[Here](https://codeforces.net/contestInvitation/87feaad2fec270d40d123ed13a33564196bcc2a1) is the contest link.↵
↵
#### [A. Free Ice Cream for the Kids](https://codeforces.net/gym/596141/problem/A)↵
↵
<spoiler summary = "Solution">↵
To solve this problem efficiently, we simulate the sequence of operations while maintaining two values: the current number of ice cream packs and the count of distressed children. We start with $x$ packs and iterate through $n$ operations. If a carrier ($+d$) arrives, we increase our stock by $d$. If a child ($-d$) requests d packs, we check if we have enough ice cream. If yes, we serve the request and decrease our stock; otherwise, we increment the count of distressed kids. By processing all operations sequentially, we determine the final ice cream count and the number of distressed kids. This approach runs in $O(n)$ time, making it efficient given the constraints.↵
</spoiler>↵
↵
<spoiler summary="Code">↵
```python3↵
def solve():↵
n, cur = map(int, input().split())↵
distress = 0↵
↵
for _ in range(n):↵
op, d = input().split()↵
d = int(d)↵
↵
if op == '+':↵
cur += d↵
elif cur >= d:↵
cur -= d↵
else:↵
distress += 1↵
↵
print(cur, distress)↵
↵
solve()↵
```↵
</spoiler>↵
↵
#### [B. Increasing Sequence](https://codeforces.net/gym/596141/problem/B)↵
↵
<spoiler summary = "Solution">↵
To solve this problem, we need to construct a sequence $b$ that satisfies the given conditions. The key idea is to ensure that $b$ is strictly increasing and does not contain any elements from $a$.↵
↵
Steps to Solve:↵
<ol>↵
<li> Initialize a variable `Last_elem` to $0$, which will keep track of the last value added to the sequence $b$.</li>↵
<li> Iterate through each element in the sequence $a$.</li>↵
<li> For each element $a_i$:↵
<ul>↵
<li> Increment `Last_elem` by $1$.</li>↵
<li> If `Last_elem` is equal to $a_i$, increment `Last_elem` by $1$ again to ensure $b_i \neq a_i$.</li>↵
</ul>↵
</li>↵
<li> After processing all elements, `Last_elem` will hold the minimum value of $b_n$.</li>↵
</ol>↵
</spoiler>↵
↵
<spoiler summary="Code">↵
```python3↵
t = int(input())↵
for _ in range(t):↵
n = int(input())↵
a = list(map(int, input().split()))↵
last_elem = 0↵
for num in a:↵
last_elem += 1↵
if last_elem == num:↵
last_elem += 1↵
print(last_elem)↵
```↵
</spoiler>↵
↵
#### [C. Penalty Shootout](https://codeforces.net/gym/596141/problem/C)↵
↵
<spoiler summary = "Problem Summary">↵
In this problem, we are given a sequence of penalty kicks between two teams. Each team takes turns kicking, with the first team starting first. The goal is to determine the earliest possible moment when the penalty phase can be stopped. The phase stops if one team has scored more goals than the other team could possibly reach with their remaining kicks.↵
↵
Each kick is represented by a character in a 10-character string. A `1` means the kick is a goal, a `0` means the kick is missed, and a `?` means the outcome is uncertain, it could be either a goal or a miss. Our task is to find the minimum number of kicks required before the game can be stopped, considering all possible ways the `?` values could be filled.↵
</spoiler>↵
↵
<spoiler summary ="Hint 1 (For Solution 1)">↵
If we know the result of every kick in advance, how would we simulate the penalty phase?↵
</spoiler>↵
↵
<spoiler summary ="Hint 2 (For Solution 1)">↵
Since `?` can be either `0` or `1`, how does this affect the number of possible outcomes?↵
</spoiler>↵
↵
<spoiler summary="Solution 1">↵
One way to solve this is by replacing every `?` with both possible values (`0` and `1`) and testing all resulting sequences. For each sequence, we simulate the match, tracking the score of both teams. At each step, we check if one team has gained an unbeatable lead, meaning the other team cannot catch up even if they score on all remaining kicks. The moment this happens, the game is stopped. The answer is the smallest number of kicks needed across all possible sequences.↵
