Given an array a of n positive integers. The task is to find the maximum of j - i subjected to the constraint of a[i] < a[j] and i < j.

This problem can be solved using an efficient approach that involves two pointers and two auxiliary arrays. Here are the steps:

1. Create two auxiliary arrays, LMin[] and RMax[].
2. Traverse the input array from the beginning and fill LMin[] such that LMin[i] contains the smallest element from 0 to i in the array.
3. Traverse the input array from the end and fill RMax[] such that RMax[j] contains the largest element from j to end in the array.
4. Initialize i and j as 0 (pointers to LMin[] and RMax[] arrays).
5. Run a loop while i < n and j < n. If LMin[i] < RMax[j], then calculate the difference j - i and update the maximum difference if it's greater than the current maximum difference. Then, increment j. If LMin[i] is greater than or equal to RMax[j], then increment i.
6. Return the maximum difference.

This approach ensures that the elements a[i] and a[j] are such that i < j and a[i] < a[j], and the difference j - i is maximized. The time complexity of this approach is O(n), and it uses O(n) extra space.