Problem: Given a non-empty array of integers nums, every element appears exactly twice except for one. Find that unique number. A solution must be implemented with linear time complexity and using only constant extra space.
Initial approach: An intuitive solution is to count each number and return the one with a count of one. In C#, this can be achieved with LINQ using GroupBy and OrderBy Count to group and sort by frequency. Example of a LINQ approach, although it does not strictly meet the constant space constraint:
public int SingleNumber(int[] nums) { return nums.GroupBy(x => x).OrderBy(g => g.Count()).First().Key; }
Optimal solution and explanation: The ideal way that meets linear time and constant space uses the bitwise XOR operation. Useful properties of XOR: a xor a = 0 and a xor 0 = a, and it is also commutative and associative. If we apply cumulative XOR over all elements, the numbers that appear twice cancel each other out, leaving the unique number. Complexity O(n) and space O(1).
public int SingleNumber(int[] nums) { int result = 0; foreach (int n in nums) result ^= n; return result; }
Practical notes: The XOR version is very efficient and clear for this classic problem. If you need a solution that also records or debugs occurrences, you can combine techniques, but remember that any auxiliary structure such as dictionaries will increase memory usage.
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