↵
This approach ensures we check every possible scenario, but it is slow because the number of cases grows exponentially with the number of `?` characters.↵
↵
**Time Complexity:**↵
If there are at most $10$ characters and each `?` can be either `0` or `1`, the worst case has $2^{10}=1024$ possible cases. Each case requires checking all $10$ kicks, leading to a worst-case complexity of $O(10 \times 1024)$ per test case. This is slow but manageable for $t \leq 1000$.↵
↵
**Space Complexity:**↵
Since we only use a few integer variables to track scores, and also the recursion depth will not go more than 10, the space complexity is $O(1)$.↵
</spoiler>↵
↵
<spoiler summary="Code 1">↵
```python3↵
import sys↵
from math import ceil↵
↵
↵
def input():↵
return sys.stdin.readline().strip()↵
↵
↵
def earliest_decision_point(s):↵
length = len(s)↵
ans = length↵
↵
def recurse(index, cur1, cur2):↵
nonlocal ans↵
if index >= length:↵
return↵
↵
remaining = length - index↵
if index % 2:↵
if cur2 + ceil(remaining / 2) < cur1 or cur1 + remaining // 2 < cur2:↵
ans = min(ans, index)↵
return↵
if s[index] == "1":↵
recurse(index + 1, cur1, cur2 + 1)↵
elif s[index] == "0":↵
recurse(index + 1, cur1, cur2)↵
else:↵
recurse(index + 1, cur1, cur2 + 1)↵
recurse(index + 1, cur1, cur2)↵
else:↵
if cur1 + ceil(remaining / 2) < cur2 or cur2 + remaining // 2 < cur1:↵
ans = min(ans, index)↵
return↵
if s[index] == "1":↵
recurse(index + 1, cur1 + 1, cur2)↵
elif s[index] == "0":↵
recurse(index + 1, cur1, cur2)↵
else:↵
recurse(index + 1, cur1 + 1, cur2)↵
recurse(index + 1, cur1, cur2)↵
↵
recurse(0, 0, 0)↵
return ans↵
↵
↵
t = int(input().strip())↵
for _ in range(t):↵
s = input().strip()↵
print(earliest_decision_point(s))↵
```↵
</spoiler>↵
↵
<spoiler summary ="Hint (For Solution 2)">↵
Instead of considering all possible ways to fill in the `?` values, can we determine the earliest stopping point by only looking at the best or worst-case scenarios for each team?↵
</spoiler>↵
↵
<spoiler summary="Solution 2">↵
A more efficient approach is to determine the earliest possible stopping point without checking all possibilities. We do this by considering two extreme cases:↵
↵
1. The best case for the first team: Assume all `?`s taken by the first team result in goals (`1`), and those taken by the second team result in misses (`0`).↵
2. The best case for the second team: Assume all `?`s taken by the second team result in goals (`1`), and those taken by the first team result in misses (`0`).↵
↵
For each case, we simulate the match while keeping track of the maximum possible score each team can still achieve. At each step, if one team’s score is greater than the highest possible score the other team could still reach, the game can be stopped. The answer is the minimum number of kicks required in these two cases.↵
↵
This method is much faster because instead of trying all possibilities, we only check two scenarios, reducing unnecessary calculations.↵
↵
**Time Complexity:**↵
Since we only go through the string twice (once for each extreme case), the complexity is $O(10)$, which is actually $O(1)$ per test case.↵
↵
**Space Complexity:**↵
We only use a few integer variables to keep track of scores, so the space complexity is $O(1)$.↵
</spoiler>↵
↵
<spoiler summary="Code 2">↵
```python3↵
import sys↵
from math import ceil↵
↵
↵
def input():↵
return sys.stdin.readline().strip()↵
↵
↵
def earliest_decision_point(s):↵
length = len(s)↵
x1, x2, cur1, cur2 = 0, 0, 0, 0↵
↵
for i in range(length):↵
remaining = length - i↵
↵
if i % 2:↵
if (↵
cur2 + ceil(remaining / 2) < cur1 + x1↵
or cur1 + remaining // 2 < cur2 + x2↵
):↵
return i↵
if s[i] == "1":↵
cur2 += 1↵
elif s[i] == "?":↵
x2 += 1↵
else:↵
if (↵
cur1 + ceil(remaining / 2) < cur2 + x2↵
or cur2 + remaining // 2 < cur1 + x1↵
):↵
return i↵
if s[i] == "1":↵
cur1 += 1↵
elif s[i] == "?":↵
x1 += 1↵
↵
return length↵
↵
↵
