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This approach leverages the properties of XOR bitwise operation. The XOR of a number with itself is 0, and the XOR of a number with 0 is the number itself. Thus, XORing all elements in the array results in getting the single number because all other numbers cancel themselves out.
Time Complexity: O(n) - We go through the array once.
Space Complexity: O(1) - We only use a single integer to store the result.
1#include <iostream>
2#include <vector>
3
4int singleNumber(std::vector<int>& nums) {
5 int result = 0;
6 for (int num : nums) {
7 result ^= num;
8 }
9 return result;
10}
11
12int main() {
13 std::vector<int> nums = {2, 2, 1};
14 std::cout << singleNumber(nums) << std::endl; // Output: 1
15 return 0;
16}singleNumber uses a range-based for loop to XOR through the vector, similarly resulting in canceling out duplicate numbers and returning the single number.
In this approach, we use a hash map (or dictionary) to count occurrences of each number. The single number will have a count of 1. This isn't the optimal solution in terms of extra space but is valid if space was not constrained.
Time Complexity: O(n) - We traverse the array twice (once for filling the map, once for checking the counts).
Space Complexity: O(n) - Extra space proportional to the input range.
1
The JavaScript function uses an object as a map to count each number, returning the number that appears only once.