Given a number N in decimal format, the task is to convert it to the hexadecimal representation of N as a string. Negative numbers are stored in 2’s complement form.
Examples:
Input: N = 134
Output: 86
Explanation:
134 = 00000000000000000000000010001000 in 32 bit representation. Grouping in four-size chunks and converting each chunk to equivalent hexadecimal yields 88. Also, we can see 8*16 + 6 = 134. We will also get the same result by remainder technique discussed in other post.Input: N = -1
Output: ffffffff
Approach: The idea is to store negative numbers in a bigger size to trick the compiler to read it as positive instead of negative and then use the normal remainder technique. Store num in a u_int, size of u_it is greater, it will be positive since MSB is 0.
For Java there is no unsigned int data type. So in case of java, convert number to long and make the 32 higher bits as all zeroes. Idea remains the same as above.
num = (long)Math.pow(2, 32) + num ;
Below is the implementation of the above approach:
// C++ program to convert decimal // to hexadecimal covering negative numbers #include <bits/stdc++.h> using namespace std;
// Function to convert decimal no. // to hexadecimal number string Hex( int num)
{ // map for decimal to hexa, 0-9 are
// straightforward, alphabets a-f used
// for 10 to 15.
map< int , char > m;
char digit = '0' ;
char c = 'a' ;
for ( int i = 0; i <= 15; i++) {
if (i < 10) {
m[i] = digit++;
}
else {
m[i] = c++;
}
}
// string to be returned
string res = "" ;
// check if num is 0 and directly return "0"
if (!num) {
return "0" ;
}
// if num>0, use normal technique as
// discussed in other post
if (num > 0) {
while (num) {
res = m[num % 16] + res;
num /= 16;
}
}
// if num<0, we need to use the elaborated
// trick above, lets see this
else {
// store num in a u_int, size of u_it is greater,
// it will be positive since msb is 0
u_int n = num;
// use the same remainder technique.
while (n) {
res = m[n % 16] + res;
n /= 16;
}
}
return res;
} // Driver Code int main()
{ int x = 134, y = -1, z = -234;
cout << "Hexa representation for" << endl;
cout << x << " is " << Hex(x) << endl;
cout << y << " is " << Hex(y) << endl;
cout << z << " is " << Hex(z) << endl;
return 0;
} |
// Java program to convert decimal // to hexadecimal covering negative numbers import java.util.*;
import java.util.HashMap;
import java.util.Map;
class GFG
{ // Function to convert decimal no. // to hexadecimal number static String Hex( int num)
{ // map for decimal to hexa, 0-9 are
// straightforward, alphabets a-f used
// for 10 to 15.
HashMap<Integer, Character> m = new HashMap<Integer, Character>();
char digit = '0' ;
char c = 'a' ;
for ( int i = 0 ; i <= 15 ; i++) {
if (i < 10 ) {
m.put(i, digit);
digit++;
}
else {
m.put(i, c);
c++;
}
}
// string to be returned
String res = "" ;
// check if num is 0 and directly return "0"
if (num == 0 ) {
return "0" ;
}
// if num>0, use normal technique as
// discussed in other post
if (num > 0 ) {
while (num != 0 ) {
res = m.get(num % 16 ) + res;
num /= 16 ;
}
}
// if num<0, we need to use the elaborated
// trick above, lets see this
else {
// store num in a u_int, size of u_it is greater,
// it will be positive since msb is 0
long n = num;
n = ( long )Math.pow( 2 , 32 ) + num ;
// use the same remainder technique.
while (n != 0 ) {
res = m.get(n % 16 ) + res;
n /= 16 ;
}
}
return res;
} // Driver Code public static void main(String []args)
{ int x = 134 , y = - 1 , z = - 234 ;
System.out.println( "Hexa representation for" );
System.out.println(x + " is " + Hex(x));
System.out.println( y + " is " + Hex(y));
System.out.println( z + " is " + Hex(z));
} } // This code is contributed by chitranayal |
# Python3 program to convert decimal # to hexadecimal covering negative numbers # Function to convert decimal no. # to hexadecimal number def Hex (num) :