t = int(input().strip())↵
for _ in range(t):↵
s = input().strip()↵
print(earliest_decision_point(s))↵
```↵
</spoiler>↵
↵
#### [D. Creating an a-Good String](https://codeforces.net/gym/596141/problem/D)↵
↵
<spoiler summary="Hint">Try breaking the problem into **smaller subproblems** by dividing the string into two halves and ensuring one half follows the rule while the other is recursively solved.</spoiler>↵
↵
<spoiler summary="Solution">↵
↵
We define a recursive function `calc(l, r, c)`, where:↵
<ul>↵
<li> `l` and `r` represent the start and end indices of the current segment. </li> ↵
<li> `c` represents the character we are trying to make this segment follow. </li> ↵
<li> The function returns the minimum number of moves needed to convert the substring ↵
`s[l:r]` into a c-good string.</li> ↵
</ul>↵
↵
**Base Case:**↵
If the length of the segment is 1 (`r = l`):↵
<ul>↵
<li> If `s[l] = c`, `return 0` (no changes needed).</li> ↵
<li> Otherwise, `return 1` (we must change `s[l]` to `c`).</li> ↵
</ul>↵
↵
**Recursive Case:**↵
We split the segment into two halves and consider two possible transformations:↵
<ul>↵
<li> <strong> case 1: </strong> The first half should be fully filled with c, and the second half should be (c+1)-good.</li>↵
<li> <strong> case 2: </strong> The second half should be fully filled with c, and the first half should be (c+1)-good.</p>↵
</li>↵
</ul>↵
↵
For each case, we compute the number of changes needed:↵
<ul>↵
<li> `left_result`: Changes required to make the first half fully filled with c `+` recursive call on the second half to make it (c+1)-good.</li>↵
<li> `right_result`: Changes required to make the second half fully filled with c `+` recursive call on the first half to make it (c+1)-good.</li>↵
↵
The final result for this segment is:↵
`min(left_result,right_result)`↵
</ul>↵
↵
**Time Complexity:**↵
The function processes each segment by counting characters (in $O(n)$ time) and then recursively splitting it into two halves. Since each level of recursion does a total of $O(n)$ work and there are $\log n$ levels (because the segment size halves each time), the total complexity is $O(n \log n)$.↵
↵
**Space complexity:**↵
The only space used in this solution is the call stack due to recursion. Since the recursion divides the problem into two halves at each step, the maximum depth of the recursion tree is $\log n$. Therefore, the space complexity is determined by the depth of the recursion, which is $O(\log n)$.↵
</spoiler>↵
↵
<spoiler summary="Code">↵
```python3↵
def calc(l, r, c):↵
if l == r:↵
if s[l] == c:↵
return 0↵
return 1↵
↵
mid = (l + r) // 2↵
↵
left_changes = (mid - l + 1) - s[l:mid + 1].count(c) ↵
left_result = left_changes + calc(mid + 1, r, chr(ord(c) + 1))↵
↵
right_changes = (r - mid) - s[mid + 1:r + 1].count(c)↵
right_result = right_changes + calc(l, mid, chr(ord(c) + 1))↵
↵
return min(left_result, right_2result)↵
↵
for _ in range(int(input())):↵
n = int(input())↵
s = input()↵
↵
print(calc(0, len(s) - 1, 'a'))↵
```↵
</spoiler>↵
↵
#### [E. Strange Mirroring](https://codeforces.net/gym/596141/problem/E)↵
↵
<spoiler summary="Hint">↵
Instead of building the whole transformed string, find a way to track where each character comes from in the original string. Notice that each step doubles the string and follows a clear pattern.↵
</spoiler>↵
↵
<spoiler summary="Solution">↵
The problem asks us to find the $k^{th}$ character after repeatedly transforming a string. Since the string grows extremely fast, it is impossible to generate the final version directly. Instead, we need a smarter way to determine the answer.↵
↵
Each transformation doubles the string: the first half stays the same, and the second half is the same as the first but with uppercase and lowercase letters swapped. This means that any character in the final string can be traced back to its original position in the first version.↵
↵
