Time Display Stickers

Research note: I checked the official Codeforces statement and the official analysis PDF. The intended idea is to split displays by hour type and greedily assign stickers from most restrictive positions to least restrictive positions. Below is the same core idea, tightened into a direct binary search on the answer.

Each display has format HH:MM, but the colon is free. Tiny mercy. Only four stickers matter.

There are only two possible hour shapes:

• Type A: hour is 00 to 09. The first hour digit must be 0, and the second hour digit can be anything from 0 to 9.
• Type B: hour is 10 or 11. The first hour digit must be 1, and the second hour digit must be 0 or 1.

For any display, the minute tens digit must be in 0..5, and the minute ones digit can be anything.

Let:

• $c_d$ be the number of stickers with digit $d$,
• $c_{01}=c_0+c_1$,
• $c_{\le 5}=c_0+c_1+\cdots+c_5$.

Suppose we want to make exactly $m$ displays. Let $y$ be the number of Type B displays, so the number of Type A displays is

$$x=m-y.$$

Now list the required positions from most restrictive to least restrictive:

1. Type A first hour digit: needs $x=m-y$ stickers equal to 0.
2. Type B first hour digit: needs $y$ stickers equal to 1.
3. Type B second hour digit: needs $y$ stickers from {0,1}.
4. Minute tens digits: need $m$ stickers from {0,1,2,3,4,5}.
5. Remaining flexible positions: everything else can be any digit.

This gives simple conditions on $y$.

First, Type A needs enough zeroes:

$$m-y \le c_0 \quad \Longleftrightarrow \quad y \ge m-c_0.$$

Type B first digits need enough ones:

$$y \le c_1.$$

After assigning the first hour digit for all $m$ displays, we have consumed exactly $m$ stickers from {0,1}: $m-y$ zeroes and $y$ ones. Therefore the remaining number of {0,1} stickers is $c_{01}-m$. The Type B second hour digits need $y$ of them:

$$y \le c_{01}-m.$$

After that, we have consumed $m+y$ stickers from 0..5: the $m$ first-hour stickers plus the $y$ Type B second-hour stickers. The minute tens positions need another $m$ stickers from 0..5, so:

$$c_{\le 5}-(m+y) \ge m$$

or equivalently:

$$y \le c_{\le 5}-2m.$$

Also obviously:

$$0 \le y \le m.$$

Finally, every display uses four stickers total, so we need:

$$4m \le n.$$

So for a fixed $m$, feasible $y$ values must satisfy:

$$\max(0, m-c_0) \le y \le \min(m, c_1, c_{01}-m, c_{\le 5}-2m).$$

If this interval is nonempty and $4m\le n$, then $m$ displays are possible. Otherwise, impossible.

Why is this sufficient? Because the allowed digit sets are nested after the two exact requirements:

$$\{0\},\ \{1\},\ \{0,1\},\ \{0,1,2,3,4,5\},\ \{0,1,\dots,9\}.$$

Greedily filling the smallest allowed sets first is optimal: a sticker that can serve a stricter position can also serve a looser later position, but not the other way around. If one of these aggregate inequalities fails, no clever rearrangement saves you. The math already caught the crime scene.

Feasibility is monotonic: if we can make $m$ displays, we can throw away one display and make $m-1$. Therefore we binary search the largest feasible $m$ from $0$ to $\lfloor n/4\rfloor$.

Edge cases:

• If $n<4$, the answer is immediately 0 through the binary search range.
• Digits 6..9 cannot help with hour tens, Type B hour ones, or minute tens; they only help flexible positions.
• Leading zero hours like 04 are valid. Do not accidentally reject them like a parser with trust issues.

For each test case, feasibility is $O(1)$, and binary search takes $O(\log n)$. Counting digits costs $O(n)$. Across all test cases, the total complexity is

$$O\left(\sum n + t\log n\right),$$

with $O(1)$ extra memory.