Single-Number

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 * 10^4

-3 * 10^4 <= nums[i] <= 3 * 10^4

Each element in the array appears twice except for one element which appears only once.

Solution in Java

public int singleNumber(int[] nums) {
        Map<Integer, Integer> map = new HashMap<>();

        for (int i = 0; i < nums.length; i++) {
            if (map.containsKey(nums[i])) map.remove(nums[i]);
            else map.put(nums[i], 1);
        }

        return map.keySet().iterator().next();
    }

Solution in JavaScript

var singleNumber = function(nums) {
    let singleNumber
    for (let i = 0; i < nums.length; i++) {
        if (nums.indexOf(nums[i]) === nums.lastIndexOf(nums[i])) {
            singleNumber = nums[i];
            break;
        }
    }
    
    return singleNumber;
};

Solution in Python

from collections import Counter
class Solution:
    def singleNumber(self, nums: List[int]) -> int:
        cnt = Counter(nums).items()
        for i in cnt:
            if i[1] == 1:
                return i[0]