To find the character at position $k$, we work backwards to see where it came from. If $k$ is in the first half of a transformation step, it remains unchanged. If it is in the second half, we can shift it back to the first half of the previous step and apply a case swap. We repeat this process until we reach the original string.↵
↵
For the time complexity, each query runs in $O(\log k)$, which is at most $60$ for $k \leq 10^{18}$. Since $q \leq 2 \times 10 ^ 5$, this is fast enough.↵
↵
For the space complexity, the solution uses $O(\log k)$ extra space since we do not store the full transformed string. Our only space consumption is the call stack from the recursive calls.↵
</spoiler>↵
↵
<spoiler summary="Code">↵
```python↵
import sys↵
↵
↵
def input():↵
return sys.stdin.readline().strip()↵
↵
↵
def read_list():↵
return list(map(int, input().split()))↵
↵
↵
# Recursive function to find the character at position k in the transformed sequence↵
def find_char(depth, k, m, original):↵
if depth == 1: # Base case: when depth is 1, return the original character↵
return original[m]↵
↵
# Calculate the halfway point in the current sequence↵
half_length = 2 ** (depth - 2)↵
↵
if k > half_length:↵
# If k is in the second half, move to the previous depth and swap case↵
return find_char(depth - 1, k - half_length, m, original).swapcase()↵
↵
# If k is in the first half, continue in the same depth↵
return find_char(depth - 1, k, m, original)↵
↵
↵
def solve():↵
original_str = input() # Read the original string↵
_ = int(input())↵
queries = read_list() # Read the list of queries↵
n = len(original_str) # Get the length of the original string↵
↵
results = []↵
for k in queries:↵
k -= 1 # Convert to zero-based index↵
depth = k // n + 1 # Determine the depth level↵
index = k % n # Find the corresponding index in the original string↵
results.append(find_char(61, depth, index, original_str)) # Compute the result↵
↵
print(*results) # Print all results space-separated↵
↵
↵
t = int(input())↵
for _ in range(t):↵
solve()↵
``` ↵
</spoiler>↵
↵
↵
#### [A. Free Ice Cream for the Kids](https://codeforces.net/gym/596141/problem/A)↵
↵
<spoiler summary = "Solution">↵
To solve this problem efficiently, we simulate the sequence of operations while maintaining two values: the current number of ice cream packs and the count of distressed children. We start with $x$ packs and iterate through $n$ operations. If a carrier ($+d$) arrives, we increase our stock by $d$. If a child ($-d$) requests d packs, we check if we have enough ice cream. If yes, we serve the request and decrease our stock; otherwise, we increment the count of distressed kids. By processing all operations sequentially, we determine the final ice cream count and the number of distressed kids. This approach runs in $O(n)$ time, making it efficient given the constraints.↵
</spoiler>↵
↵
<spoiler summary="Code">↵
```python3↵
def solve():↵
n, cur = map(int, input().split())↵
distress = 0↵
↵
for _ in range(n):↵
op, d = input().split()↵
d = int(d)↵
↵
if op == '+':↵
cur += d↵
elif cur >= d:↵
cur -= d↵
else:↵
distress += 1↵
↵
print(cur, distress)↵
↵
solve()↵
```↵
</spoiler>↵
↵
#### [B. Increasing Sequence](https://codeforces.net/gym/596141/problem/B)↵
↵
<spoiler summary = "Solution">↵
To solve this problem, we need to construct a sequence $b$ that satisfies the given conditions. The key idea is to ensure that $b$ is strictly increasing and does not contain any elements from $a$.↵
↵
Steps to Solve:↵
<ol>↵
<li> Initialize a variable `Last_elem` to $0$, which will keep track of the last value added to the sequence $b$.</li>↵
<li> Iterate through each element in the sequence $a$.</li>↵
<li> For each element $a_i$:↵
<ul>↵
<li> Increment `Last_elem` by $1$.</li>↵
<li> If `Last_elem` is equal to $a_i$, increment `Last_elem` by $1$ again to ensure $b_i \neq a_i$.</li>↵
</ul>↵
</li>↵
<li> After processing all elements, `Last_elem` will hold the minimum value of $b_n$.</li>↵