# map for decimal to hexa, 0-9 are
# straightforward, alphabets a-f used
# for 10 to 15.
m = dict .fromkeys( range ( 16 ), 0 );
digit = ord ( '0' );
c = ord ( 'a' );
for i in range ( 16 ) :
if (i < 10 ) :
m[i] = chr (digit);
digit + = 1 ;
else :
m[i] = chr (c);
c + = 1
# string to be returned
res = "";
# check if num is 0 and directly return "0"
if ( not num) :
return "0" ;
# if num>0, use normal technique as
# discussed in other post
if (num > 0 ) :
while (num) :
res = m[num % 16 ] + res;
num / / = 16 ;
# if num<0, we need to use the elaborated
# trick above, lets see this
else :
# store num in a u_int, size of u_it is greater,
# it will be positive since msb is 0
n = num + 2 * * 32 ;
# use the same remainder technique.
while (n) :
res = m[n % 16 ] + res;
n / / = 16 ;
return res;
# Driver Code if __name__ = = "__main__" :
x = 134 ; y = - 1 ; z = - 234 ;
print ( "Hexa representation for" );
print (x, "is" , Hex (x));
print (y, "is" , Hex (y));
print (z, "is" , Hex (z));
# This code is contributed by AnkitRai01 |
// C# program to convert decimal // to hexadecimal covering negative numbers using System;
using System.Collections.Generic;
public class GFG {
// Function to convert decimal no.
// to hexadecimal number
static string Hex( long num)
{
// map for decimal to hexa, 0-9 are
// straightforward, alphabets a-f used
// for 10 to 15.
IDictionary< long , char > m
= new Dictionary< long , char >();
char digit = '0' ;
char c = 'a' ;
for ( int i = 0; i <= 15; i++) {
if (i < 10) {
m[i] = digit;
digit++;
}
else {
m[i] = c;
c++;
}
}
// string to be returned
string res = "" ;
// check if num is 0 and directly return "0"
if (num == 0) {
return "0" ;
}
// if num>0, use normal technique as
// discussed in other post
if (num > 0) {
while (num != 0) {
res = m[num % 16] + res;
num /= 16;
}
}
// if num<0, we need to use the elaborated
// trick above, lets see this
else {
// we shall convert num to a 32 bit number
num = ( long )Math.Pow(2, 32) + num;
long n = num;
// use the same remainder technique.
while (n != 0) {
res = m[n % 16] + res;
n /= 16;
}
}
return res;
}
public static void Main( string [] args)
{
long x = 134, y = -1, z = -234;
Console.WriteLine( "Hexa representation for" );
Console.WriteLine(x + " is " + Hex(x));
Console.WriteLine(y + " is " + Hex(y));
Console.WriteLine(z + " is " + Hex(z));
}
} // this code is contributed by phasing17 |
<script> // JavaScript program to convert decimal // to hexadecimal covering negative numbers // Function to convert decimal no. // to hexadecimal number function Hex(num)
{ // map for decimal to hexa, 0-9 are
// straightforward, alphabets a-f used
// for 10 to 15.
let m = new Map();
let digit = '0' .charCodeAt(0);
let c = 'a' .charCodeAt(0);
for (let i = 0; i <= 15; i++) {
if (i < 10) {
m.set(i, String.fromCharCode(digit));
digit++;
}
else {
m.set(i, String.fromCharCode(c));
c++;
}
}
// string to be returned
let res = "" ;
// check if num is 0 and directly return "0"
if (num == 0) {
return "0" ;
}
// if num>0, use normal technique as
// discussed in other post
if (num > 0) {
while (num != 0) {
res = m.get(num % 16) + res;
num =Math.floor(num/ 16);
}
}
// if num<0, we need to use the elaborated
// trick above, lets see this
else {
// store num in a u_int, size of u_it is greater,
// it will be positive since msb is 0
let n = num+Math.pow(2,32);
// use the same remainder technique.
while (n != 0) {
res = m.get(n % 16) + res;
n = Math.floor(n/16);
}
}
return res;
} // Driver Code let x = 134, y = -1, z = -234; document.write( "Hexa representation for<br>" );
document.write(x + " is " + Hex(x)+ "<br>" );
document.write( y + " is " + Hex(y)+ "<br>" );
document.write( z + " is " + Hex(z)+ "<br>" );
// This code is contributed by rag2127 </script> |
Hexa representation for 134 is 86 -1 is ffffffff -234 is ffffff16
Time Complexity: O(log16num)
Auxiliary Space: O(1)
Method: Using look up table method
The function decToHex takes an integer num as input and returns its hexadecimal representation as a string. It first initializes a lookup table hexLookupTable which maps integers from 0 to 15 to their corresponding hexadecimal characters.
