# Simple code c++

• here is my code which is just similar to the code for previous problem 3Sum . but here we don't need to take consideration of duplicates.

threeSumClosest

``````int threeSumClosest(vector<int>& nums, int target) {
sort(nums.begin(),nums.end());
int n=nums.size(),ans;
int min=INT_MAX;
for(int i=0;i<n-2;i++){
int l=i+1, r= n-1;
while(l<r){
int sum =nums[i]+nums[l]+nums[r];
if(abs(sum-target)<min){  // updating the sum if sum  so far. is closest to target
min=abs(sum-target);
ans=sum;
}
if(sum<target) l++;        //
else if(sum>target)r--;
else {
return sum;  // we have sum equal to target which is closest so no need to check further
}
}
}
return ans;
}
``````

threeSum

``````vector<vector<int>> threeSum(vector<int>& nums) {
sort(nums.begin(),nums.end());
int n=nums.size();
vector<vector<int>> res;
for(int i=0;i<n-2;i++){
if(i>0 && (nums[i]==nums[i-1]) )continue;   // to avoid duplicates
int l=i+1, r= n-1;
while(l<r){
int sum =nums[i]+nums[l]+nums[r];

if(sum<0) l++;
else if(sum>0)r--;
else {
res.push_back(vector<int>{nums[i],nums[l],nums[r]});
while(l+1<r && nums[l]==nums[l+1])l++;    // to avoid duplicates
while(l<r-1 && nums[r]==nums[r-1]) r--;     // to avoid duplicates
l++; r--;
}
}
}

return res;
}``````

• thanks for share . this solution helped me a lot .

• thanks for share. it is very clear now.

• For the equal case, why not just return target. No breaking loops requires

• oops !! thanks for pointing it . I have made the appropriate changes .

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