I have recently started to work through the problems on Leetcode, for the sake of bettering my own skills as well as preparing for interviews. I was faced with the 3Sum problem which is: Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.Now I thought about this: the brute force solution is to iterate through the array twice for every number to find the other two numbers that make the sum of the three equal 0. This would have a time complexity of \$O(n^3)\$ which is less than ideal. So I thought maybe it would be instead good to store all the pairs in the given array.Then iterate through the array and see if the sum with any pair is equal to 0, and push the pair along with that number into a vector for the output. It is slightly working backwards, and I thought that this would give me an efficiency of \$O(n^2)\$.vector<vector<int>> threeSum(vector<int>& nums) {    vector<pair<int,int>> pairs;    vector<vector<int>> output;    if(nums.size()<3){        return output;    }    for(int i=0; i<nums.size();i++){        for(int j=i+1; j<nums.size();j++){                pairs.push_back(make_pair(nums.at(i),nums.at(j)));        }    }    sort(pairs.begin(),pairs.end());    pairs.erase(unique(pairs.begin(),pairs.end()),pairs.end());    for(int i=0; i<nums.size();i++){        for(auto x:pairs){            if(x.first!=nums.at(i) && x.second!=nums.at(i)){                if(x.first+x.second+nums.at(i)==0){                    vector<int> curr;                    curr.push_back(x.first);                    curr.push_back(x.second);                    curr.push_back(nums.at(i));                    sort(curr.begin(),curr.end());                    output.push_back(curr);                }            }        }    }    int count = 0;    for(int i:nums){        if(i==0){            count++;            if(count==3){                vector<int> zeros;                zeros.push_back(0);                zeros.push_back(0);                zeros.push_back(0);                output.push_back(zeros);                break;            }        }    }    sort(output.begin(),output.end());    output.erase(unique(output.begin(),output.end()),output.end());    return output;}At the end in the lines:    int count = 0;    for(int i:nums){        if(i==0){            count++;            if(count==3){                vector<int> zeros;                zeros.push_back(0);                zeros.push_back(0);                zeros.push_back(0);                output.push_back(zeros);                break;            }        }    }Here I am checking if there are 3 or more zeros in the given input and then pushing a vector [0,0,0] into the output to cover that case, because my condition for avoiding duplicates in the line below avoids zeros (something I haven't been able to figure out):if(x.first!=nums.at(i) && x.second!=nums.at(i)){So after running it, the Leetcode test cases with very large inputs give me a Time Limit Exceeded message, I am wondering what am I doing wrong or how I can go about to make it better? I was thinking maybe the sorting is an issue but from what I know \$O(n^2)\$ should be dominant compared to \$O(nlogn)\$.
3Sum implementation
c++;performance;array;vectors
Avoid unnecessary memory allocations.Memory allocations take time, significant time if you do them in your inner most loop.For starters, change:vector<vector<int>> output;to:vector<std::tuple<int, int, int>> output;This converts your putput vector from A vector of pointers to fixed size dynamic elements (note the oxymoron, fixed size dynamic elements, there's your problem) to A vector of fixed size elements. Not only will you be avoiding lots and lots of memory allocations but you will also improve your cache performance by miles because your entire output vector is now contiguous and cache friendly.Then change:for(int i=0; i<nums.size();i++){    for(auto x:pairs){        if(x.first!=nums.at(i) && x.second!=nums.at(i)){            if(x.first+x.second+nums.at(i)==0){                vector<int> curr;                curr.push_back(x.first);                curr.push_back(x.second);                curr.push_back(nums.at(i));                sort(curr.begin(),curr.end());                output.push_back(curr);            }        }    }}to:for(int i=0; i<nums.size();i++){    for(auto& x:pairs){        if(x.first!=nums.at(i) && x.second!=nums.at(i)){            if(x.first+x.second+nums.at(i)==0){                int a = x.first;                int b = x.second;                int c = nums.at(i);                // You could write out the if-else tree to                 // possibly optimise this if your compiler doesn't                 // already do that.                if(a > b)                    swap(a,b);                if(a > c)                    swap(a,c);                if(b > c)                    swap(b,c);                output.emplace_back(a,b,c);            }        }    }}Return valueYou are returning a vector by value, a naive compiler may cause the entire vector contents to be copied and if you have a vector of vectors, this is really bad. However most modern compilers perform Return Value Optimization(RVO). However if you don't want to be at the mercy of a possibly dodgy compiler, just pass a reference to a vector where you want to store the results as an argument to the function.Time complexitySo you do this:for(int i=0; i<nums.size();i++){    for(int j=i+1; j<nums.size();j++){            pairs.push_back(make_pair(nums.at(i),nums.at(j)));which means that pairs.size() == n^2. Then proceed with sorting:sort(pairs.begin(),pairs.end());Because you are sorting m = n^2 entries, the time complexity becomes O(m*log(m)) = O(n^2*log(n)).But then you do this...for(int i=0; i<nums.size();i++){    for(auto x:pairs){        if(x.first!=nums.at(i) && x.second!=nums.at(i)){Okay so this becomes O(n*n^2). Note that O(n^2) is easily achievable by other algorithms.This is why you get TLE.Better algorithmWe can easily reach O(n^2) time complexity.We need to find all a, b, c such that a+b+c=0. Note that this is equivalent to c = -(a+b). Hence if we can check if c exists in the input in O(1) time, then we just need to try each pair of a and b and see if a matching c exists. Since there are O(n^2) pairs we have O(n^2*1) time.A hash set provides the necessary O(1) check if c is present in the input. (I'm not going to handle the three zeros, you can figure that out). Pseudocode:unordered_map<int> hashset;for(auto& x : input){ hashset.put(x); }for(int i = 0; i < input.size(); ++i){     for(int j = i+1; j < input.size(); j++){        auto c = -(input[i] + input[j]);        if(hashset.contains(c)){            output.addTuple(input[i], input[j], - c);        }    }}