Generate all binary numbers in range [L, R] with same length
Given two positive integer numbers L and R. The task is to convert all the numbers from L to R to binary number. The length of all binary numbers should be same.
Examples:
Input: L = 2, R = 4
Output:
010
011
100
Explanation: The binary representation of the numbers: 2 = 10, 3 = 11 and 4 = 100.
For the numbers to have same length one preceding 0 is added to the binary representation of both 3 and 4.
Input: L = 2, R = 8
Output:
0010
0011
0100
0101
0110
0111
1000
Approach: Follow the approach mentioned below to solve the problem.
- To determine the length of resultant binary numbers, take log of R+1 to the base 2.
- Then traverse from L to R and convert every number to binary of determined length.
- Store each number and print it at the end.
Below is the implementation of the above approach
C++
#include <bits/stdc++.h>
using namespace std;
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);
num = num / 2;
}
return bits;
}
vector<vector< int > > getAllBinary( int l,
int r)
{
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;
}
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
import java.util.*;
public class GFG
{
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 ;
num = num / 2 ;
}
return bits;
}
static ArrayList<ArrayList<Integer> > getAllBinary( int l,
int r)
{
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;
}
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();
}
}
}
|
Python3
import math
def convertToBinary(num, length):
bits = [ 0 ] * (length)
if (num = = 0 ):
return bits
i = length - 1
while (num ! = 0 ):
bits[i] = (num % 2 )
i - = 1
num = num / / 2
return bits
def getAllBinary(l, r):
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
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 ()
|
C#
using System;
using System.Collections.Generic;
public class GFG
{
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;
num = num / 2;
}
return bits;
}
static List<List< int >> getAllBinary( int l, int r)
{
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;
}
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( "" );
}
}
}
|
Javascript
<script>
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);
num = parseInt(num / 2);
}
return bits;
}
const getAllBinary = (l, r) => {
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;
}
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/>" );
}
</script>
|
Output
0010
0011
0100
0101
0110
0111
1000
Time Complexity: O(N * logR) where N = (R – L + 1)
Auxiliary Space: O(N * logR)
Last Updated :
23 Dec, 2021
Like Article
Save Article
Share your thoughts in the comments
Please Login to comment...