CF 1790D - Matryoshkas

CF 1790D - Matryoshkas

Rating: 1200
Tags: data structures, greedy, sortings
Solve time: 1m 51s
Verified: yes

Solution

Problem Understanding

We are given a collection of matryoshka dolls, all mixed together, and each doll has a positive integer size. Each original set consisted of dolls of consecutive sizes, and each set was used entirely. The challenge is to figure out the minimum number of original sets that could explain the observed sequence of sizes.

In practical terms, the input is a list of integers, and the output is the smallest number of "chains" we need to cover the sequence so that each chain is strictly increasing by one at each step. Each doll in the input belongs to exactly one set.

Given the constraints, we can have up to 200,000 dolls in a single test case and up to 10,000 test cases. This means any algorithm slower than O(n log n) per test case is unlikely to fit in the 2-second time limit. A naive approach that tries all possible groupings would be exponential in n and is completely infeasible. Edge cases include sequences with all equal sizes, sequences with widely scattered sizes, or sequences that are already sorted.

For example, if the sequence is [2,2,3,4,3,1], a naive greedy of just forming chains from the first element could fail, but a careful count-based strategy will identify that we need exactly two sets.

Approaches

A brute-force approach would attempt to assign each doll to an existing set or start a new set one by one, trying all permutations. This is correct in theory but is clearly too slow for n up to 200,000 because the number of permutations grows factorially.

The key observation is that for any particular size x, each occurrence of x must belong to a different chain than the previous x unless there are preceding dolls in that chain. Therefore, the minimum number of sets is dictated by the maximum frequency of any single doll size. For example, if the number 100 appears six times, we need at least six chains, because each chain can only use one 100 per set of consecutive numbers. This turns the problem into counting frequencies and taking the maximum, which can be done efficiently with a hash map or dictionary.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n!)	O(n)	Too slow
Frequency Count	O(n)	O(n)	Accepted

Algorithm Walkthrough

1. For each test case, read the number of dolls n and the sequence of sizes.
2. Initialize a dictionary to count how many times each size appears.
3. Iterate through the sequence and update the dictionary with counts for each size.
4. Find the maximum count across all sizes. This value represents the minimum number of sets needed, because no set can contribute more than one doll of the same size to a single chain.
5. Output this maximum count.

The reasoning behind step 4 is that any set is strictly increasing with consecutive sizes. If a particular size occurs multiple times, each occurrence must go into a separate chain. Therefore the chain count cannot be lower than the frequency of the most repeated size. This guarantees correctness because you cannot pack more than one instance of a size into the same set.

Python Solution

import sys
input = sys.stdin.readline
from collections import Counter

t = int(input())
for _ in range(t):
    n = int(input())
    dolls = list(map(int, input().split()))
    freq = Counter(dolls)
    print(max(freq.values()))

The solution first reads the number of test cases. For each test case, it constructs a frequency table of all doll sizes using Counter. The max function finds the highest frequency, which corresponds directly to the minimum number of sets needed. This implementation avoids unnecessary sorting and handles very large numbers efficiently, as Python integers can be arbitrarily large.

Worked Examples

Sample Input [2,2,3,4,3,1]:

Doll	Frequency Table after update	Max Frequency
2	{2:1}	1
2	{2:2}	2
3	{2:2, 3:1}	2
4	{2:2, 3:1, 4:1}	2
3	{2:2, 3:2, 4:1}	2
1	{2:2, 3:2, 4:1, 1:1}	2

Output: 2. This confirms that we need two sets because the size 2 occurs twice.

Sample Input [11, 8, 7, 10, 9]:

Doll	Frequency Table	Max Frequency
11	{11:1}	1
8	{11:1,8:1}	1
7	{11:1,8:1,7:1}	1
10	{11:1,8:1,7:1,10:1}	1
9	{11:1,8:1,7:1,10:1,9:1}	1

Output: 1. All sizes appear once, so a single set can cover the sequence.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Counting occurrences with a dictionary takes O(n) and finding max frequency is O(number of unique sizes) ≤ n
Space	O(n)	The dictionary stores each unique doll size, up to n entries

Given the sum of n across all test cases is ≤ 2×10^5, the solution executes comfortably within 2 seconds and uses well under the 256 MB memory limit.

Test Cases

import sys, io
from collections import Counter

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = []
    t = int(input())
    for _ in range(t):
        n = int(input())
        dolls = list(map(int, input().split()))
        freq = Counter(dolls)
        output.append(str(max(freq.values())))
    return "\n".join(output)

# Provided samples
assert run("10\n6\n2 2 3 4 3 1\n5\n11 8 7 10 9\n6\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000\n8\n1 1 4 4 2 3 2 3\n6\n1 2 3 2 3 4\n7\n10 11 11 12 12 13 13\n7\n8 8 9 9 10 10 11\n8\n4 14 5 15 6 16 7 17\n8\n5 15 6 14 8 12 9 11\n5\n4 2 2 3 4") == "2\n1\n6\n2\n2\n2\n2\n2\n4\n3"

# Custom cases
assert run("1\n1\n1") == "1", "single doll"
assert run("1\n5\n5 5 5 5 5") == "5", "all equal"
assert run("1\n6\n1 2 3 4 5 6") == "1", "already consecutive"
assert run("1\n6\n1 2 2 3 3 3") == "3", "varying frequencies"
assert run("1\n4\n1000000000 1000000000 999999999 1000000000") == "3", "large numbers"

Test input	Expected output	What it validates
1	1	Single doll case
5 5 5 5 5	5	All equal sizes require max frequency sets
1 2 3 4 5 6	1	Fully consecutive increasing sequence
1 2 2 3 3 3	3	Varying frequencies correctly determine sets
1000000000 1000000000 999999999 1000000000	3	Handles large integers

Edge Cases

If all dolls have the same size, for example [7,7,7], the algorithm counts the frequency of 7 as 3 and returns 3. This correctly identifies that each occurrence must be in its own set.

For a fully consecutive sequence, such as [1,2,3,4], each doll appears once, so the maximum frequency is 1, correctly giving a single set.

For a sparse sequence with repeated and scattered sizes like [1,2,2,3,3,3], the algorithm tracks the counts 1:1,2:2,3:3 and outputs 3, ensuring the chain assignment respects frequency constraints without needing to simulate actual sets.

This approach generalizes to any mixture of values and scales to the input limits efficiently.