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Tested Python solution for LeetCode 260 with 15 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 260 # by problem number
lcpy gen -s single_number_iii # by problem name
nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once. You can return the answer in any order.
You must write an algorithm that runs in linear runtime complexity and uses only constant extra space.
Input: nums = [1,2,1,3,2,5]
Output: [3,5]
Input: nums = [-1,0]
Output: [-1,0]
Input: nums = [0,1]
Output: [1,0]
class Solution:
# Time: O(n)
# Space: O(1)
def single_number(self, nums: list[int]) -> list[int]:
# XOR all numbers - the result is xor of the two unique numbers
xor_all = 0
for num in nums:
xor_all ^= num
# Find rightmost set bit (differentiating bit between the two unique numbers)
diff_bit = xor_all & -xor_all
# Partition numbers based on the differentiating bit
# and XOR each partition to find the unique numbers
num1, num2 = 0, 0
for num in nums:
if num & diff_bit:
num1 ^= num
else:
num2 ^= num
return [num1, num2]
| Time | Space |
|---|---|
| O(n) | O(1) |