# 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;
}
}``````