Throughout this editorial, we refer to the starting alphabetical value of ~S~ as ~cur~, and the number of asterisks as ~ast~.

First, let us check for impossibility. There are two cases where it is impossible to achieve an alphabetical value of ~N~.

1. ~cur + ast > N~, as even replacing all the asterisks with a will exceed ~N~.
2. ~cur + 26ast < N~, as even replacing all the asterisks with z will come up short of ~N~.

All other cases are possible, as we can simply change a replaced letter to the next or previous letter in the alphabet to adjust the total alphabetical value by ~1~. We now provide some insights that lead us to the solution.

Consider the relative positions of the asterisks in ~S~. The asterisks at lower indices will contribute more to the total lexicographical order than those at later indices. Thus, the problem is equivalent to finding a string with size ~ast~ and a total alphabetical score of ~N-cur~. Intuitively, we want the letters to the right to be as large as possible, since we want the letters to be as small as possible as we move to the left. Let us illustrate this by making a string of length ~3~ with an alphabetical score of ~6~. bbb is lexicographically larger than abc, which is in turn lexicographically larger than aad. Notice how it is optimal for the rightmost letters to be as large as possible so that we can get the maximum number of as at the left.

With this in mind, we can implement a simple greedy algorithm that iterates through ~S~ from right to left. We want to replace the asterisks with the largest letter possible while still keeping enough remaining alphabetical value to finish filling the remaining asterisks with a. Knowledge of ASCII values is helpful for efficient implementation.

Time Complexity: ~\mathcal{O}(|S|)~