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++
// C++ code to implement the approach#include <bits/stdc++.h>using namespace std;// Function to convert a number to binaryvector<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 lengthvector<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 codeint 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 approachimport 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 approachimport math# Function to convert a number to binarydef 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 lengthdef 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 codeif __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 approachusing 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> |
Output
0010 0011 0100 0101 0110 0111 1000
Time Complexity: O(N * logR) where N = (R – L + 1)
Auxiliary Space: O(N * logR)
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