3Sum Smaller

3Sum Smaller two pointers

Given an array of n integers nums and a target, find the number of index triplets i, j, k with 0 <= i < j < k < n that satisfy the condition nums[i] + nums[j] + nums[k] < target.

For example, given nums = [-2, 0, 1, 3], and target = 2.

Return 2. Because there are two triplets which sums are less than 2:

[-2, 0, 1]
[-2, 0, 3]

Follow up: Could you solve it in O(n2) runtime?

3Sum Smaller Solution

public class 3SumSmaller {
    public int threeSumSmaller(int[] nums, int target) {
        //initialize total count to zero
        int count = 0;

        //sort the array
        Arrays.sort(nums);

        //loop through entire array
        for(int i = 0; i < nums.length - 2; i++) {
            //set left to i + 1
            int left = i + 1;

            //set right to end of array
            int right = nums.length - 1;

            //while left index < right index
            while(left < right) {
                //if the 3 indices add to less than the target increment count
                if(nums[i] + nums[left] + nums[right] < target) {
                    //increment the count by the distance between left and right because the array is sorted
                    count += right - left;

                    //decrement right pointer
                    left++;
                } else {
                    //if they sum to a value greater than target...
                    //increment left pointer
                    right--;
                }
            }
        }

        return count;
    }
}

Last modified October 4, 2020