) .........
Tous les nombres peuvent être générés dans un ordre croissant en fonction du plus élevé des deux bits définis. L'idée est de fixer les bits supérieurs de deux bits un par un. Pour le bit actuellement défini le plus élevé, considérez tous les bits inférieurs et imprimez les nombres formés.



C++ 
// C++ program to print first n numbers // with exactly two set bits #include    using namespace std; // Prints first n numbers with two set bits void printTwoSetBitNums(int n) {  // Initialize higher of two sets bits  int x = 1;  // Keep reducing n for every number  // with two set bits.  while (n > 0)  {  // Consider all lower set bits for  // current higher set bit  int y = 0;  while (y < x)  {  // Print current number  cout << (1 << x) + (1 << y) << ' ';  // If we have found n numbers  n--;  if (n == 0)  return;  // Consider next lower bit for current  // higher bit.  y++;  }  // Increment higher set bit  x++;  } } // Driver code int main() {  printTwoSetBitNums(4);  return 0; }
Java 
// Java program to print first n numbers // with exactly two set bits import java.io.*; class GFG  {  // Function to print first n numbers with two set bits  static void printTwoSetBitNums(int n)  {  // Initialize higher of two sets bits  int x = 1;    // Keep reducing n for every number  // with two set bits  while (n > 0)  {  // Consider all lower set bits for  // current higher set bit  int y = 0;  while (y < x)  {  // Print current number  System.out.print(((1 << x) + (1 << y)) +' ');    // If we have found n numbers  n--;  if (n == 0)  return;    // Consider next lower bit for current  // higher bit.  y++;  }    // Increment higher set bit  x++;  }  }    // Driver program  public static void main (String[] args)   {  int n = 4;  printTwoSetBitNums(n);  } } // This code is contributed by Pramod Kumar
Python3 
# Python3 program to print first n  # numbers with exactly two set bits  # Prints first n numbers  # with two set bits  def printTwoSetBitNums(n) : # Initialize higher of # two sets bits  x = 1 # Keep reducing n for every  # number with two set bits.  while (n > 0) : # Consider all lower set bits  # for current higher set bit  y = 0 while (y < x) : # Print current number  print((1 << x) + (1 << y) end = ' ' ) # If we have found n numbers  n -= 1 if (n == 0) : return # Consider next lower bit  # for current higher bit.  y += 1 # Increment higher set bit  x += 1 # Driver code  printTwoSetBitNums(4) # This code is contributed  # by Smitha
C# 
// C# program to print first n numbers // with exactly two set bits using System; class GFG   {    // Function to print first n  // numbers with two set bits  static void printTwoSetBitNums(int n)  {    // Initialize higher of   // two sets bits  int x = 1;    // Keep reducing n for every  // number with two set bits  while (n > 0)  {    // Consider all lower set bits   // for current higher set bit  int y = 0;  while (y < x)  {    // Print current number  Console.Write(((1 << x) +  (1 << y)) +' ');    // If we have found n numbers  n--;  if (n == 0)  return;    // Consider next lower bit   // for current higher bit.  y++;  }    // Increment higher set bit  x++;  }  }    // Driver program  public static void Main()   {  int n = 4;  printTwoSetBitNums(n);  } }   // This code is contributed by Anant Agarwal.
JavaScript 
<script> // Javascript program to print first n numbers // with exactly two set bits // Prints first n numbers with two set bits function printTwoSetBitNums(n) {  // Initialize higher of two sets bits  let x = 1;  // Keep reducing n for every number  // with two set bits.  while (n > 0)  {    // Consider all lower set bits for  // current higher set bit  let y = 0;  while (y < x)  {    // Print current number  document.write((1 << x) + (1 << y) + ' ');  // If we have found n numbers  n--;  if (n == 0)  return;  // Consider next lower bit for current  // higher bit.  y++;  }  // Increment higher set bit  x++;  } } // Driver code printTwoSetBitNums(4); // This code is contributed by Mayank Tyagi </script>
PHP 
 // PHP program to print  // first n numbers with  // exactly two set bits // Prints first n numbers  // with two set bits function printTwoSetBitNums($n) { // Initialize higher of // two sets bits $x = 1; // Keep reducing n for  // every number with  // two set bits. while ($n > 0) { // Consider all lower set  // bits for current higher  // set bit $y = 0; while ($y < $x) { // Print current number echo (1 << $x) + (1 << $y) ' '; // If we have found n numbers $n--; if ($n == 0) return; // Consider next lower  // bit for current  // higher bit. $y++; } // Increment higher set bit $x++; } } // Driver code printTwoSetBitNums(4); // This code is contributed by Ajit ?>

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retour à la ligne CSS 
3 5 6 9


Complexité temporelle : Sur)

angle aigu

Espace auxiliaire : O(1)



Approche n°2 : Utiliser while et join


L'approche consiste à partir de l'entier 3 et à vérifier si le nombre de bits définis dans sa représentation binaire est égal à 2 ou non. S'il a exactement 2 bits définis, ajoutez-le à la liste des nombres avec 2 bits définis jusqu'à ce que la liste contienne n éléments.

