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This approach uses bit manipulation to solve the problem with constant space and linear runtime. The idea is to count the number of times each bit is set in the given numbers modulo 3 using two integers. This effectively filters out the bits that appear in numbers repeated thrice and leaves the bits of the unique number.
Time Complexity: O(n) - where n is the number of elements in the array.
Space Complexity: O(1) - as only a fixed amount of space is used regardless of the input size.
1public class Solution {
2 public int singleNumber(int[] nums) {
3 int ones = 0, twos = 0;
4 for (int num : nums) {
5 twos |= ones & num;
6 ones ^= num;
7 int common = ones & twos;
8 ones &= ~common;
9 twos &= ~common;
10 }
11 return ones;
12 }
13
14 public static void main(String[] args) {
15 int[] nums = {2, 2, 3, 2};
16 Solution sol = new Solution();
17 System.out.println(sol.singleNumber(nums));
18 }
19}In this Java solution, the logic is translated similarly using bitwise operations to manage sets of bits. The ones and twos variables are updated in each iteration to ensure bits appearing thrice are removed from consideration, leaving behind the unique element.
This approach involves using a data structure to count occurrences of each number. Although this uses linear time complexity, it requires additional memory to store frequencies, which violates the constant space constraint.
Time Complexity: O(n log n) - primarily due to the sorting step.
Space Complexity: O(1) - if in-place sort is assumed.
1function
Using an object as a dictionary to store number occurrences offers effective counting but does not meet the problem's constraints for extra space, making it less suited for large datasets where space is limited.