Number Complement | Multiple Approaches | Leetcode 476 | codestorywithMIK
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This is the 52nd Video of our Playlist "Leetcode Easy : Popular Interview Problems" by codestorywithMIK
In this video we will try to solve an easy problem : Number Complement | Multiple Approaches | Leetcode 476 | codestorywithMIK
I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY.
We will do live coding after explanation and see if we are able to pass all the test cases.
Also, please note that my Github solution link below contains both C++ as well as JAVA code.
Problem Name : Number Complement | Multiple Approaches | Leetcode 476 | codestorywithMIK
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Summary :
Approach 1 (Using bit magic of XOR):
This approach calculates the total number of bits required to represent the given number num and then flips each bit one by one using the XOR operation with 1. This flips the bits, effectively giving the complement.
Time Complexity: O(log₂(num))
Space Complexity: O(1)
Approach 2 (Using XOR mask):
This method creates a mask with all bits set to 1 that matches the bit length of num. Then, it XORs num with this mask, flipping all bits to get the complement. The mask is generated by left-shifting and OR-ing bits until it covers all bits of num.
Time Complexity: O(log₂(num))
Space Complexity: O(1)
Approach 3 (Iterate digit by digit and create complement):
In this approach, you iterate through the bits of num and construct the complement bit by bit. If the current bit of num is 0, you set the corresponding bit in the result to 1. This continues until all bits are processed.
Time Complexity: O(log₂(num))
Space Complexity: O(1)
Each approach leverages different bit manipulation techniques to compute the complement efficiently.
✨ Timelines✨
00:00 - Introduction
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