136_Single_Number.py

# 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.


class Solution:
    def singleNumber(self, nums: list[int]) -> int:
        dict_num = {}
        for num in nums:
            if dict_num.get(num) == None:
                dict_num.setdefault(num, 1)
            else:
                dict_num[num] += 1
        for key in dict_num.keys():
            if dict_num[key] == 1:
                return key