Short Solution Using XOR [AC]


  • 0
    S

    XOR with a number itself returns 0. So, we will eliminate all the duplicates using xor. The final element left will be our result.

    class Solution {
    public:
        int singleNumber(vector<int>& nums) {
            int n = nums.size(), ans = nums[0];
            for(int i=1;i<n;i++) ans = ans^nums[i];
            return ans;
        }
    };
    

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