If num is positive, it uses the same technique as discussed in the previous post to convert it to hexadecimal. If num is negative, it converts it to an unsigned 32-bit integer n using a static cast, and then uses the same technique to convert n to hexadecimal. Finally, it returns the resulting hexadecimal string.
#include <iostream> #include <unordered_map> using namespace std;
// Function to convert decimal no. to hexadecimal number string decToHex( int num) {
// Lookup table for hexadecimal values
unordered_map< int , char > hexLookupTable{
{0, '0' }, {1, '1' }, {2, '2' }, {3, '3' },
{4, '4' }, {5, '5' }, {6, '6' }, {7, '7' },
{8, '8' }, {9, '9' }, {10, 'A' }, {11, 'B' },
{12, 'C' }, {13, 'D' }, {14, 'E' }, {15, 'F' }
};
string res = "" ;
// check if num is 0 and directly return "0"
if (num == 0) {
return "0" ;
}
// if num>0, use normal technique as discussed in other post
if (num > 0) {
while (num) {
res = hexLookupTable[num % 16] + res;
num /= 16;
}
}
// if num<0, we need to use the elaborated trick above
else {
// convert num to 32-bit unsigned integer
uint32_t n = static_cast <uint32_t>(num);
// use the same remainder technique
while (n) {
res = hexLookupTable[n % 16] + res;
n /= 16;
}
}
return res;
} int main() {
int x = 134, y = -1, z = -234;
cout << "Hexa representation for" << endl;
cout << x << " is " << decToHex(x) << endl;
cout << y << " is " << decToHex(y) << endl;
cout << z << " is " << decToHex(z) << endl;
return 0;
} |
import java.util.HashMap;
public class GFG {
// Function to convert a decimal number to its
// hexadecimal representation
public static String decToHex( int num) {
// Create a lookup table to map decimal digits to
// their corresponding hexadecimal characters
HashMap<Integer, Character> hexLookupTable = new HashMap<>();
hexLookupTable.put( 0 , '0' );
hexLookupTable.put( 1 , '1' );
hexLookupTable.put( 2 , '2' );
hexLookupTable.put( 3 , '3' );
hexLookupTable.put( 4 , '4' );
hexLookupTable.put( 5 , '5' );
hexLookupTable.put( 6 , '6' );
hexLookupTable.put( 7 , '7' );
hexLookupTable.put( 8 , '8' );
hexLookupTable.put( 9 , '9' );
hexLookupTable.put( 10 , 'A' );
hexLookupTable.put( 11 , 'B' );
hexLookupTable.put( 12 , 'C' );
hexLookupTable.put( 13 , 'D' );
hexLookupTable.put( 14 , 'E' );
hexLookupTable.put( 15 , 'F' );
StringBuilder res = new StringBuilder();
// Special cases
if (num == 0 ) {
return "0" ;
}
// If the number is positive, convert it to hexadecimal
if (num > 0 ) {
while (num > 0 ) {
res.insert( 0 , hexLookupTable.get(num % 16 ));
num /= 16 ;
}
}
// If the number is negative, convert its two's complement
// to hexadecimal
else {
// Convert the negative number to a 32-bit unsigned
// integer
long n = ( long ) num & 0xffffffffL;
while (n > 0 ) {
res.insert( 0 , hexLookupTable.get(( int ) (n % 16 )));
n /= 16 ;
}
}
return res.toString();
}
public static void main(String[] args) {
int x = 134 , y = - 1 , z = - 234 ;
System.out.println( "Hexadecimal representation for:" );
System.out.println(x + " is " + decToHex(x));
System.out.println(y + " is " + decToHex(y));
System.out.println(z + " is " + decToHex(z));
}
} |
def dec_to_hex(num):
# Lookup table for hexadecimal values
hex_lookup_table = {
0 : '0' , 1 : '1' , 2 : '2' , 3 : '3' ,
