Single Element in a Sorted Array – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
You are given a sorted array consisting of only integers where every element appears exactly twice, except for one element which appears exactly once.
Return the single element that appears only once.
Your solution must run in O(log n)
time and O(1)
space.
Example 1:
Input: nums = [1,1,2,3,3,4,4,8,8] Output: 2
Example 2:
Input: nums = [3,3,7,7,10,11,11] Output: 10
Constraints:
1 <= nums.length <= 105
0 <= nums[i] <= 105
C++ Single Element in a Sorted Array LeetCode Solution
class Solution {
public:
int singleNonDuplicate(vector<int>& nums) {
int left = 0, right = nums.size() - 1;
while(left < right){
int mid = (left + right)/2;
if((mid % 2 == 0 && nums[mid] == nums[mid + 1]) || (mid % 2 == 1 && nums[mid] == nums[mid - 1]))
left = mid + 1;
else
right = mid;
}
return nums[left];
}
};
Java Single Element in a Sorted Array LeetCode Solution
class Solution {
public int singleNonDuplicate(int[] nums) {
int left = 0, right = nums.length-1;
while(left < right){
int mid = (left + right)/2;
if( (mid % 2 == 0 && nums[mid] == nums[mid +1]) || (mid %2 == 1 && nums[mid] == nums[mid - 1]) )
left = mid + 1;
else
right = mid;
}
return nums[left];
}
}
Python 3 Single Element in a Sorted Array LeetCode Solution
class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
left, right = 0, len(nums)-1
while left < right:
mid = int((left + right)/2)
if (mid % 2 == 1 and nums[mid - 1] == nums[mid]) or (mid%2 == 0 and nums[mid] == nums[mid + 1]):
left = mid + 1
else:
right = mid
return nums[left]
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