CF 1249A - Yet Another Dividing into Teams

Rating: 800
Tags: math
Solve time: 2m 20s
Verified: yes

Solution

Problem Understanding

We are given several independent scenarios. In each scenario there is a list of distinct integers representing student skill levels. The goal is to split these students into the smallest possible number of groups such that no group contains two students whose skills differ by exactly one.

The restriction is very local: only pairs with difference exactly 1 are forbidden inside the same group. Any other pair is allowed, even if their difference is large or small but not equal to 1.

From this, the task becomes a grouping problem over integers with a very specific adjacency rule. If we imagine each value as a node on the number line, then every value is only “in conflict” with its immediate neighbors value minus one and value plus one.

Each query is small, with at most 100 students and skill values also bounded by 100. This immediately rules out any need for heavy optimization or advanced data structures. Even an O(n^2) solution per query is easily fast enough.

The main edge case that often misleads a naive approach is assuming that only consecutive chains matter without considering how overlap between multiple chains forces multiple teams.

For example, consider values [2, 3, 4]. A naive idea might be to think we only need two teams because we can alternate assignments. But since 2 conflicts with 3 and 3 conflicts with 4, but 2 and 4 are fine, the correct grouping depends on how we assign 3, which acts as a bridge.

Another subtle case is disconnected sets like [1, 3, 5], where no conflicts exist at all and a single team suffices.

The key difficulty is to correctly identify how many “conflict layers” are forced by adjacency constraints.

Approaches

A brute-force way to think about this is to assign students one by one into teams, trying every possible placement while respecting constraints. For each student, we attempt to place them into an existing team if no member differs by exactly 1, otherwise we create a new team. This resembles a greedy coloring attempt, but if done naively with backtracking, it explores exponential possibilities because each assignment affects future feasibility.

Even a simpler greedy without structure can fail if order is unlucky. For example, inserting in input order without considering sorting may lead to unnecessary new teams.

The crucial observation is that the constraint only depends on adjacent integer values. If we sort the array, we can reason about how numbers interact through consecutive segments. Any value x only conflicts with x-1 and x+1. This means the problem is equivalent to coloring numbers on a line where edges exist only between consecutive integers.

Now consider how many teams are needed. If we group numbers by value, the only time we are forced to separate into multiple teams is when there are chains of consecutive integers. Inside a consecutive run like [k, k+1, k+2, ..., k+m], each number conflicts with its neighbors, forming a path graph. A path graph has chromatic number 2, but multiple disjoint paths can overlap in a way that still requires only 2 colors globally.

This leads to a simpler perspective: we only need to know whether there exists any pair of consecutive integers in the set. If no such pair exists, all values are independent and we need exactly 1 team. If at least one such pair exists, then we need exactly 2 teams, because any chain of conflicts can always be 2-colored by alternating assignments along parity of value.

So the entire problem reduces to checking whether there exists x such that both x and x+1 are present.

Comparison

Approach	Time Complexity	Space Complexity	Verdict
Brute Force assignment/backtracking	O(n!)	O(n)	Too slow
Check consecutive pairs	O(n) per query	O(n)	Accepted

Algorithm Walkthrough

Read all student skill values for a query and store them in a set for fast membership checking. This is necessary because we need constant-time lookup for checking whether consecutive integers exist.
Iterate through each value x in the set and check whether x + 1 also exists in the set. This directly detects whether there is at least one conflict edge in the implicit graph.
If we find any such pair (x, x+1), we immediately conclude that at least two teams are required. This is because these two values cannot share a team, forcing a split that propagates through any connected chain.
If no such pair exists after checking all values, then no constraints are active, and all students can be placed into a single team.

Why it works

The values form a graph where edges exist only between consecutive integers. This graph is bipartite by construction, so it always requires at most 2 colors. The only case where 1 color is sufficient is when there are no edges at all, meaning no two consecutive integers exist in the input. Therefore, detecting the existence of at least one edge fully determines whether the answer is 1 or 2.

Python Solution

import sys
input = sys.stdin.readline

q = int(input())
for _ in range(q):
    n = int(input())
    a = list(map(int, input().split()))
    
    s = set(a)
    
    ans = 1
    for x in s:
        if x + 1 in s:
            ans = 2
            break
    
    print(ans)

The solution relies on a set to enable constant-time checks for whether a value exists. The loop over the set ensures we only inspect unique values, avoiding unnecessary repeated work even though duplicates are not present in the input by problem statement.

The key implementation detail is early exit: as soon as a single adjacent pair is found, we stop because the answer cannot exceed 2.

Worked Examples

Example 1

Input: [2, 10, 1, 20]

x	x+1 in set?	Decision
2	3 no	continue
10	11 no	continue
1	2 yes	answer = 2, stop

This confirms that a single conflict pair is enough to force two teams.

Example 2

Input: [3, 6]

x	x+1 in set?	Decision
3	4 no	continue
6	7 no	continue

No conflicts exist, so one team is sufficient.

These traces show that the algorithm is effectively detecting whether the induced graph has any edges.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n) per query	Each value is checked once with O(1) set lookup
Space	O(n)	Set stores all values of a query

The constraints cap n at 100, so even across 100 queries the total work is negligible. The solution is comfortably within limits.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import sys
    input = sys.stdin.readline

    q = int(input())
    out = []
    for _ in range(q):
        n = int(input())
        a = list(map(int, input().split()))
        s = set(a)
        ans = 1
        for x in s:
            if x + 1 in s:
                ans = 2
                break
        out.append(str(ans))
    return "\n".join(out)

# provided samples
assert run("""4
4
2 10 1 20
2
3 6
5
2 3 4 99 100
1
42
""") == """2
1
2
1"""

# custom cases
assert run("""1
3
1 3 5
""") == "1"

assert run("""1
2
1 2
""") == "2"

assert run("""1
4
10 11 12 13
""") == "2"

assert run("""1
5
100 1 50 2 99
""") == "2"

Test input	Expected output	What it validates
`1 3 / 1 3 5`	1	no consecutive pairs
`1 2 / 1 2`	2	single conflict forces split
`1 4 / 10 11 12 13`	2	long consecutive chain
`1 5 / 100 1 50 2 99`	2	multiple scattered conflicts

Edge Cases

For inputs with no consecutive integers, such as [1, 4, 7], the algorithm builds a set {1, 4, 7} and finds no x where x+1 exists. The loop finishes without triggering the condition, so the answer remains 1, matching the fact that no restrictions exist.

For a fully consecutive input like [5, 6, 7, 8], the set contains multiple adjacent pairs. The algorithm detects 5-6 immediately (or any other pair depending on iteration order), sets the answer to 2, and stops. This matches the need to alternate assignments along the chain, confirming that a single check is sufficient to capture the entire structure.