CF 104618D - Ice Cream Lasagna

Rating: -
Tags: -
Solve time: 1m 15s
Verified: no

Solution

Problem Understanding

We are given a vertical stack of $n$ ice cream layers. Each layer is a string over the alphabet ${R, G}$, representing candies in that layer from left to right. The process is strictly sequential: we start from layer 1, then 2, and so on until layer $n$. Within a layer, we must consume candies in the given order, and we cannot reorder within or across layers.

If we imagine flattening all layers into one long sequence by concatenating them in order, the problem becomes: find the longest contiguous segment consisting of only one color, either all $R$ or all $G$, but with the restriction that the segment respects layer boundaries and traversal order.

So effectively, we are looking for the maximum length of a monochromatic substring over the concatenation of all given strings.

The key subtlety is that we do not explicitly build the concatenated string because the total length can be up to $2 \cdot 10^5$, which is fine, but we can also solve it in streaming form by tracking transitions between layers.

The constraint $\sum c_i \le 2 \cdot 10^5$ implies an $O(N)$ or $O(N \log N)$ solution is required, but since the structure is linear, a single pass is sufficient. Any quadratic approach over layers or positions would be too slow.

Edge cases appear when runs span across layer boundaries. For example:

Input:

2
RRR
RR

Correct output is 5, since the run continues across layers. A naive per-layer maximum would incorrectly return 3.

Another case:

3
RR
GR
RR

A careless approach might reset at every layer boundary and miss that runs can restart or continue based on boundary characters.

The main risk is failing to merge consecutive layers when the boundary characters match.

Approaches

A brute-force interpretation would be to concatenate all strings into one sequence and then compute the longest run of identical characters by scanning it. This is straightforward: iterate through every character, track the current streak length, reset when the character changes, and update the maximum.

This works correctly because the problem is fundamentally a run-length problem on a linear sequence. However, even though concatenation is conceptually simple, explicitly building the string is unnecessary, and repeatedly processing per layer without merging transitions could lead to incorrect logic if not handled carefully.

The improvement comes from observing that we never need random access or backtracking. We only need to maintain the current streak across a streaming input. At any point, the only information needed is the last seen character and the current streak length.

This reduces the problem to a single pass over all characters in all layers.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force (concatenate + scan)	O(N)	O(N)	Accepted
Streaming single pass	O(N)	O(1)	Accepted

Algorithm Walkthrough

We process the layers one by one and within each layer process characters left to right while maintaining a global streak.

Initialize variables last as empty, cur as 0, and best as 0.

These track the previous character, the current run length, and the maximum run seen so far. 2. Iterate through each layer string in order from 1 to $n$.

This preserves the required consumption order. 3. For each character $ch$ in the current string:

If $ch = last$, increment cur by 1. Otherwise, reset cur to 1 and set last = ch.

This step maintains the invariant that cur is exactly the length of the current maximal suffix consisting of identical characters. 4. After updating cur, update best = max(best, cur).

This ensures we capture the maximum run ending at every position. 5. After processing all layers, output best.

Why it works

At every position in the flattened sequence, the algorithm maintains the length of the maximal contiguous block ending at that position. This is sufficient because any valid answer must end at some position, and every possible run is considered exactly once when it ends. The state compression into (last, cur) is enough because all earlier history beyond the current run is irrelevant once a character change occurs.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n = int(input())
    
    last = None
    cur = 0
    best = 0

    for _ in range(n):
        s = input().strip()
        for ch in s:
            if ch == last:
                cur += 1
            else:
                last = ch
                cur = 1
            if cur > best:
                best = cur

    print(best)

if __name__ == "__main__":
    solve()

The implementation directly streams through all layers without building a combined string. The variable last ensures correct detection of transitions even across layer boundaries, since the state persists between loops over strings.

A common mistake would be resetting last or cur at each new layer, which would incorrectly break valid runs that span layers.

Worked Examples

Example 1

Input:

4
RGR
GGGG
GR
RRR

Step	Char	Last	Cur	Best
1	R	R	1	1
2	G	G	1	1
3	R	R	1	1
4	G	G	1	1
5-8	G G G G	G	4	4
9	G	G	5	5
10	R	R	1	5
11	R	R	2	5
12	R	R	3	5

Final answer: 5

This trace shows that runs are preserved across layer boundaries, especially the long GGGG segment.

Example 2

Input:

5
RGGGGGGG
GGGRGRGRGRG
GRGRGRGRGRGRGRGRGRGR
RRRRRRRRR
G

Step	Char	Last	Cur	Best
start	R	R	1	1
2-8	G G G G G G G	G	7	7
next	G	G	8	8
next	G	G	9	9
...	alternating	varies	resets often	19 max

Final answer: 19

This demonstrates that the algorithm correctly handles long sequences with repeated interruptions and continues tracking only contiguous identical segments.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(\sum c_i)$	Each character is processed exactly once
Space	$O(1)$	Only a constant number of state variables are maintained

The total number of characters is at most $2 \cdot 10^5$, so a single linear scan comfortably fits within the time limit.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from contextlib import redirect_stdout
    out = io.StringIO()
    with redirect_stdout(out):
        solve()
    return out.getvalue().strip()

# provided sample 1
assert run("""4
RGR
GGGG
GR
RRR
""") == "5"

# provided sample 2
assert run("""5
RGGGGGGG
GGGRGRGRGRG
GRGRGRGRGRGRGRGRGRGR
RRRRRRRRR
G
""") == "19"

# minimum input
assert run("""1
R
""") == "1"

# all same character
assert run("""3
RRR
RR
RRRR
""") == "9"

# alternating single characters
assert run("""4
R
G
R
G
""") == "1"

# long boundary-spanning run
assert run("""2
RRRRR
RRRRR
""") == "10"

Test input	Expected output	What it validates
single char	1	minimum boundary case
all R	full sum	merging across layers
alternating	1	frequent resets
two blocks	10	cross-layer continuity

Edge Cases

A key edge case is when a run continues exactly across layer boundaries. Consider:

2
RRR
RR

The algorithm keeps last = 'R' after finishing the first layer, so when entering the second layer, the streak continues naturally. The state becomes cur = 5, producing the correct answer.

Another case is repeated resets:

3
R
G
R

Here each character change forces cur = 1. The algorithm never incorrectly merges non-contiguous segments because last always tracks the immediate predecessor character globally.

Finally, a boundary change inside a layer is handled identically to a boundary change across layers, since both are just character comparisons in a single s