$$$N/10$$$ will give you the digit at tenth place.
$$$N\%10$$$ will give you the digit at ones place.

We can bruteforce all the possible rounds.

The possible pairings are -

• $$$A_1 - B_1$$$ and $$$A_2 - B_2$$$
• $$$A_2 - B_2$$$ and $$$A_1 - B_1$$$
• $$$A_2 - B_1$$$ and $$$A_1 - B_2$$$
• $$$A_1 - B_2$$$ and $$$A_2 - B_1$$$

Alex can shower on one of the following times
- Before the first task
- Between any two tasks
- After all tasks

We should check if any of these break is $$$\ge S$$$

Solve for the case when there are no ?.
392. Is Subsequence

If you look closely at algorithm in hint 1, notice that whenever we get ? in S, we can replace it with the char we want to match in T.

For each no count the no of $$$\lfloor \frac{y}{3} \rfloor$$$ operations we will require to make it 0.

Make one of the nos 0. Then chose that no as $$$X$$$. $$$3*0 = 0$$$

Let $$$C$$$ be the no of operations we have done to get first $$$0$$$. Notice that we have multipled some other nos by $$$3$$$, $$$C$$$ times. So to make them equal to original no, we will have to do $$$C$$$ more operations with them as $$$Y$$$, and $$$X=0$$$

Notice that no of operations required to make $$$L$$$ equal to 0 $$$\le$$$ Notice that no of operations required to make $$$L+1$$$ equal to 0

So in optimal case, we should make $$$L$$$ equal to $$$0$$$ first, then make all the other nos $$$0$$$

To count the sum of no of operations required to make everything in range $$$L$$$ to $$$R$$$ equal to 0, use prefix sum

What is the fastest way to check if a particular no exists in a sorted array or not?

Binary search FTW

Also notice that $$$log_2(999) = 10$$$.
Notorious coincidence?
I think not.

Lets query for area of of $$$A*A$$$ square instead of a rectangle.

If the checker returns $$$A*A$$$, we know that the missing no $$$\gt A$$$
If the checker returns $$$(A+1)*(A+1)$$$, we know that the missing no $$$\le A$$$

Looks similar to how we search in a sorted array?

Also notice that $$$log_3(999) = 7$$$.
Notorious coincidence?
I think not.

Solution is similar to ternary search

Instead of one mid in binary search lets have 2 mids.
Divide $$$[L,R]$$$ range into almost 3 equal parts and see in which range missing no lies.

If we have 2 nos $$$A,B$$$ such that $$$L \lt A \lt B \lt R$$$.
Area of rectange will be $$$(A+1)*(B+1)$$$ is missing no lies between $$$[0,A]$$$
Area of rectange will be $$$A*(B+1)$$$ is missing no lies between $$$[A+1,B]$$$
Area of rectange will be $$$A*B)$$$ is missing no lies between $$$[B+1,1000]$$$