Table of Contents
Single Number – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
Given a non-empty array of integers nums, every element appears twice except for one. Find that single one.
You must implement a solution with a linear runtime complexity and use only constant extra space.
Example 1:
Input: nums = [2,2,1] Output: 1
Example 2:
Input: nums = [4,1,2,1,2] Output: 4
Example 3:
Input: nums = [1] Output: 1
Constraints:
1 <= nums.length <= 3 * 104-3 * 104 <= nums[i] <= 3 * 104- Each element in the array appears twice except for one element which appears only once.
C++ Single Number LeetCode Solution
class Solution {
public:
int singleNumber(vector<int>& arr) {
int n = arr.size(); // taking the size of the array
unordered_map<int, int> mp; // unordered map to store the frequency
// storing frequency in the map
for(int i = 0; i < n; i++)
{
mp[arr[i]]++;
}
int ans; // variable to store our answer
for(auto x: mp) // traverse from the map
{
if(x.second == 1) //if frequency of any elemennt is 1
{
ans = x.first; // store in our answer
break; // break the loop, as we got our answer now
}
}
return ans; // return ans
}
};
Java Single Number LeetCode Solution
class Solution {
public int singleNumber(int[] nums) {
int res = 0;
for(int i=0; i< nums.length; i++) {
res = res ^ nums[i];
}
return res;
}
}
Python 3 Single Number LeetCode Solution
class Solution:
def singleNumber(self, nums: List[int]) -> int:
xor = 0
for num in nums:
xor ^= num
return xor
Array-1180
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