leetcode_gym/_problems/number_compliment.md

layout, title, last_solved, category, leetcode_url, status

layout	title	last_solved	category	leetcode_url	status
page	Number Complement		binary	https://leetcode.com/problems/number-complement	Attempted

Problem

Given a positive integer num, output its complement number. The complement strategy is to flip the bits of its binary representation.

 

Example 1:

Input: num = 5
Output: 2
Explanation: The binary representation of 5 is 101 (no leading zero bits), and its complement is 010. So you need to output 2.
Example 2:

Input: num = 1
Output: 0
Explanation: The binary representation of 1 is 1 (no leading zero bits), and its complement is 0. So you need to output 0.

Solution

Using binary string.

{% highlight python %}

def findComplement(self, num: int) -> int:
    
    binString = bin(num).replace('0b', '')
        
    output = ''
    for c in binString:
        
        if c=='0':
            output+='1'
        else:
            output+='0'
    
    return int(output, 2)

{% endhighlight %}

---

using bit manipulation

{% highlight python %}

def findComplement(self, num: int) -> int:
    ## RC ##
	## APPROACH : BIT MANIPULATION ##
	## LOGIC ##
	#	1. 1 xor 1 = 0 and 0 or 1 = 1, from these we can say that XORing with 1 will give its complement bit
	#	2. so, for 5 ==> 101, bitmask is 111, for 6 ==> 110, bitmask is 111, so bitmask is (length number of 1's)
	#	3. How do we length of given bit ? we use log function
	#	4. (ex: for 5), we take 1 and left shift by length obtained from 3, i.e (1000) now, substract 1, it will give (111) ==> bitmask
	
    ## TIME COMPLEXICITY : O(1) ## (coz make n= 32)
	## SPACE COMPLEXICITY : O(1) ##
    
    n = floor(log2(num)) + 1        # n is a length of num in binary representation
    bitmask = (1 << n) - 1          # bitmask has the same length as num and contains only ones 1...1
    return bitmask ^ num

{% endhighlight %}