single element in a sorted array leetcode | single element in a sorted array leetcode python

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Really appreciate it brother. Idk why but i was unable to understand any other solution that were using the odd-even index concept. Just wasnt able to understand it and your approach really clicked instantly. Thanks a lot!

dhruvmaindola

bro I used dictionary and it became an easy one for me
great job btw.

dic = {}
for i in range (len(nums)) :
if nums[i] in dic :
dic[nums[i]] = dic.get(nums[i]) + 1
else :
dic[nums[i]] = 1
for key in dic.keys() :
if dic.get(key) == 1 :
return key

return -1

saugatgarwa

There is a closed-form mathematical solution : def singleNonDuplicate(self, nums: List[int]) -> int:
actualSum = sum(nums)
projectedSum = sum(list(set(nums))) *2
return projectedSum-actualSum

kkveena

I really appreciate the time and effort you put into explanation and visualization. Looking forward for more solutions for medium problems.

tolegenazamat

Here is my code.
Can you please tell me the run time complexity and if its good?
Hashtable is my goto method for such type of questions.
Please let me know your input.

class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:

if not nums:
return

for ele in count:
if count[ele]==1:
return ele

shlokjain

Can you please what is the reason behind this logic like how you came up with this

code_school

class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
return sum(set(nums))*2-sum(nums)
Optimised Solution

viratsharma

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