Dots skaitlis n vispirms izdrukā n pozitīvus veselus skaitļus ar tieši diviem kopas bitiem to binārajā attēlojumā.
Piemēri:
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
A Vienkāršs risinājums ir ņemt vērā visus pozitīvos veselos skaitļus pa vienam, sākot no 1. Katram skaitlim pārbaudiet, vai tajā ir tieši divas bitu kopas. Ja skaitlim ir tieši divi iestatīti biti, izdrukājiet to un palieliniet šādu skaitļu skaitu.
An Efektīvs risinājums ir tieši ģenerēt šādus skaitļus. Ja mēs skaidri novērojam skaitļus, mēs varam tos pārrakstīt, kā norādīts zemāk pow(21)+pow(20) pow(22)+pow(20) pow(22)+pow(21) pow(23)+pow(20) pow(23)+pow(21) pow(23)+pow(22) .........
Visus skaitļus var ģenerēt augošā secībā atbilstoši lielākajam no diviem bitiem. Ideja ir noteikt augstāku no diviem bitiem pa vienam. Pašreizējai augstākas kopas bitiem ņemiet vērā visus zemākos bitus un izdrukājiet izveidotos skaitļus.
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 ?> Izvade:
c# satur virkni
3 5 6 9
Laika sarežģītība: O(n)
pavedienu sinhronizācija
Palīgtelpa: O(1)
2. pieeja: kamēr un pievienošanās izmantošana
Pieeja ir sākt no vesela skaitļa 3 un pārbaudīt, vai iestatīto bitu skaits tā binārajā attēlojumā ir vienāds ar 2 vai nē. Ja tajā ir precīzi 2 biti, pievienojiet to skaitļu sarakstam ar 2 bitiem, līdz sarakstā ir n elementi.
Algoritms
1. Inicializējiet tukšu sarakstu, lai saglabātu veselus skaitļus ar tieši diviem iestatītiem bitiem.
2. Inicializējiet veselu skaitļu mainīgo no i līdz 3.
3. Kamēr saraksta res garums ir mazāks par n, rīkojieties šādi:
a. Pārbaudiet, vai iestatīto bitu skaits i binārajā attēlojumā ir vienāds ar 2, izmantojot virknes metodi count().
b. Ja iestatīto bitu skaits ir vienāds ar 2, pievienojiet i sarakstam res.
c. Palieliniet i par 1.
4. Atgrieziet sarakstu res.
#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
Izvade
3 5 6
Laika sarežģītība: O(n log n), kur n ir veselu skaitļu skaits ar tieši diviem iestatītiem bitiem. Tas ir tāpēc, ka mēs pārbaudām iestatīto bitu skaitu katra vesela skaitļa binārajā attēlojumā, kas aizņem O(log n) laiku.
Telpas sarežģītība: O(n), kur n ir veselu skaitļu skaits ar tieši diviem iestatītiem bitiem. Tas ir tāpēc, ka mēs atmiņā saglabājam veselu skaitļu sarakstu ar diviem iestatītiem bitiem.
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