Given an array of integers, every element appears twice except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
class Solution(object):
def singleNumber(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
a = 0
for i in nums:
a ^= i
return a
Concept
If we take XOR of zero and some bit, it will return that bit
a⊕0=a
If we take XOR of two same bits, it will return 0
a⊕a=0
a⊕b⊕a=(a⊕a)⊕b=0⊕b=b
So we can XOR all bits together to find the unique number.
Complexity Analysis:
Time complexity : O(n). We only iterate through nums, so the time complexity is the number of elements in nums.
Space complexity : O(1).
class Solution(object):
def singleNumber(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
hash_table = {}
for i in nums:
try:
hash_table.pop(i)
except:
hash_table[i] = 1
return hash_table.popitem()[0]
Complexity Analysis:
Time complexity : O(n∗1)=O(n). Time complexity of for loop is O(n). Time complexity of hash table(dictionary in python) operation pop is O(1).
Space complexity : O(n). The space required by hash_table is equal to the number of elements in nums