</ol>↵
</spoiler>↵
↵
<spoiler summary="Code">↵
```python3↵
t = int(input())↵
for _ in range(t):↵
n = int(input())↵
a = list(map(int, input().split()))↵
last_elem = 0↵
for num in a:↵
last_elem += 1↵
if last_elem == num:↵
last_elem += 1↵
print(last_elem)↵
```↵
</spoiler>↵
↵
#### [C. Penalty Shootout](https://codeforces.net/gym/596141/problem/C)↵
↵
<spoiler summary = "Problem Summary">↵
In this problem, we are given a sequence of penalty kicks between two teams. Each team takes turns kicking, with the first team starting first. The goal is to determine the earliest possible moment when the penalty phase can be stopped. The phase stops if one team has scored more goals than the other team could possibly reach with their remaining kicks.↵
↵
Each kick is represented by a character in a 10-character string. A `1` means the kick is a goal, a `0` means the kick is missed, and a `?` means the outcome is uncertain, it could be either a goal or a miss. Our task is to find the minimum number of kicks required before the game can be stopped, considering all possible ways the `?` values could be filled.↵
</spoiler>↵
↵
<spoiler summary ="Hint 1 (For Solution 1)">↵
If we know the result of every kick in advance, how would we simulate the penalty phase?↵
</spoiler>↵
↵
<spoiler summary ="Hint 2 (For Solution 1)">↵
Since `?` can be either `0` or `1`, how does this affect the number of possible outcomes?↵
</spoiler>↵
↵
<spoiler summary="Solution 1">↵
One way to solve this is by replacing every `?` with both possible values (`0` and `1`) and testing all resulting sequences. For each sequence, we simulate the match, tracking the score of both teams. At each step, we check if one team has gained an unbeatable lead, meaning the other team cannot catch up even if they score on all remaining kicks. The moment this happens, the game is stopped. The answer is the smallest number of kicks needed across all possible sequences.↵
↵
This approach ensures we check every possible scenario, but it is slow because the number of cases grows exponentially with the number of `?` characters.↵
↵
**Time Complexity:**↵
If there are at most $10$ characters and each `?` can be either `0` or `1`, the worst case has $2^{10}=1024$ possible cases. Each case requires checking all $10$ kicks, leading to a worst-case complexity of $O(10 \times 1024)$ per test case. This is slow but manageable for $t \leq 1000$.↵
↵
**Space Complexity:**↵
Since we only use a few integer variables to track scores, and also the recursion depth will not go more than 10, the space complexity is $O(1)$.↵
</spoiler>↵
↵
<spoiler summary="Code 1">↵
```python3↵
import sys↵
from math import ceil↵
↵
↵
def input():↵
return sys.stdin.readline().strip()↵
↵
↵
def earliest_decision_point(s):↵
length = len(s)↵
ans = length↵
↵
def recurse(index, cur1, cur2):↵
nonlocal ans↵
if index >= length:↵
return↵
↵
remaining = length - index↵
if index % 2:↵
if cur2 + ceil(remaining / 2) < cur1 or cur1 + remaining // 2 < cur2:↵
ans = min(ans, index)↵
return↵
if s[index] == "1":↵
recurse(index + 1, cur1, cur2 + 1)↵
elif s[index] == "0":↵
recurse(index + 1, cur1, cur2)↵
else:↵
recurse(index + 1, cur1, cur2 + 1)↵
recurse(index + 1, cur1, cur2)↵
else:↵
if cur1 + ceil(remaining / 2) < cur2 or cur2 + remaining // 2 < cur1:↵
ans = min(ans, index)↵
return↵
if s[index] == "1":↵
recurse(index + 1, cur1 + 1, cur2)↵
elif s[index] == "0":↵
recurse(index + 1, cur1, cur2)↵
else:↵
recurse(index + 1, cur1 + 1, cur2)↵
recurse(index + 1, cur1, cur2)↵
↵
recurse(0, 0, 0)↵
return ans↵
↵
↵
t = int(input().strip())↵
for _ in range(t):↵
s = input().strip()↵
print(earliest_decision_point(s))↵
```↵
</spoiler>↵
↵
<spoiler summary ="Hint (For Solution 2)">↵
Instead of considering all possible ways to fill in the `?` values, can we determine the earliest stopping point by only looking at the best or worst-case scenarios for each team?↵
</spoiler>↵
↵
