Genere todos los números binarios en el rango [L, R] con la misma longitud

Posted on by Rudeus Greyrat

Dados dos números enteros positivos L y R . La tarea es convertir todos los números de L a R en números binarios. La longitud de todos los números binarios debe ser la misma.

Ejemplos:

Entrada: L = 2, R = 4
Salida:
010
011
100
Explicación: La representación binaria de los números: 2 = 10, 3 = 11 y 4 = 100.
Para que los números tengan la misma longitud, se agrega un 0 anterior al binario representación de 3 y 4.

Entrada: L = 2, R = 8
Salida:
0010
0011
0100
0101
0110
0111
1000

 

Enfoque: siga el enfoque mencionado a continuación para resolver el problema.

  • Para determinar la longitud de los números binarios resultantes, lleve el logaritmo de R+1 a la base 2.
  • Luego, recorra de L a R y convierta cada número a binario de longitud determinada.
  • Guarde cada número e imprímalo al final.

A continuación se muestra la implementación del enfoque anterior.

C++

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to convert a number to binary
vector<int> convertToBinary(int num,
                            int length)
{
    vector<int> bits(length, 0);
    if (num == 0) {
        return bits;
    }
 
    int i = length - 1;
    while (num != 0) {
        bits[i--] = (num % 2);
 
        // Integer division
        // gives quotient
        num = num / 2;
    }
    return bits;
}
 
// Function to convert all numbers
// in range [L, R] to binary of
// same length
vector<vector<int> > getAllBinary(int l,
                                  int r)
{
 
    // Length of the binary numbers
    int n = (int) ceil(log(r+1) / log (2));
 
    vector<vector<int> > binary_nos;
 
    for (int i = l; i <= r; i++) {
        vector<int> bits =
            convertToBinary(i, n);
        binary_nos.push_back(bits);
    }
 
    return binary_nos;
}
 
// Driver code
int main()
{
    int L = 2, R = 8;
    vector<vector<int> > binary_nos =
        getAllBinary(L, R);
    for (int i = 0; i < binary_nos.size();
        i++) {
        for (int j = 0; j <
            binary_nos[i].size(); j++)
            cout << binary_nos[i][j];
        cout << endl;
    }
    return 0;
}

Java

// Java code to implement the approach
import java.util.*;
public class GFG
{
 
  // Function to convert a number to binary
  static ArrayList<Integer> convertToBinary(int num,
                                            int length)
  {
    ArrayList<Integer> bits= new ArrayList<Integer>();
    for(int i = 0; i < length; i++) {
      bits.add(0);
    }
    if (num == 0) {
      return bits;
    }
 
    int i = length - 1;
    while (num != 0) {
      bits.set(i, (num % 2));
      i = i - 1;
 
      // Integer division
      // gives quotient
      num = num / 2;
    }
 
    return bits;
  }
 
  // Function to convert all numbers
  // in range [L, R] to binary of
  // same length
  static ArrayList<ArrayList<Integer> > getAllBinary(int l,
                                                     int r)
  {
 
    // Length of the binary numbers
    double x = Math.log(r+1);
    double y = Math.log (2);
    int n = (int) Math.ceil(x / y);
 
    ArrayList<ArrayList<Integer> > binary_nos =
      new ArrayList<ArrayList<Integer> >();
 
    for (int i = l; i <= r; i++) {
      ArrayList<Integer> bits =
        convertToBinary(i, n);
      binary_nos.add(bits);
    }
 
    return binary_nos;
  }
 
  // Driver code
  public static void main(String args[])
  {
    int L = 2, R = 8;
    ArrayList<ArrayList<Integer> > binary_nos =
      getAllBinary(L, R);
    for (int i = 0; i < binary_nos.size(); i++) {
      for (int j = 0; j < binary_nos.get(i).size(); j++) {
        System.out.print(binary_nos.get(i).get(j));
      }
      System.out.println();
    }
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Python 3 code to implement the approach
import math
 
