16. 3Sum Closest

//DAY 13 PROBLEM 1
//Same approach as 3Sum ... that we fix one element and then move the two pointers in the rest of the array to check on every triplet
//Just the difference is that, here we keep a track of min_diff and keep on updating it whenever we get a smaller absolute value of difference than we previously had stored

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
      
        sort(nums.begin(),nums.end());
        int min_sum=0,min_diff=INT_MAX;
    
        for(int index=0;index<nums.size()-2;index++)
        {
           int end=nums.size()-1,start=index+1;
            while(start<end)
            {
                 
                 if(nums[index]+nums[start]+nums[end]==target)
                 {
                    if(abs((nums[index]+nums[start]+nums[end])-target)<min_diff)
                    {
                     min_sum=(nums[index]+nums[start]+nums[end]);
                         min_diff=abs((nums[index]+nums[start]+nums[end])-target);
                        // cout<<min_sum<<endl;
                        // cout<<nums[index]<<" "<<nums[start]<<" "<<nums[end]<<endl;
                        }
                     break;
                 }
               else if(nums[index]+nums[start]+nums[end]<target)
                {
                   
                     if(abs((nums[index]+nums[start]+nums[end])-target)<min_diff)
                    {
                     min_sum=(nums[index]+nums[start]+nums[end]);
                         min_diff=abs((nums[index]+nums[start]+nums[end])-target);
                           // cout<<min_sum<<endl;
                        // cout<<nums[index]<<" "<<nums[start]<<" "<<nums[end]<<endl;
                        }
                   start++;
                }
                else  if(nums[index]+nums[start]+nums[end]>target)
                {
                   
                   
                  if(abs((nums[index]+nums[start]+nums[end])-target)<min_diff)
                    {
                     min_sum=(nums[index]+nums[start]+nums[end]);
                         min_diff=abs((nums[index]+nums[start]+nums[end])-target);
                        // cout<<min_sum<<endl;
                        // cout<<nums[index]<<" "<<nums[start]<<" "<<nums[end]<<endl;
                        }
                    end--;
                }
                
              
            }
            //This is important to check, as in ideal position the difference between the triplets sum and the target is 0 i.e. they are same .
            //In this condition, we break out of the while loop, but we also have to leave the for loop as we have found the answer... so we check if min_diff==0
            //after the while loop ends.
            //Remember, we can't use abs((nums[index]+nums[start]+nums[end])-target)==0 condition as the end and the start can change in one of the else if conditions
            //And, we may get false combinations
         if(min_diff==0)
                  {
                      min_sum=target;
                      break;
                  }
        }
        return min_sum;
    }
};