Single Number III LeetCode Solution | Easy Approach

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Single Number III Given an integer array 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.

Example 1:

Input: nums = [1,2,1,3,2,5]
Output: [3,5]
Explanation:  [5, 3] is also a valid answer.

Example 2:

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

Example 3:

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

Constraints:

  • 2 <= nums.length <= 3 * 104
  • -231 <= nums[i] <= 231 - 1
  • Each integer in nums will appear twice, only two integers will appear once.

Single Number III Solutions

Time: O(n)
Space: O(1)

C++

class Solution {
 public:
  vector<int> singleNumber(vector<int>& nums) {
    const int xors = accumulate(begin(nums), end(nums), 0, bit_xor<>());
    const int lowbit = xors & -xors;
    vector<int> ans(2);

    // seperate nums into two groups by the lowbit
    for (const int num : nums)
      if (num & lowbit)
        ans[0] ^= num;
      else
        ans[1] ^= num;

    return ans;
  }
};

Java

 class Solution {
  public int[] singleNumber(int[] nums) {
    final int xors = Arrays.stream(nums).reduce((a, b) -> a ^ b).getAsInt();
    final int lowbit = xors & -xors;
    int[] ans = new int[2];

    // seperate nums into two groups by the lowbit
    for (final int num : nums)
      if ((num & lowbit) > 0)
        ans[0] ^= num;
      else
        ans[1] ^= num;

    return ans;
  }
}

Python


class Solution:
  def singleNumber(self, nums: List[int]) -> List[int]:
    xors = reduce(operator.xor, nums)
    lowbit = xors & -xors
    ans = [0, 0]

    # seperate nums into two groups by the lowbit
    for num in nums:
      if num & lowbit:
        ans[0] ^= num
      else:
        ans[1] ^= num

    return ans

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