Similar to 3-sum problem, we sort the array first. Again, similar to 3-sum problem, we use two pointers (`lo`

and `hi`

) to check if the sum satisfies the condition. The only trick here is that if we found out

```
nums[i] + nums[lo] + nums[hi] < target
```

then for all `hi`

in (lo, hi] satisfy the condition. That's why we have

```
count += hi-lo;
```

in the code.

Code in Java:

```
public class Solution {
public int threeSumSmaller(int[] nums, int target) {
int L = nums.length;
Arrays.sort(nums);
int count = 0;
for(int i=0; i<L-2; i++) {
int lo = i+1;
int hi = L-1;
while(lo<hi) {
if(nums[i] + nums[lo] + nums[hi] < target) {
count += hi-lo;
lo++;
}
else
hi--;
}
}
return count;
}
}
```