Algorithme

1. Initialisez une liste vide res pour stocker les entiers avec exactement deux bits définis.
2. Initialisez une variable entière i à 3.
3. Alors que la longueur de la liste res est inférieure à n, procédez comme suit :
un. Vérifiez si le nombre de bits définis dans la représentation binaire de i est égal à 2 ou non en utilisant la méthode count() de la chaîne.
b. Si le nombre de bits définis est égal à 2, ajoutez i à la liste res.
c. Incrémenter i de 1.
4. Renvoyez la liste res.

listes en java
C++ 
#include    #include  using namespace std; int countSetBits(int num) {  int count = 0;  while (num > 0) {  count += num & 1;  num >>= 1;  }  return count; } vector<int> numbersWithTwoSetBits(int n) {  vector<int> res;  int i = 3;  while (res.size() < n) {  if (countSetBits(i) == 2) {  res.push_back(i);  }  i++;  }  return res; } int main() {  int n = 3;  vector<int> result = numbersWithTwoSetBits(n);  cout << 'Result: ';  for (int i = 0; i < result.size(); i++) {  cout << result[i] << ' ';  }  cout << endl;  return 0; }
Java 
// Java program for the above approach import java.util.ArrayList; import java.util.List; public class GFG {  // Function to count the number of set bits (binary 1s)  // in an integer  static int countSetBits(int num)  {  int count = 0;  while (num > 0) {  count += num & 1; // Increment count if the last  // bit is set (1)  num >>= 1; // Right shift to check the next bit  }  return count;  }  // Function to generate 'n' numbers with exactly two set  // bits in their binary representation  static List<Integer> numbersWithTwoSetBits(int n)  {  List<Integer> res = new ArrayList<>();  int i = 3; // Start from 3 as the first number with  // two set bits  while (res.size() < n) {  if (countSetBits(i)  == 2) { // Check if the number has exactly  // two set bits  res.add(  i); // Add the number to the result list  }  i++; // Move to the next number  }  return res;  }  public static void main(String[] args)  {  int n = 3; // Number of numbers with two set bits to  // generate  List<Integer> result = numbersWithTwoSetBits(  n); // Get the generated numbers  for (int num : result) {  System.out.print(  num + ' '); // Display the generated numbers  }  System.out.println();  } } // This code is contributed by Susobhan Akhuli
Python3 
def numbersWithTwoSetBits(n): res = [] i = 3 while len(res) < n: if bin(i).count('1') == 2: res.append(i) i += 1 return res n = 3 result = numbersWithTwoSetBits(n) output_string = ' '.join(str(x) for x in result) print(output_string)
C# 
using System; using System.Collections.Generic; class Program {  // Function to count the number of set bits (binary 1s) in an integer  static int CountSetBits(int num)  {  int count = 0;  while (num > 0)  {  count += num & 1; // Increment count if the last bit is set (1)  num >>= 1; // Right shift to check the next bit  }  return count;  }  // Function to generate 'n' numbers with exactly two set bits in their binary representation  static List<int> NumbersWithTwoSetBits(int n)  {  List<int> res = new List<int>();  int i = 3; // Start from 3 as the first number with two set bits  while (res.Count < n)  {  if (CountSetBits(i) == 2) // Check if the number has exactly two set bits  {  res.Add(i); // Add the number to the result list  }  i++; // Move to the next number  }  return res;  }  static void Main(string[] args)  {  int n = 3; // Number of numbers with two set bits to generate  List<int> result = NumbersWithTwoSetBits(n); // Get the generated numbers  Console.Write('Result: ');  foreach (int num in result)  {  Console.Write(num + ' '); // Display the generated numbers  }  Console.WriteLine();  } }
JavaScript 
// Javascript program for the above approach // Function to count the number of set bits (binary 1s) // in an integer function countSetBits(num) {  let count = 0;  while (num > 0) {  count += num & 1; // Increment count if the last  // bit is set (1)  num >>= 1; // Right shift to check the next bit  }  return count; } // Function to generate 'n' numbers with exactly two set // bits in their binary representation function numbersWithTwoSetBits(n) {  let res = [];  let i = 3; // Start from 3 as the first number with  // two set bits  while (res.length < n) {  if (countSetBits(i) === 2) { // Check if the number has exactly  // two set bits  res.push(i); // Add the number to the result list  }  i++; // Move to the next number  }  return res; } // Number of numbers with two set bits to generate let n = 3; // Get the generated numbers let result = numbersWithTwoSetBits(n); // Display the generated numbers console.log(result.join(' ')); // This code is contributed by Susobhan Akhuli

Sortir 
3 5 6

Complexité temporelle : O(n log n) où n est le nombre d'entiers avec exactement deux bits définis. En effet, nous vérifions le nombre de bits définis dans la représentation binaire de chaque entier, ce qui prend un temps O(log n).



Complexité spatiale : O(n) où n est le nombre d'entiers avec exactement deux bits définis. En effet, nous stockons la liste des entiers avec deux bits définis en mémoire.