Daily Leetcode Challenge |NOV 8 | HINDI | Maximum XOR for Each Query

[@darshankumar5546]

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):
binaryResult=bin(xorResult)[2:]
print('xor result=',xorResult,'In binary:',binaryResult)
if(len(binaryResult)

[@darshankumar5546]

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[@darshankumar5546]

# 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
# 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