✅Python || 💥C++ ||✔ Clean and Concise Solution💥||🚀 O(logN) Binary Search
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:O(log n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code Upvote please\uD83D\uDE4C\uD83D\uDE09\n```\nclass Solution(object):\n    def singleNonDuplicate(self, nums):\n        """\n        :type nums: List[int]\n        :rtype: int\n        """\n        start , end = 0 , len(nums) -1\n        while start < end :\n            mid = (start + end) // 2\n            if(nums[mid]==nums[mid+1]):\n                mid = mid-1\n            \n            if((mid-start+1)%2!=0):\n                end = mid\n            \n            else:\n                start = mid+1\n        return nums[start]\n            \n```\n\n# UPVOTE ME \uD83D\uDE09\uD83E\uDD1E\uD83D\uDE22
['Array', 'Binary Search', 'Python']