【Incrementing triplet】Based on array B

Increment triplet

for (; j < N; j++) {
    
    
	while (i < N && A[i] < B[j])
		i++;
	while (k < N && C[k] <= B[j])
		k++;

	res += ((long)i * (N - k));
}

Key: Iterating over the array B. Because Astrictly less than B, Cstrictly greater than B.
If A is used as the benchmark, the elements in B are found to be just greater than A, and the elements of C are just greater than B. What about now?
Example:

1 2 3
1 2 3
1 2 3

Based on 1 in A, B found 2, and C found 3. It cannot be calculated/determined in one step, that is, the elements of B have to be traversed, because the next element of B may not be smaller than the element of C at this time.

However, with B as the benchmark, it can be calculated/determined in one step. Find a number in A that is exactly smaller than B, and find a number in C that is just larger than B, and this a*cis the result.