Daily Leetcode Challenge | Nov 8 | Maximum XOR for Each Query

Daily Leetcode Challenge- November 2024 - Day 8 - Maximum XOR for Each Query

Approach: bit manipulation ( xor properties )

optimized approach beats 90% of the test cases

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[darshankumar]

class Solution:
def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
xorResult=nums[0]
for i in nums[1:]:
xorResult=xorResult^i

n=len(nums)
#print(xorResult, bin(xorResult)[2:])

answer=[]
for i in range(n):

print('xor result=', xorResult, 'In binary:', binaryResult)


print('writing xorResult in maximumBits ', binaryResult)
#print(binaryResult)
k=''
for bit in binaryResult:
if(bit=='1'):
k+='0'
else:
k+='1'
print('after negating xor result: ', k)
answer.append(int(k, 2))
#remove last element
lastEle=nums[-i-1]
xorResult=xorResult^lastEle
print(answer)
print('removed element:', lastEle)

return answer

[darshankumar]

let a=101
a xor b should be max
ie a xor b = 111
then b= 010

a^b=k
k^b=a
k^a=b

#example 2 tracing-DRY RUN

xor result= 2 In binary: 10
writing xorResult in maximumBits 010
after negating xor result: 101
[5]
removed element: 7
xor result= 5 In binary: 101
writing xorResult in maximumBits 101
after negating xor result: 010
[5, 2]
removed element: 4
xor result= 1 In binary: 1
writing xorResult in maximumBits 001
after negating xor result: 110 =4+2+0=6
[5, 2, 6]
removed element: 3
xor result= 2 In binary: 10
writing xorResult in maximumBits 010
after negating xor result: 101
[5, 2, 6, 5]
removed element: 2