Why using unordered_set is much slower than using two pointers?

  • 1

    Inspired by 2sum, I have an O(n^2) algorithm but it takes several hundreds of ms to pass all testcases. Anyone has any idea why that happens? Is it because of the repeat calls of unordered_set constructor and desctructor?

    EDIT: I tried to use a vector of unordered_set to reduce the # calls of destructor, and it's still very slow.

        vector<vector<int>> threeSum(vector<int>& nums) {
            vector<vector<int>> res;
            sort(nums.begin(), nums.end());
            for (int i = 0; i < nums.size(); ++i) {
                if (i > 0 && nums[i] == nums[i-1]) continue;
                long a = nums[i];
                if (a > 0) break;
                unordered_set<int> dp;
                dp.reserve(nums.size() - i);
                unordered_set<int> used;
                for (int j = i + 1; j < nums.size(); ++j) {
                    if (used.find(nums[j]) != used.end()) continue;
                    long b = a + nums[j];
                    if (b + nums[i+1] > 0) break;
                    if (dp.find(-b) != dp.end()) {
                        res.push_back(vector<int>({nums[i], -b, nums[j]}));
            // sort(res.begin(), res.end());
            return res;

  • 1

    I am wondering also. I using hashmap lead to TLE but two pointers can pass. It seems hashmap need O(n^2). And two pointers require O(n^2), how's the difference make one that slow? Wish someone can explain.

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