The program is written here to convert an octal number into its equivalent binary 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 binary.
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 binary by dividing it with 2 until the quotient of the division becomes zero and then take the remainders in reversed order.
The binary equivalent is
INPUT: An octal number
OUTPUT: Equivalent binary number
PROCESS:
Step 1: [Taking the input]
Read n [an octal number]
Step 2: [Converting from octal to binary]
Set d<-0
Set p<-1
Set i<-0
While n>0 repeat
[Calculating the power of 8 and multiplied with the octal digit]
Set d<-d+(n mod 10)×p
[Increasing the power]
Set p<-p×8
Set n<-n/10
[End of ‘while’ loop]
[Converting from decimal to binary]
While d>0 repeat
Set binary[i]<-character of (d mod 2+48)
Set d<-d/2
[End of ‘while’ loop]
Set binary[i]<-'\0’
[printing the binary number]
Reverse the string stored in the variable ‘binary’
Print the value of the variable ‘binary’
Step 3: Stop.
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 binary
while(d>0)----------------------------------------O(m)
{
binary[i++]=(char)(d%2+48);
d=d/2;
}
The time complexity of converting an octal number to binary 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.
The space complexity is O(1) as it requires a constant number of spaces to execute for any given input.