Single Number II – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

Given an integer array nums where every element appears three times except for one, which appears exactly once. Find the single element and return it.

You must implement a solution with a linear runtime complexity and use only constant extra space.

Example 1:

Input: nums = [2,2,3,2]
Output: 3

Example 2:

Input: nums = [0,1,0,1,0,1,99]
Output: 99

Constraints:

• 1 <= nums.length <= 3 * 104
• -231 <= nums[i] <= 231 - 1
• Each element in nums appears exactly three times except for one element which appears once.

C++ Single Number II LeetCode Solution

class Solution {

public:

int singleNumber(vector& nums) {

if(nums.size()==1) return nums[0];

sort(nums.begin(),nums.end());

for(int i=0;i<nums.size();i++)

{

int count=1;

while(i<nums.size()-1 and nums[i]==nums[i+1]){i++;count++;}

if(count==1) return nums[i];
 
    }
    return 0;
}

 
};

Java Single Number II LeetCode Solution

class Solution {
public int singleNumber(int[] nums) {
    HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
    for(int num : nums) {
        map.put(num,map.getOrDefault(num, 0)+1);
    }
    
     int result = 0;
     for (Map.Entry<Integer,Integer> entry : map.entrySet()) 
     {
         if(entry.getValue()==1)
            result = entry.getKey();
     }
    
     return result;
}

 
}
 

Python 3 Single Number II LeetCode Solution

class Solution:
# approach 1
	def singleNumber(self, nums: List[int]) -> int:
        one,two=0,0
        for i in nums:
            one=~two&(one^i)
            two=~one&(two^i)
        return one
		
# approach 2

def singleNumber(self, nums: List[int]) -> int:
        d={}
        for i in nums:
            if i in d:
                d[i]+=1
            else:
                d[i]=1
        for k,v in d.items():
            if v==1:
                return k
        return -1 

Array-1180
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