feat: add solutions to lc problem: No.3289

[yanglbme]

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[giruuuuj]

Problem Understanding
The problem is to find two numbers that appear twice in an array containing numbers from 0 to n-1, where the array length is n+2.

Bit Manipulation Solution Analysis
Core Logic
python
class Solution:
def getSneakyNumbers(self, nums: List[int]) -> List[int]:
n = len(nums) - 2
xx = nums[n] ^ nums[n + 1]
for i in range(n):
xx ^= i ^ nums[i]
k = xx.bit_length() - 1
ans = [0, 0]
for x in nums:
ans[x >> k & 1] ^= x
for i in range(n):
ans[i >> k & 1] ^= i
return ans
Step-by-Step Explanation
Initial XOR Setup

python
n = len(nums) - 2
xx = nums[n] ^ nums[n + 1]
Calculate n as the expected range (0 to n-1)

Start XOR with the last two elements

Compute Total XOR

python
for i in range(n):
xx ^= i ^ nums[i]
XOR all numbers from 0 to n-1 with corresponding array elements

The result xx contains the XOR of the two duplicate numbers

Find Differentiating Bit

python
k = xx.bit_length() - 1
Find the highest set bit where the two duplicates differ

Separate and Find Duplicates

python
ans = [0, 0]
for x in nums:
ans[x >> k & 1] ^= x
for i in range(n):
ans[i >> k & 1] ^= i
return ans
Group numbers based on the k-th bit

XOR operations cancel out non-duplicate numbers

Final result contains the two duplicates

Time & Space Complexity
Time Complexity: O(n) - Single pass through the array

Space Complexity: O(1) - Constant extra space

Key Insight
The solution cleverly uses XOR properties:

a ^ a = 0

a ^ 0 = a

XOR is commutative and associative