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Octal to Hexadecimal
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The program is written here to convert an octal number into its equivalent hexadecimal number. The octal number is taken as input then it is first converted into its equivalent decimal number and then the decimal number is converted into its equivalent hexadecimal.
For example, if an octal number is
Now, the equivalent decimal number will be calculated by multiplying each digit of the octal number with the increasing power of 8 (the power will starts from 0 and will increase from right to left).
Therefore, the equivalent decimal number:
5× + 7×
= 40 + 7
=
Now, this decimal number will be converted into its equivalent hexadecimal by dividing it with 16 until the quotient of the division becomes zero and then take the remainders in reversed order.
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The hexadecimal equivalent is
Algorithm
INPUT: An octal number
OUTPUT: Equivalent hexadecimal number
PROCESS:
Step 1: [Taking the input]
Read n [the octal number]
Step 2: [Converting the octal number to hexadecimal]
Set d<-0
Set p<-1
While n>0 repeat
[Calculating the power of 8 and multiplied with the octal digit]
Set d<-d+(n%10)×p
[Increasing the power]
Set p<-p×8
Set n<-n/10
[End of ‘while’ loop]
[Converting from decimal to hexadecimal]
Set p<-0
While d>0 repeat
Set r<-d mod 16
If r≤9 then
Set hex[p]<-character of (r+48)
Set p<-p+1
else if r=10 then
Set hex[p]<-'A'
Set p<-p+1
else if r=11 then
Set hex[p]<-'B'
Set p<-p+1
else if r=12 then
Set hex[p]<-'C'
Set p<-p+1
else if r=13 then
Set hex[p]<-'D'
Set p<-p+1
else if r=14 then
Set hex[p]<-'E'
Set p<-p+1
else if r=15 then
Set hex[p]<-'F'
Set p<-p+1
[End of ‘if-else if’]
Set d<-d/16
[End of ‘while’ loop]
Reverse the string stored in ‘hex’
[Printing the hexadecimal number]
Print the string stored in ‘hex’
Step 3: Stop.
Code
TIME COMPLEXITY:
while(n>0)-----------------------------------O(k)
{
//calculating the power of 8 and multiplied
//with the octal digit
d=d+(n%10)*p;
//increasing the power
p=p*8;
n=n/10;
}
//converting from decimal to hexadecimal
p=0;
while(d>0)-------------------------------O(m)
{
r=d%16;
if(r<=9)
hex[p++]=(char)(r+48);
else if(r==10)
hex[p++]='A';
else if(r==11)
hex[p++]='B';
else if(r==12)
hex[p++]='C';
else if(r==13)
hex[p++]='D';
else if(r==14)
hex[p++]='E';
else if(r==15)
hex[p++]='F';
d=d/16;
}
The time complexity of this program is O(k+m) where ‘k’ is the number of digits of the octal number and ‘m’ is the number of digits of the decimal number.
SPACE COMPLEXITY:
The space complexity of this program is O(1) as it requires a constant number of memory spaces to execute for any given input.