# Function to convert a number to binary
def convertToBinary(num, length):
 
    bits = [0]*(length)
    if (num == 0):
        return bits
 
    i = length - 1
    while (num != 0):
        bits[i] = (num % 2)
        i -= 1
 
        # Integer division
        # gives quotient
        num = num // 2
 
    return bits
 
# Function to convert all numbers
# in range [L, R] to binary of
# same length
def getAllBinary(l, r):
 
    # Length of the binary numbers
    n = int(math.ceil(math.log(r+1)/ math.log(2)))
 
    binary_nos = []
 
    for i in range(l, r + 1):
        bits = convertToBinary(i, n)
        binary_nos.append(bits)
 
    return binary_nos
 
# Driver code
if __name__ == "__main__":
 
    L = 2
    R = 8
    binary_nos = getAllBinary(L, R)
    for i in range(len(binary_nos)):
        for j in range(len(binary_nos[i])):
            print(binary_nos[i][j], end="")
        print()
 
        # This code is contributed by ukasp.

C#

// C# code to implement the approach
using System;
using System.Collections.Generic;
public class GFG
{
 
  // Function to convert a number to binary
  static List<int> convertToBinary(int num, int length)
  {
    List<int> bits = new List<int>();
    int i;
    for (i = 0; i < length; i++)
    {
      bits.Add(0);
    }
    if (num == 0)
    {
      return bits;
    }
 
    i = length - 1;
    while (num != 0)
    {
      bits[i] = (num % 2);
      i = i - 1;
 
      // Integer division
      // gives quotient
      num = num / 2;
    }
 
    return bits;
  }
 
  // Function to convert all numbers
  // in range [L, R] to binary of
  // same length
  static List<List<int>> getAllBinary(int l, int r)
  {
 
    // Length of the binary numbers
    double x = Math.Log(r + 1);
    double y = Math.Log(2);
    int n = (int)Math.Ceiling(x / y);
 
    List<List<int>> binary_nos = new List<List<int>>();
 
    for (int i = l; i <= r; i++)
    {
      List<int> bits = convertToBinary(i, n);
      binary_nos.Add(bits);
    }
 
    return binary_nos;
  }
 
  // Driver code
  public static void Main()
  {
    int L = 2, R = 8;
    List<List<int>> binary_nos = getAllBinary(L, R);
    for (int i = 0; i < binary_nos.Count; i++)
    {
      for (int j = 0; j < binary_nos[i].Count; j++)
      {
        Console.Write(binary_nos[i][j]);
      }
      Console.WriteLine("");
    }
  }
}
 
// This code is contributed by Saurabh Jaiswal

Javascript

<script>
    // JavaScript code to implement the approach
 
    // Function to convert a number to binary
    const convertToBinary = (num, length) => {
        let bits = new Array(length).fill(0);
        if (num == 0) {
            return bits;
        }
 
        let i = length - 1;
        while (num != 0) {
            bits[i--] = (num % 2);
 
            // Integer division
            // gives quotient
            num = parseInt(num / 2);
        }
        return bits;
    }
 
    // Function to convert all numbers
    // in range [L, R] to binary of
    // same length
    const getAllBinary = (l, r) => {
 
        // Length of the binary numbers
        let n = Math.ceil(Math.log(r + 1) / Math.log(2));
 
        let binary_nos = [];
 
        for (let i = l; i <= r; i++) {
            let bits = convertToBinary(i, n);
            binary_nos.push(bits);
        }
 
        return binary_nos;
    }
 
    // Driver code
 
    let L = 2, R = 8;
    let binary_nos = getAllBinary(L, R);
    for (let i = 0; i < binary_nos.length;
        i++) {
        for (let j = 0; j <
            binary_nos[i].length; j++)
            document.write(binary_nos[i][j]);
        document.write("<br/>");
    }
     
    // This code is contributed by rakeshsahni
 
</script>
Producción

0010
0011
0100
0101
0110
0111
1000

Complejidad de Tiempo: O(N * logR) donde N = (R – L + 1)
Espacio Auxiliar: O(N * logR)

Publicación traducida automáticamente

Artículo escrito por Code_r y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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