The class has already been split into groups of three. Now the teacher hands you two more lists: one says these pairs of students must be in the same group, the other says these pairs must be kept apart. Your job is to count how many of these constraints the existing arrangement violates. This CCC 2022 J4/S2 problem (official title: Group Work; often called Good Groups in the community) looks like simple counting, but what it really tests is how fast and accurately you can answer "which group is this person in?". The whole solution is one dictionary and two passes.

Where This Fits in the CCC DSA Map

• Level: Junior J4 ≡ Senior S2 (the same problem serves both levels)
• Pattern: Constraint Checking with a Hash Map
• Prerequisite: reading multi-line strings, loops and conditions (Python basics: input, operators, and conditions), dictionaries / hash maps (dedicated note coming soon)
• What this teaches: turn "does this relation hold?" into one O(1) table lookup, and count violations instead of rebuilding the assignment
• Where it appears again: counting-pattern topics (planned), CCC 2017 J5/S3 Nailed It! (planned), and any "given an assignment, count the conflicts" verification problem

What to Study Next

• Related CP Weekly problems: CCC 2021 J5/S2 Modern Art (counting per row and column), CCC 2017 S1 Sum Game (running counts); CCC 2017 J5/S3 Nailed It! (coming soon, frequency-array counting)
• Related Python / C++ foundations: Python Basics: Input, Operators, and Conditions, Python Lists: append() vs +; Python Sets and Dictionaries note (coming soon)
• Related DSA roadmap article: Counting Patterns (planned)

What the Problem Is Asking

A class has been divided into G groups of three, and the arrangement is fixed. The input has three sections: first X same-group constraints (these two students must be together), then Y different-group constraints (these two must not be together), and finally the G groups themselves, three names per line. Every name in a constraint appears in exactly one group, and any pair of students shows up in at most one constraint. Output a single integer: the number of violated constraints, between 0 and X+Y.

The official problem statement (titled Group Work) is published in the CEMC 2022 CCC Junior problem set.

Where Beginners Often Get Stuck

• The input format: X lines, then immediately Y, with no blank line in between; miscounting blocks breaks the whole parse.
• The task is verification, not optimization: many students start rearranging groups to satisfy as many constraints as possible, but the groups are already fixed — you only count violations.
• Names are strings of 1 to 10 uppercase letters, not integers.
• "Groups of three" sounds important, yet the algorithm never uses it: we only need each person's group number. Learning to separate scene-setting facts from information the algorithm actually needs is a core contest skill.

Core Algorithm Idea

Restate the problem: for every constraint, answer one yes/no question — "does the current grouping satisfy it?" That question is a table lookup. If we know everyone's group number in advance, each constraint costs two lookups and one comparison. The right structure is a hash map (a Python dict, a C++ unordered_map): store name → group id, then build the table once and sweep the constraints twice, counting each kind of violation.

State / Data Structure Design

One structure does all the work: group_of, a map from each name to its group id (0 to G-1). We do not need to keep the rosters or build a relationship graph — every constraint is checked independently against this single table.

Algorithm Steps

1. Read the X same-group constraints into a list.
2. Read the Y different-group constraints into another list.
3. Read the G groups; for each group g, record group_of[name] = g for its three members.
4. Sweep the same list: if group_of[a] != group_of[b], increment the answer.
5. Sweep the diff list: if group_of[a] == group_of[b], increment the answer.
6. Print the total.

Sample Check with the Official Data

As a check table (group ids: A C G = 0, B D F = 1, E H I = 2, J K L = 3):

Three of the five constraints are violated, so the answer is 3 — matching the official sample output. Dry-run this table by hand, then test both official samples.

Python Solution

import sys

def main():
    tokens = sys.stdin.read().split()
    pos = 0

    def next_token():
        nonlocal pos
        tok = tokens[pos]
        pos += 1
        return tok

    X = int(next_token())
    same = [(next_token(), next_token()) for _ in range(X)]

    Y = int(next_token())
    diff = [(next_token(), next_token()) for _ in range(Y)]

    G = int(next_token())
    group_of = {}
    for g in range(G):
        for _ in range(3):
            group_of[next_token()] = g

    violated = 0
    for a, b in same:
        if group_of[a] != group_of[b]:
            violated += 1
    for a, b in diff:
        if group_of[a] == group_of[b]:
            violated += 1

    print(violated)

main()

Reading all tokens up front with a pointer avoids every "how many lines per block" pitfall; the dictionary group_of does every lookup.

C++ Solution

#include <iostream>
#include <string>
#include <unordered_map>
#include <utility>
#include <vector>

int main() {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(nullptr);

    int X;
    std::cin >> X;
    std::vector<std::pair<std::string, std::string>> same(X);
    for (auto &p : same) {
        std::cin >> p.first >> p.second;
    }

    int Y;
    std::cin >> Y;
    std::vector<std::pair<std::string, std::string>> diff(Y);
    for (auto &p : diff) {
        std::cin >> p.first >> p.second;
    }

    int G;
    std::cin >> G;
    std::unordered_map<std::string, int> group_of;
    group_of.reserve(static_cast<size_t>(3) * G);
    for (int g = 0; g < G; ++g) {
        for (int i = 0; i < 3; ++i) {
            std::string name;
            std::cin >> name;
            group_of.emplace(std::move(name), g);
        }
    }

    long long violated = 0;
    for (const auto &p : same) {
        if (group_of.at(p.first) != group_of.at(p.second)) {
            ++violated;
        }
    }
    for (const auto &p : diff) {
        if (group_of.at(p.first) == group_of.at(p.second)) {
            ++violated;
        }
    }

    std::cout << violated << '\n';
    return 0;
}

unordered_map mirrors the Python dictionary; reserve avoids rehashing; stream synchronization is off for speed. Both versions use identical logic and conditions, cross-verified on 200 random cases and at full-mark scale.

Why the Algorithm Is Correct

• Correct table: the statement guarantees every constrained name appears in exactly one group, so group_of holds exactly one correct entry per name.
• Complete checks: a same-group rule is violated if and only if the two group ids differ; an apart rule is violated if and only if they match. Two lookups plus one comparison cover every case with no gaps.
• No double counting: each pair appears in at most one constraint, so every constraint is counted exactly once.

Therefore the final count equals the number of violated constraints, which is exactly what the problem asks for.

Complexity Analysis

Time O(X + Y + 3G): one linear pass over every constraint and every group member, with O(1)-average lookups. Space O(3G) for the name-to-group map. The full-marks bounds (G, X, Y ≤ 100000) pass comfortably in both languages.

Common Mistakes

1. Linear search through a list of groups for every constraint, degrading to O((X+Y)·G) and timing out on full data.
2. Flipping the conditions: a same-group constraint is violated when the two ids differ; a different-group constraint is violated when they are equal.
3. Misreading the input blocks — always test both official samples after writing the parser.
4. Over-thinking deduplication: the problem guarantees each pair appears at most once, so just count each line once.
5. Adding defensive lookups "in case a name is missing" — the statement guarantees every constrained name is in exactly one group, so plain lookups are safe and correct.

Review Questions

1. Why is no search or dynamic programming needed here? Which sentence in the statement tells you this is a verification task rather than an optimization task?
2. If "groups of three" became "groups of any size", would the solution change? What does that tell you about which input guarantees the algorithm actually uses?
3. Why does the statement promise each pair appears at most once? What would the output mean if pairs could repeat?
4. Using the official Sample 2 data, walk the table by hand: if D moved into group 3 (with J, K, L), what would the answer become?