aboutsummaryrefslogtreecommitdiff
path: root/maximum_xor_query.py
diff options
context:
space:
mode:
authorJack Sangdahl <[email protected]>2024-11-03 00:01:49 -0600
committerJack Sangdahl <[email protected]>2024-11-03 00:01:49 -0600
commit64fa827c19c72e7a2f490e75856870a41d6b4c5f (patch)
tree43b6f938b62f0c7d4ff4f558586f998a9b6960dc /maximum_xor_query.py
parent20fb571e363b19bec8d6f595d471de8fdb01a95b (diff)
downloadleetcode-64fa827c19c72e7a2f490e75856870a41d6b4c5f.tar.gz
leetcode-64fa827c19c72e7a2f490e75856870a41d6b4c5f.zip
1829. maximum xor for each query
Diffstat (limited to 'maximum_xor_query.py')
-rw-r--r--maximum_xor_query.py31
1 files changed, 31 insertions, 0 deletions
diff --git a/maximum_xor_query.py b/maximum_xor_query.py
new file mode 100644
index 0000000..1790c62
--- /dev/null
+++ b/maximum_xor_query.py
@@ -0,0 +1,31 @@
+# 1829. Maximum XOR for Each Query
+
+"""
+you are given a sorted array 'nums' of 'n' non negative integers and an
+integer 'maximumBit'. you want to perform the following query 'n' times: find
+a non-negative integer 'k < 2^maximumBit' such that 'nums[0] ^ nums[1] ^ ...
+^ nums[n]' is maximised. 'k' is the answer to the i'th query. remove the last
+element from the current array 'nums'. return an array 'answer' where
+'answer[i]' is the answer to the i'th query.
+"""
+
+
+class Solution(object):
+ def getMaximumXor(self, nums, maximumBit):
+ """
+ :type nums: List[int]
+ :type maximumBit: int
+ :rtype: List[int]
+ """
+ max_xor, arr_xor, ans = (1 << maximumBit) - 1, reduce(ixor, nums), []
+ for i in range(len(nums) - 1, -1, -1):
+ ans.append(arr_xor ^ max_xor)
+ arr_xor ^= nums[i]
+ return ans
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.getMaximumXor(nums=[0, 1, 1, 3], maximumBit=2))
+ print(obj.getMaximumXor(nums=[2, 3, 4, 7], maximumBit=3))
+ print(obj.getMaximumXor(nums=[0, 1, 2, 2, 5, 7], maximumBit=3))