540. Single Element in a Sorted Array LeetCode Solution
In this guide, you will get 540. Single Element in a Sorted Array LeetCode Solution with the best time and space complexity. The solution to Single Element in a Sorted Array problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Single Element in a Sorted Array
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
Complexity Analysis
Time Complexity: O(\log n)
Space Complexity: O(1)
540. Single Element in a Sorted Array LeetCode Solution in C++
class Solution {
public:
int singleNonDuplicate(vector<int>& nums) {
int l = 0;
int r = nums.size() - 1;
while (l < r) {
int m = (l + r) / 2;
if (m % 2 == 1)
--m;
if (nums[m] == nums[m + 1])
l = m + 2;
else
r = m;
}
return nums[l];
}
}; /* code provided by PROGIEZ */
540. Single Element in a Sorted Array LeetCode Solution in Java
class Solution {
public int singleNonDuplicate(int[] nums) {
int l = 0;
int r = nums.length - 1;
while (l < r) {
int m = (l + r) / 2;
if (m % 2 == 1)
--m;
if (nums[m] == nums[m + 1])
l = m + 2;
else
r = m;
}
return nums[l];
}
} // code provided by PROGIEZ
540. Single Element in a Sorted Array LeetCode Solution in Python
class Solution:
def singleNonDuplicate(self, nums: list[int]) -> int:
l = 0
r = len(nums) - 1
while l < r:
m = (l + r) // 2
if m % 2 == 1:
m -= 1
if nums[m] == nums[m + 1]:
l = m + 2
else:
r = m
return nums[l] # code by PROGIEZ
