(Analysis by Suhas Nagar)

Subtask 1: $T,C \leq 1000$

We can first attempt a brute force solution. Given a sequence, we can determine the number of targets that are hit by simulating Bessie's position and keeping track of hit targets in an array. We can attempt to exhaustively change every character of the sequence to another, simulate the number of targets hit, and then take the maximum across all sequences. This gives us a solution of $O(TC)$ for the first subtask.

Benjamin Qi's Python solution:

T, C = map(int, input().split())
targets = list(map(int, input().split()))

def test(cand):
avail = [0] * (2*C+1)
for t in targets:
avail[t+C] = 1
ans = 0
cur = C
for c in cand:
if c == 'L':
cur -= 1
elif c == 'R':
cur += 1
else:
ans += avail[cur]
avail[cur] = 0
return ans

commands = input()
max_ans = 0
for i in range(len(commands)):
for c in "LFR":
max_ans = max(max_ans, test(commands[:i] + c + commands[i+1:]))

print(max_ans)

Full Credit:

To improve our runtime, we make a few observations. Firstly, since we can only change a single character, Bessie's position cannot change by more than 2 from where she originally was (ie changing an L to an R). From here, we can consider an alternate solution where we calculate which targets are hit for all suffixes for displacements of $-2, -1, 0, 1$, and $2$ from Bessie's original position. If we know this information, we can iterate from left to right, try changing every character of the sequence, determine the displacement this causes for the suffix, and use our precomputation to determine the total amount of hit targets. If we naively calculate the number of targets hit for every suffix, we are duplicating work since suffixes share many common elements. However, this leads us to the optimal solution.

Let $pref[i]$ be the targets hit with the first $i$ characters of the string. Let $suff[i][k]$ be the targets hit with the last $i$ characters of the string at a displacement of $k$. We can precompute all $pref[i]$ using our simulation solution from before on the unchanged sequence. Now, we can iterate through our sequence in reverse order. Assuming no overlap between the prefix and suffix arrays, we can compute the number of targets we hit by changing the $i$-th character as $pref[i-1]+suff[i+1][k]+hit[i]$ where $k$ is the displacement of changing the current character and $hit[i]$ is $1$ if we change the current character to $F$ and there is an unhit target at our current position. Once we process the current character, we can update our suffix target count to include the current position under each displacement. This gives us a solution in $O(T+C)$ to compute the number of targets when we change each character of the sequence.

There is one caveat however. We want our suffix targets to not double count targets that have already been hit in the prefix. If we want to add a target to our suffix count but it has been hit in the prefix already, we can add it to a buffer and only move it into our suffix array when we remove that target from the prefix array.

import java.io.*;
import java.util.*;

public class TargetPractice {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int nloc = sc.nextInt();
int ncommands = sc.nextInt();
boolean[] locations = new boolean[2*ncommands+5];
for (int i = 0; i < nloc; i++){
locations[sc.nextInt()+ncommands] = true;
}
String command = sc.next();
HashMap<Integer, Integer> currentHit = new HashMap<>();
HashMap<Integer,Integer> whenHit = new HashMap<>();
int cur_pos = ncommands;
for (int i = 0; i < ncommands; i++){
if (command.charAt(i) == 'F'){
if (locations[cur_pos] && !currentHit.containsKey(cur_pos)){
currentHit.put(cur_pos, i);
whenHit.put(i,cur_pos);
}
}
cur_pos += (command.charAt(i) == 'L') ? -1 : 0;
cur_pos += (command.charAt(i) == 'R') ? 1 : 0;
}
int max = currentHit.size();
HashSet[] rightSide = new HashSet[5];
for (int i = 0; i < 5; i++) rightSide[i] = new HashSet<Integer>();
for (int i = 0; i < 5; i++) toBeAdded[i] = new HashSet<Integer>();
for (int i = ncommands-1; i >= 0; i--){
if (whenHit.containsKey(i)){
currentHit.remove(whenHit.get(i));
whenHit.remove(i);
for (int j = 0; j < 5; j++){
}
}
}
cur_pos += (command.charAt(i) == 'L') ? 1 : 0;
cur_pos += (command.charAt(i) == 'R') ? -1 : 0;
switch (command.charAt(i)){
case 'L':
// try F and add all displacement 1
int addL = locations[cur_pos] && !currentHit.containsKey(cur_pos) && !rightSide[3].contains(cur_pos)? 1 : 0;
// try changing to R
max = Math.max(max, whenHit.size()+rightSide[4].size());
break;
case 'R':
// try F and add all displacement 1
int addR = locations[cur_pos] && !currentHit.containsKey(cur_pos) && !rightSide[1].contains(cur_pos) ? 1 : 0;
// try changing to L
max = Math.max(max, whenHit.size()+rightSide[0].size());
break;
case 'F':
// Try changing to L
max = Math.max(max, whenHit.size()+rightSide[1].size());
// Try changing to R
max = Math.max(max, whenHit.size()+rightSide[3].size());
break;
}
if (command.charAt(i) == 'F') {
for (int j = cur_pos - 2; j <= cur_pos + 2; j++) {
if (j < 0 || j >= locations.length) continue;
if (locations[j]){
if (currentHit.containsKey(j)) {
}else{