Maximum XOR for Each Query | Simple Explanation | Leetcode 1829 | codestorywithMIK
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This is the 21st Video of our Playlist "Bit-Manipulation : Popular Interview Problems" by codestorywithMIK
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In this video we will try to solve a medium-easy bit manipulation related problem : Maximum XOR for Each Query | Simple Explanation | Leetcode 1829 | 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 : Maximum XOR for Each Query | Simple Explanation | Leetcode 1829 | codestorywithMIK
Company Tags : will update
My solutions on Github(C++ & JAVA) - https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/Bit_Magic/Maximum XOR for Each Query.cpp
Leetcode Link : https://leetcode.com/problems/maximum-xor-for-each-query
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Summary :
Calculate Total XOR: First, compute the XOR of all elements in the array nums, which gives us a cumulative XOR value.
Create a Mask for Maximum XOR: Using maximumBit, create a mask with all bits set to 1 up to maximumBit (e.g., if maximumBit = 3, the mask would be 111 in binary, or 7 in decimal). This mask helps in flipping bits to get the maximum possible XOR.
Iteratively Find Maximum XOR: For each element in reverse order, compute the maximum XOR possible by XOR-ing the cumulative XOR with the mask. This gives the flipped version of the cumulative XOR, yielding the highest possible value for each position.
Update XOR: After computing each result, update the cumulative XOR by removing the contribution of the element at the end of the current range.
This results in the desired array of maximum XOR values at each step.
✨ Timelines✨
00:00 - Introduction
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