<spoiler summary="Solution 2">↵
A more efficient approach is to determine the earliest possible stopping point without checking all possibilities. We do this by considering two extreme cases:↵
↵
1. The best case for the first team: Assume all `?`s taken by the first team result in goals (`1`), and those taken by the second team result in misses (`0`).↵
2. The best case for the second team: Assume all `?`s taken by the second team result in goals (`1`), and those taken by the first team result in misses (`0`).↵
↵
For each case, we simulate the match while keeping track of the maximum possible score each team can still achieve. At each step, if one team’s score is greater than the highest possible score the other team could still reach, the game can be stopped. The answer is the minimum number of kicks required in these two cases.↵
↵
This method is much faster because instead of trying all possibilities, we only check two scenarios, reducing unnecessary calculations.↵
↵
**Time Complexity:**↵
Since we only go through the string twice (once for each extreme case), the complexity is $O(10)$, which is actually $O(1)$ per test case.↵
↵
**Space Complexity:**↵
We only use a few integer variables to keep track of scores, so the space complexity is $O(1)$.↵
</spoiler>↵
↵
<spoiler summary="Code 2">↵
```python3↵
import sys↵
from math import ceil↵
↵
↵
def input():↵
return sys.stdin.readline().strip()↵
↵
↵
def earliest_decision_point(s):↵
length = len(s)↵
x1, x2, cur1, cur2 = 0, 0, 0, 0↵
↵
for i in range(length):↵
remaining = length - i↵
↵
if i % 2:↵
if (↵
cur2 + ceil(remaining / 2) < cur1 + x1↵
or cur1 + remaining // 2 < cur2 + x2↵
):↵
return i↵
if s[i] == "1":↵
cur2 += 1↵
elif s[i] == "?":↵
x2 += 1↵
else:↵
if (↵
cur1 + ceil(remaining / 2) < cur2 + x2↵
or cur2 + remaining // 2 < cur1 + x1↵
):↵
return i↵
if s[i] == "1":↵
cur1 += 1↵
elif s[i] == "?":↵
x1 += 1↵
↵
return length↵
↵
↵
t = int(input().strip())↵
for _ in range(t):↵
s = input().strip()↵
print(earliest_decision_point(s))↵
```↵
</spoiler>↵
↵
#### [D. Creating an a-Good String](https://codeforces.net/gym/596141/problem/D)↵
↵
<spoiler summary="Hint">Try breaking the problem into **smaller subproblems** by dividing the string into two halves and ensuring one half follows the rule while the other is recursively solved.</spoiler>↵
↵
<spoiler summary="Solution">↵
↵
We define a recursive function `calc(l, r, c)`, where:↵
<ul>↵
<li> `l` and `r` represent the start and end indices of the current segment. </li> ↵
<li> `c` represents the character we are trying to make this segment follow. </li> ↵
<li> The function returns the minimum number of moves needed to convert the substring ↵
`s[l:r]` into a c-good string.</li> ↵
</ul>↵
↵
**Base Case:**↵
If the length of the segment is 1 (`r = l`):↵
<ul>↵
<li> If `s[l] = c`, `return 0` (no changes needed).</li> ↵
<li> Otherwise, `return 1` (we must change `s[l]` to `c`).</li> ↵
</ul>↵
↵
**Recursive Case:**↵
We split the segment into two halves and consider two possible transformations:↵
<ul>↵
<li> <strong> case 1: </strong> The first half should be fully filled with c, and the second half should be (c+1)-good.</li>↵
<li> <strong> case 2: </strong> The second half should be fully filled with c, and the first half should be (c+1)-good.</p>↵
</li>↵
</ul>↵
↵
For each case, we compute the number of changes needed:↵
<ul>↵
<li> `left_result`: Changes required to make the first half fully filled with c `+` recursive call on the second half to make it (c+1)-good.</li>↵
<li> `right_result`: Changes required to make the second half fully filled with c `+` recursive call on the first half to make it (c+1)-good.</li>↵
↵
The final result for this segment is:↵
`min(left_result,right_result)`↵
</ul>↵
↵
**Time Complexity:**↵
The function processes each segment by counting characters (in $O(n)$ time) and then recursively splitting it into two halves. Since each level of recursion does a total of $O(n)$ work and there are $\log n$ levels (because the segment size halves each time), the total complexity is $O(n \log n)$.↵
