nums
.count
to store the number of triplets found.i
from 0 to n - 3
.
a. For each i
, initialize left
to i + 1
and right
to n - 1
.
b. While left < right
, check if nums[i] + nums[left] + nums[right] < target
.
i. If true, increment count
by right - left
(all elements between left
and right
will form valid triplets with nums[i]
), and move left
to the right.
ii. If false, move right
to the left.count
of valid triplets.