16. 3Sum Closest (medium)

Brute force

• We can find all the possible combinations of three numbers in the array.
• Then we can find the closest sum by comparing the sum of the three numbers with the target sum.
• TC:O(n^3)
• SC:O(1)

Optimized

• We use similar idea as 3Sum.
• First sort the array.
• then we will fix first number and find other 2 numbers with the two pointer method.
• If the sum of current 3 elements is equal to the target sum, we return the sum OR target.
• If absolute difference between the sum of current 3 elements and target sum abs(sum - target) is less than the current closest difference and target sum abs(ans - target), we update the closest difference.
• If the sum is less than target then move right pointer by 1 else move left pointer by 1.
• Return the closest difference(ans).
• TC:O(n^2)
• SC:O(1)

Code

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target)
    {
        int n = nums.size();
        sort(nums.begin(), nums.end());

        int ans = nums[0] + nums[1] + nums[2];
        for (int i = 0; i < n; ++i) {
            int left = i + 1, right = n - 1;
            while (left < right) {
                int sum = nums[i] + nums[left] + nums[right];

                if (sum == target)
                    return target;
                if (abs(sum - target) < abs(ans - target))
                    ans = sum;
                if (sum < target)
                    ++left;
                else
                    --right;
            }
        }
        return ans;
    }
};