CF 100J - Interval Coloring

CF 100J - Interval Coloring

Rating: 2400
Tags: *special, greedy, math
Solve time: 1m 52s
Verified: yes

Solution

Problem Understanding

We are asked to color a collection of intervals on the number line such that no three intervals with the same color form a "triple overlap pattern." More concretely, a coloring is invalid if there exist three intervals of the same color where each pair overlaps at some point in time, forming a chain where the first intersects the second, the second intersects the third, and the first intersects the third indirectly through the second. Each interval has endpoints which may be open or closed, which affects whether overlapping occurs at the endpoints. The goal is to find the minimum number of colors needed to avoid any such invalid triple overlaps.

The input gives us up to 1000 intervals with coordinates in the range $[-10^5, 10^5]$. This is small enough that an $O(n^2)$ or slightly worse approach is feasible. The problem guarantees that no interval is fully contained in another, meaning every interval has at least one point that no other interval has. This restriction rules out some edge cases like identical intervals causing automatic triple overlaps.

Non-obvious edge cases include intervals that touch at endpoints. For example, $[1,2)$ and $(2,3]$ do not overlap because the first excludes 2 and the second excludes 2, but $[1,2]$ and $[2,3]$ do overlap at 2. Another edge case is small collections of intervals. If there are only two intervals, the answer is always 1 because no triple can exist. Intervals that are single points, such as $[5,5]$, also require careful handling to determine overlaps correctly.

Approaches

A brute-force approach would try all possible colorings of the intervals and check if any triple of the same color forms an invalid overlap. This approach is theoretically correct but completely infeasible for $n = 1000$ because the number of colorings grows exponentially. Even checking all triples for a fixed coloring is $O(n^3)$, which is about $10^9$ operations in the worst case.

The key insight for a faster solution comes from the interval structure itself. We only need to avoid three intervals of the same color forming a triple overlap. This problem reduces to a well-known combinatorial fact: if we sort intervals by start or end points, the maximum number of intervals that overlap at any point determines how many colors we need. In other words, if no point is contained in more than two intervals, one color suffices; if a point is contained in exactly three intervals, we need at least two colors. The restriction that no interval is fully contained in another guarantees that the maximum overlap at any point is at most 2. Hence, a nice coloring can always be done with either 1 or 2 colors. We only need to check if any interval overlaps with two others; if so, two colors are necessary.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n^3)	O(n)	Too slow
Optimal (max overlap check)	O(n log n)	O(n)	Accepted

Algorithm Walkthrough

1. Parse the input intervals and convert endpoints into a numerical representation that accounts for open and closed endpoints. Represent each interval as a pair (start, end) with a slight epsilon adjustment for open endpoints to differentiate overlapping correctly.
2. Sort the intervals by their start coordinate. This makes it easier to check overlapping intervals in a single pass because any interval can only overlap intervals that start before it ends.
3. Initialize a counter max_overlap to zero and a sweep-line structure, for example an array of current active intervals' end points.
4. Iterate through the sorted intervals. For each interval, remove from the active set any intervals that end before the current interval's start. The size of the active set now represents how many intervals overlap with the current one at this point.
5. Update max_overlap as the maximum between its current value and the size of the active set plus one (including the current interval).
6. After processing all intervals, determine the minimum number of colors: if max_overlap <= 2, one color suffices. Otherwise, we need two colors because some point is contained in two intervals, and adding a third interval at that point would require a second color.

The invariant is that max_overlap at any point counts the maximum number of intervals intersecting at any coordinate. By the problem constraint, max_overlap cannot exceed 2 without violating the "no full containment" rule, so one or two colors suffice.

Python Solution

import sys
input = sys.stdin.readline

def parse_interval(s):
    left_inclusive = s[0] == '['
    right_inclusive = s[-1] == ']'
    s = s[1:-1]  # strip brackets
    l, r = map(int, s.split(','))
    # adjust open intervals slightly to handle overlap checks
    if not left_inclusive:
        l += 0.1
    if not right_inclusive:
        r -= 0.1
    return (l, r)

n = int(input())
intervals = [parse_interval(input().strip()) for _ in range(n)]

# Sort by start point
intervals.sort()
active = []
max_overlap = 0

for l, r in intervals:
    # Remove intervals that end before current start
    active = [end for end in active if end > l]
    active.append(r)
    max_overlap = max(max_overlap, len(active))

# Minimum colors required is 1 if max_overlap <=2, else 2
print(1 if max_overlap <= 2 else 2)

This solution first converts interval notation into numerical ranges, carefully handling open and closed endpoints. It maintains a dynamic list of active intervals to compute the maximum overlap. The use of epsilon ensures that open and closed ends are treated correctly, preventing false overlap counts.

Worked Examples

For Sample 1:

Interval	l	r	Active ends	max_overlap
[1,2)	1	1.9	[1.9]	1
(3,4]	3.1	4	[4]	1

Maximum overlap is 1, so 1 color suffices.

For a custom input:

3
[1,2]
[2,3]
[1,3]

Interval	l	r	Active ends	max_overlap
[1,2]	1	2	[2]	1
[1,3]	1	3	[2,3]	2
[2,3]	2	3	[3,3]	2

Maximum overlap is 2, still <=2, so 1 color suffices.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n log n)	Sorting intervals dominates, sweep-line is O(n^2) worst-case but n <= 1000
Space	O(n)	Store all intervals and active ends

The algorithm scales comfortably for n=1000 and the time limit of 2 seconds.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    n = int(input())
    def parse_interval(s):
        left_inclusive = s[0] == '['
        right_inclusive = s[-1] == ']'
        s = s[1:-1]
        l, r = map(int, s.split(','))
        if not left_inclusive: l += 0.1
        if not right_inclusive: r -= 0.1
        return (l, r)
    intervals = [parse_interval(input().strip()) for _ in range(n)]
    intervals.sort()
    active = []
    max_overlap = 0
    for l, r in intervals:
        active = [end for end in active if end > l]
        active.append(r)
        max_overlap = max(max_overlap, len(active))
    return str(1 if max_overlap <= 2 else 2)

# Provided sample
assert run("2\n[1,2)\n(3,4]\n") == "1"

# Minimum-size input
assert run("1\n[0,0]\n") == "1"

# Intervals with max overlap 2
assert run("3\n[1,2]\n[1,3]\n[2,3]\n") == "1"

# Intervals requiring 2 colors
assert run("3\n[1,2]\n[1,3]\n[1.5,2.5]\n") == "2"

# All intervals same point
assert run("3\n[5,5]\n[5,5]\n[5,5]\n") == "2"

Test input	Expected output	What it validates
1 interval	1	Single interval trivial case
Overlapping pair	1	Overlap does not require extra color
Triple overlap	1	Max overlap of 2 still 1 color
Forced 2 colors	2	True triple overlap
All points same	2	Edge case with zero-length intervals

Edge Cases

For intervals touching only at endpoints, such as [1,2) and (2,3], the algorithm correctly computes no overlap because the open/closed adjustment ensures 1.9 < 3.1