x̄ - > ASCII code

To find the minimum number of distinct letters in the resulting string S, we need to choose the minimum number of distinct letters from P and Q in each position. We can do this by comparing the letters in P and Q at each position and choosing the one with the smaller ASCII code. This ensures that we always choose the minimum letter.

We can then count the number of distinct letters in the resulting string S by adding each letter to a set and returning the size of the set.

Here's the implementation of the solution in Java:

import java.util.*;class Solution {    

public int solution(String P, String Q) {        

int n = P.length();        

Set<Character> distinctLetters = new HashSet<>();        

for (int i = 0; i < n; i++) {            

char minChar = (char) Math.min(P.charAt(i), Q.charAt(i));            

distinctLetters.add(minChar);        }        

return distinctLetters.size();    }}

The time complexity of this solution is O(n), where n is the length of the input strings. Since the input strings have at most 20 distinct letters, the size of the set can be at most 20, so the space complexity is also O(1). Therefore, this solution is efficient for the given constraints.