**Hint 1**

Consider the strategy of forming teams based on the power of the players. How can this strategy be optimized to maximize the number of wins?

**Hint 2**

How can sorting help you come up with a strategy?

**Solution**

To maximize the number of wins, we can sort the candidate players in decreasing order of power and form teams by picking a player with the highest power and pairing them with the player with the lowest power. This strategy ensures that we use the highest power available while minimizing the number of players needed to form a team with a particular power.

We can then leverage Pak Chanek's ability to change the power of players in a team by increasing their power to the maximum available power in the team. This increases the chances of winning a match by ensuring that the team's total power is as high as possible.

To determine the maximum number of wins, we can iterate through the formed teams and count the number of teams that can defeat an enemy team with a specified power. This can be done by comparing the total power of the team with the power of the enemy team and checking if it is strictly greater.

**Code**

```
n, d = map(int, input().split())
players = sorted(list(map(int, input().split())), reverse=True)
num_wins = 0
right = n - 1
for left in range(n):
if left > right:
break
curr = players[left]
while right > left and curr <= d:
curr += players[left]
right -= 1
num_wins += int(curr > d)
print(num_wins)
```

**Hint 1**

Simulate the program.

**Solution**

Simulate the program by simulating the recursion stack. Since we only have "add" operations and a for loop we can use simply multiply nest loop results.

Time Complexity — O(N) Space Complexity — O(N)

**Code**

```
from sys import stdin
n = int(stdin.readline().strip())
stack = [[1, 0]]
for _ in range(n):
command = stdin.readline().strip()
if command == "add":
stack[-1][1] += 1
elif command == "end":
iterations, add = stack.pop()
stack[-1][1] += iterations * add
else:
_, iterations = command.split()
stack.append([int(iterations), 0])
if stack[-1][1] > (2**32 — 1):
print("OVERFLOW!!!")
else:
print(stack[-1][1])
```

**Hint**

Try sorting.

**Solution**

After sorting we can observe that the target rectangle area should be $$$a_{i} \cdot a_{4n}$$$. Then we just need to check for each $$${i}$$$ from 1 to n that $$$a_{2i-1}=a_{2i}$$$ and $$$a_{4n-2i+1}=a_{4n-2i+2}$$$ and $$$a_{2i-1}\cdot a_{4n-2i+2}=area$$$. If all conditions are satisfied for all i then the answer is "YES". Otherwise, the answer is "NO".

**Code**

```
from sys import stdin, stdout
t = int(stdin.readline().strip())
for _ in range(t):
n = int(stdin.readline().strip())
arr = list(map(int, stdin.readline().split()))
arr.sort()
i, j = 0, len(arr) - 1
ok = True
area = arr[i] * arr[j]
while i < j:
if arr[i] == arr[i + 1] and arr[j] == arr[j - 1] and arr[i] * arr[j] == area:
i += 2
j -= 2
else:
ok = False
break
print("YES" if ok else "NO")
```

**Hint**

Look for an optimal way to arrange the emotes.

**Solution**

The optimal way to arrange the emotes is to choose a chunk of k of the most potent emote separated by a single emote of the second most potent emote. Eg: Let's say "a" is the most potent emote and "b" is the second most potent. The best way to arrange the emotes is like: $$${"a_1, a_2, a_3 .... a_k, b, a_1, .... a_k, b , a_1..."}$$$

**Code**

```
from sys import stdin, stdout
n, m, k = map(int, stdin.readline().split())
arr = list(map(int, stdin.readline().split()))
arr.sort(reverse=True)
chunk_len = k + 1
full = (arr[0] * k + arr[1]) * (m // chunk_len)
rem = m % chunk_len
part = arr[0] * min(rem, k)
rem -= min(rem, k)
part += arr[1] * rem
print(full + part)
```

**Hint**

What happens as we increase or decrease "k" in regard to the damage done to the dragon?

**Solution**

Use binary search to look for the minimum k that deals at least h damage to the dragon.

**Code**

```
from sys import stdin
t = int(stdin.readline().strip())
for _ in range(t):
n, h = map(int, stdin.readline().split())
arr = list(map(int, stdin.readline().split()))
left = 1
best = right = h
while left <= right:
mid = (left + right) // 2
curr_max = -1
damage = 0
for time in arr:
damage += mid - max(curr_max - time, 0)
curr_max = time + mid
if damage >= h:
best = mid
right = mid - 1
else:
left = mid + 1
print(best)
```

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