Gegeven een getal n, print dan eerst n positieve gehele getallen met precies twee ingestelde bits in hun binaire representatie.
Voorbeelden:
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 Eenvoudige oplossing is om alle positieve gehele getallen één voor één te beschouwen, beginnend bij 1. Controleer voor elk getal of het precies twee sets bits heeft. Als een getal precies twee ingestelde bits heeft, drukt u het af en verhoogt u het aantal van dergelijke getallen.
Een Efficiënte oplossing is om dergelijke cijfers rechtstreeks te genereren. Als we de getallen duidelijk waarnemen, kunnen we ze herschrijven zoals hieronder aangegeven pow(21)+pow(20) pow(22)+pow(20) pow(22)+pow(21) pow(23)+pow(20) pow(23)+pow(21) pow(23)+pow(22) .........
Alle getallen kunnen in oplopende volgorde worden gegenereerd volgens de hoogste van twee ingestelde bits. Het idee is om de hoogste van twee bits één voor één te repareren. Voor de huidige hogere ingestelde bits, overweeg alle lagere bits en druk de gevormde getallen af.
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 ?> Uitgang:
upcasting
3 5 6 9
Tijdcomplexiteit: Op)
sneltoets met alleen hoofdletters excel
Hulpruimte: O(1)
Aanpak #2: gebruik while en join
De aanpak is om te beginnen vanaf het gehele getal 3 en te controleren of het aantal ingestelde bits in de binaire representatie gelijk is aan 2 of niet. Als het precies 2 set bits heeft, voeg het dan toe aan de lijst met getallen met 2 set bits totdat de lijst n elementen bevat.
Algoritme
1. Initialiseer een lege lijst res om de gehele getallen met precies twee ingestelde bits op te slaan.
2. Initialiseer een geheel getalvariabele i tot en met 3.
3. Terwijl de lengte van de lijst res kleiner is dan n, doe je het volgende:
A. Controleer of het aantal ingestelde bits in de binaire representatie van i gelijk is aan 2 of niet met behulp van de count()-methode van de string.
B. Als het aantal ingestelde bits gelijk is aan 2, voeg dan i toe aan de lijst res.
C. Verhoog i met 1.
4. Retourneer de lijstres.
#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
Uitvoer
3 5 6
Tijdcomplexiteit: O(n log n) waarbij n het aantal gehele getallen is met precies twee ingestelde bits. Dit komt omdat we het aantal ingestelde bits in de binaire representatie van elk geheel getal controleren, wat O(log n) tijd kost.
Ruimtecomplexiteit: O(n) waarbij n het aantal gehele getallen is met precies twee ingestelde bits. Dit komt omdat we de lijst met gehele getallen met twee ingestelde bits in het geheugen opslaan.
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