4 : '4' , 5 : '5' , 6 : '6' , 7 : '7' ,
8 : '8' , 9 : '9' , 10 : 'A' , 11 : 'B' ,
12 : 'C' , 13 : 'D' , 14 : 'E' , 15 : 'F'
}
res = ""
# Check if num is 0 and directly return "0"
if num = = 0 :
return "0"
# If num > 0, use normal technique
if num > 0 :
while num:
res = hex_lookup_table[num % 16 ] + res
num / / = 16
# If num < 0, use the same remainder technique
else :
# Convert num to 32-bit unsigned integer
n = num & 0xFFFFFFFF
while n:
res = hex_lookup_table[n % 16 ] + res
n / / = 16
return res
x = 134
y = - 1
z = - 234
print ( "Hexa representation for" )
print (x, "is" , dec_to_hex(x))
print (y, "is" , dec_to_hex(y))
print (z, "is" , dec_to_hex(z))
|
using System;
using System.Collections.Generic;
namespace DecimalToHexConversionExample
{ class Program
{
// Function to convert decimal number to hexadecimal number
static string DecToHex( int num)
{
// Lookup table for hexadecimal values
Dictionary< int , char > hexLookupTable = new Dictionary< int , char >
{
{0, '0' }, {1, '1' }, {2, '2' }, {3, '3' },
{4, '4' }, {5, '5' }, {6, '6' }, {7, '7' },
{8, '8' }, {9, '9' }, {10, 'A' }, {11, 'B' },
{12, 'C' }, {13, 'D' }, {14, 'E' }, {15, 'F' }
};
string res = "" ;
// Check if num is 0 and directly return "0"
if (num == 0)
{
return "0" ;
}
// If num > 0, use normal technique as discussed in other post
if (num > 0)
{
while (num > 0)
{
res = hexLookupTable[num % 16] + res;
num /= 16;
}
}
// If num < 0, we need to use the elaborated trick above
else
{
// Convert num to 32-bit unsigned integer
uint n = ( uint )Math.Abs(num);
// Use the same remainder technique
while (n > 0)
{
res = hexLookupTable[( int )(n % 16)] + res;
n /= 16;
}
}
return res;
}
// Main code to test the function
static void Main( string [] args)
{
int x = 134, y = -1, z = -234;
Console.WriteLine( "Hexa representation for" );
Console.WriteLine($ "{x} is {DecToHex(x)}" );
Console.WriteLine($ "{y} is {DecToHex(y)}" );
Console.WriteLine($ "{z} is {DecToHex(z)}" );
}
}
} |
function decToHex(num) {
// Lookup table for hexadecimal values
const hexLookupTable = {
0: '0' , 1: '1' , 2: '2' , 3: '3' ,
4: '4' , 5: '5' , 6: '6' , 7: '7' ,
8: '8' , 9: '9' , 10: 'A' , 11: 'B' ,
12: 'C' , 13: 'D' , 14: 'E' , 15: 'F'
};
let res = "" ;
// Check if num is 0 and directly return "0"
if (num === 0) {
return "0" ;
}
// If num > 0, use normal technique
if (num > 0) {
while (num > 0) {
res = hexLookupTable[num % 16] + res;
num = Math.floor(num / 16);
}
}
// If num < 0, use the elaborated trick
else {
// Convert num to 32-bit unsigned integer
let n = Math.abs(num);
// Use the same remainder technique
while (n > 0) {
res = hexLookupTable[n % 16] + res;
n = Math.floor(n / 16);
}
}
return res;
} // Test the function const x = 134, y = -1, z = -234; console.log( "Hexadecimal representation for:" );
console.log(`${x} is ${decToHex(x)}`); console.log(`${y} is ${decToHex(y)}`); console.log(`${z} is ${decToHex(z)}`); |
Hexa representation for 134 is 86 -1 is FFFFFFFF -234 is FFFFFF16
Time complexity: The time complexity of this function is O(log(n)), where num is the magnitude of the input integer.
Auxiliary space: The auxiliary space used by this function is O(log(n)), where num is the magnitude of the input integer.