↵
**Space complexity:**↵
The only space used in this solution is the call stack due to recursion. Since the recursion divides the problem into two halves at each step, the maximum depth of the recursion tree is $\log n$. Therefore, the space complexity is determined by the depth of the recursion, which is $O(\log n)$.↵
</spoiler>↵
↵
<spoiler summary="Code">↵
```python3↵
def calc(l, r, c):↵
if l == r:↵
if s[l] == c:↵
return 0↵
return 1↵
↵
mid = (l + r) // 2↵
↵
left_changes = (mid - l + 1) - s[l:mid + 1].count(c) ↵
left_result = left_changes + calc(mid + 1, r, chr(ord(c) + 1))↵
↵
right_changes = (r - mid) - s[mid + 1:r + 1].count(c)↵
right_result = right_changes + calc(l, mid, chr(ord(c) + 1))↵
↵
return min(left_result, right_2result)↵
↵
for _ in range(int(input())):↵
n = int(input())↵
s = input()↵
↵
print(calc(0, len(s) - 1, 'a'))↵
```↵
</spoiler>↵
↵
#### [E. Strange Mirroring](https://codeforces.net/gym/596141/problem/E)↵
↵
<spoiler summary="Hint">↵
Instead of building the whole transformed string, find a way to track where each character comes from in the original string. Notice that each step doubles the string and follows a clear pattern.↵
</spoiler>↵
↵
<spoiler summary="Solution">↵
The problem asks us to find the $k^{th}$ character after repeatedly transforming a string. Since the string grows extremely fast, it is impossible to generate the final version directly. Instead, we need a smarter way to determine the answer.↵
↵
Each transformation doubles the string: the first half stays the same, and the second half is the same as the first but with uppercase and lowercase letters swapped. This means that any character in the final string can be traced back to its original position in the first version.↵
↵
To find the character at position $k$, we work backwards to see where it came from. If $k$ is in the first half of a transformation step, it remains unchanged. If it is in the second half, we can shift it back to the first half of the previous step and apply a case swap. We repeat this process until we reach the original string.↵
↵
For the time complexity, each query runs in $O(\log k)$, which is at most $60$ for $k \leq 10^{18}$. Since $q \leq 2 \times 10 ^ 5$, this is fast enough.↵
↵
For the space complexity, the solution uses $O(\log k)$ extra space since we do not store the full transformed string. Our only space consumption is the call stack from the recursive calls.↵
</spoiler>↵
↵
<spoiler summary="Code">↵
```python↵
import sys↵
↵
↵
def input():↵
return sys.stdin.readline().strip()↵
↵
↵
def read_list():↵
return list(map(int, input().split()))↵
↵
↵
# Recursive function to find the character at position k in the transformed sequence↵
def find_char(depth, k, m, original):↵
if depth == 1: # Base case: when depth is 1, return the original character↵
return original[m]↵
↵
# Calculate the halfway point in the current sequence↵
half_length = 2 ** (depth - 2)↵
↵
if k > half_length:↵
# If k is in the second half, move to the previous depth and swap case↵
return find_char(depth - 1, k - half_length, m, original).swapcase()↵
↵
# If k is in the first half, continue in the same depth↵
return find_char(depth - 1, k, m, original)↵
↵
↵
def solve():↵
original_str = input() # Read the original string↵
_ = int(input())↵
queries = read_list() # Read the list of queries↵
n = len(original_str) # Get the length of the original string↵
↵
results = []↵
for k in queries:↵
k -= 1 # Convert to zero-based index↵
depth = k // n + 1 # Determine the depth level↵
index = k % n # Find the corresponding index in the original string↵
results.append(find_char(61, depth, index, original_str)) # Compute the result↵
↵
print(*results) # Print all results space-separated↵
↵
↵
t = int(input())↵
for _ in range(t):↵
solve()↵
``` ↵
</spoiler>↵
↵
