Complement of a number with any base b
In this post, a general method of finding complement with any arbitrary base is discussed.
Remember: It is recommended to go through below as follows:
Steps to find (b-1)’s complement: To find (b-1)’s complement,
- Subtract each digit of the number from the largest number in the number system with base .
- For example, if the number is a three-digit number in base 9, then subtract the number from 888 as 8 is the largest number in base 9 number system.
- The obtained result is the (b-1)’s (8’s complement) complement.
Steps to find b’s complement: To find b’s complement, just add 1 to the calculated (b-1)’s complement. Now, this holds true for any base in the number system that exists. It can be tested with familiar bases that is the 1’s and 2’s complement.
Example:
Let the number be 10111 base 2 (b=2)
Then, 1's complement will be 01000 (b-1)
2's complement will be 01001 (b)
Taking a number with Octal base:
Let the number be -456.
Then 7's complement will be 321
and 8's complement will be 322
Below is the implementation of the above idea as follows:
C++
#include <cmath>
#include <iostream>
using namespace std;
int prevComplement( int n, int b)
{
int maxDigit, maxNum = 0, digits = 0, num = n;
while (n != 0) {
digits++;
n = n / 10;
}
maxDigit = b - 1;
while (digits--) {
maxNum = maxNum * 10 + maxDigit;
}
return maxNum - num;
}
int complement( int n, int b)
{
return prevComplement(n, b) + 1;
}
int main()
{
cout << prevComplement(25, 7) << endl;
cout << complement(25, 7);
return 0;
}
|
Java
class GFG
{
static int prevComplement( int n, int b)
{
int maxDigit, maxNum = 0 ,
digits = 0 , num = n;
while (n != 0 )
{
digits++;
n = n / 10 ;
}
maxDigit = b - 1 ;
while ((digits--) > 0 )
{
maxNum = maxNum * 10 + maxDigit;
}
return maxNum - num;
}
static int complement( int n, int b)
{
return prevComplement(n, b) + 1 ;
}
public static void main(String args[])
{
System.out.println(prevComplement( 25 , 7 ));
System.out.println(complement( 25 , 7 ));
}
}
|
Python 3
def prevComplement(n, b) :
maxNum, digits, num = 0 , 0 , n
while n > 1 :
digits + = 1
n = n / / 10
maxDigit = b - 1
while digits :
maxNum = maxNum * 10 + maxDigit
digits - = 1
return maxNum - num
def complement(n, b) :
return prevComplement(n, b) + 1
if __name__ = = "__main__" :
print (prevComplement( 25 , 7 ))
print (complement( 25 , 7 ))
|
C#
class GFG
{
static int prevComplement( int n, int b)
{
int maxDigit, maxNum = 0,
digits = 0, num = n;
while (n != 0)
{
digits++;
n = n / 10;
}
maxDigit = b - 1;
while ((digits--) > 0)
{
maxNum = maxNum * 10 + maxDigit;
}
return maxNum - num;
}
static int complement( int n, int b)
{
return prevComplement(n, b) + 1;
}
public static void Main()
{
System.Console.WriteLine(prevComplement(25, 7));
System.Console.WriteLine(complement(25, 7));
}
}
|
PHP
<?php
function prevComplement( $n , $b )
{
$maxNum = 0;
$digits = 0;
$num = $n ;
while ((int) $n != 0)
{
$digits ++;
$n = $n / 10;
}
$maxDigit = $b - 1;
while ( $digits --)
{
$maxNum = $maxNum * 10 +
$maxDigit ;
}
return $maxNum - $num ;
}
function complement( $n , $b )
{
return prevComplement( $n , $b ) + 1;
}
echo prevComplement(25, 7), "\n" ;
echo (complement(25, 7));
?>
|
Javascript
<script>
function prevComplement(n , b) {
var maxDigit, maxNum = 0, digits = 0,
num = n;
while (n != 0) {
digits++;
n = parseInt(n / 10);
}
maxDigit = b - 1;
while ((digits--) > 0) {
maxNum = maxNum * 10 + maxDigit;
}
return maxNum - num;
}
function complement(n , b) {
return prevComplement(n, b) + 1;
}
document.write(prevComplement(25, 7)+ "<br/>" );
document.write(complement(25, 7));
</script>
|
Time Complexity: O(log n)
Auxiliary Space: O(1)
Last Updated :
01 Sep, 2022
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