CF 437A - The Child and Homework

Rating: 1300
Tags: implementation
Solve time: 1m 7s
Verified: yes

Solution

Problem Understanding

The task is to simulate how a child chooses an answer on a multiple-choice question with four options, labeled A through D. Each option has a textual description. The child considers one choice “great” if its description is either at least twice shorter than every other choice or at least twice longer than every other choice. If exactly one choice is great, the child selects it. Otherwise, the child defaults to C.

The input consists of four lines. Each line starts with a label like "A." or "B." followed by the description. The description is the part we measure for length; the label itself is not counted. The output is a single letter representing the child’s choice.

The maximum length of a description is 100 characters. With only four options, there is no risk of time complexity issues. Even brute force comparisons between all pairs are negligible. The challenge is correctly identifying the “great” choice according to the rules.

An edge case arises when multiple options satisfy the “great” condition. For example, if one option is twice shorter than all others and another is twice longer than all others, then there is no unique great choice, and the child should choose C. Another subtle edge case is when descriptions are very close in length but not strictly twice as short or twice as long - the comparison must use integer division or exact floating-point comparison correctly.

Approaches

A brute-force approach would compare each choice against all others. For each option, compute its length, then check whether it is at least twice shorter than all others, or at least twice longer than all others. With four options, this requires six comparisons per option, which is trivial. This approach is correct and fast enough because the input size is tiny.

The key insight for clarity is that we only need to compute the lengths once and then check each option individually against the others. Since there are only four choices, there is no need for sorting or advanced data structures. The problem is primarily about careful implementation: extracting lengths correctly, applying the “twice as short or long” condition, and handling the tie-breaking rule.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(1)	O(1)	Accepted
Optimal	O(1)	O(1)	Accepted

In this problem, brute force and optimal are essentially identical because of the constant small input size.

Algorithm Walkthrough

Read the four lines of input. For each line, strip the first two characters (the label and dot) and compute the length of the remaining string. Store these lengths in an array corresponding to choices A through D.
Initialize an empty list to track which choices are “great.”
For each choice, compare its length against the other three. A choice is great if its length is at least twice as large as each of the other three lengths or at most half of each of the other three lengths. Use strict comparison with multiplication by 2 to avoid floating-point errors.
If exactly one choice is marked as great, print its label. Otherwise, print "C".

Why it works: By iterating over each choice and checking it against the others with precise arithmetic, we guarantee that we detect all and only the “great” choices. Because the input size is fixed at four, no choice can be overlooked, and the tie-breaking rule is applied correctly.

Python Solution

import sys
input = sys.stdin.readline

choices = ['A', 'B', 'C', 'D']
lengths = []

for _ in range(4):
    line = input().strip()
    # ignore the first two characters (label and dot)
    desc_len = len(line[2:])
    lengths.append(desc_len)

great = []

for i in range(4):
    smaller = all(lengths[i] * 2 <= lengths[j] for j in range(4) if j != i)
    larger = all(lengths[i] >= 2 * lengths[j] for j in range(4) if j != i)
    if smaller or larger:
        great.append(choices[i])

if len(great) == 1:
    print(great[0])
else:
    print('C')

We first map choices to lengths, ensuring the label is excluded. Then we check each choice using all() across the other three, making the comparison precise and avoiding accidental integer division issues. The final selection logic is simple: if exactly one is great, pick it; otherwise default to C.

Worked Examples

Sample 1

Input lengths: A=39, B=35, C=37, D=15

Choice	Length	Smaller than twice others?	Larger than twice others?	Great?
A	39	False	False	False
B	35	False	False	False
C	37	False	False	False
D	15	True	False	True

The table confirms D is the unique great choice. Output: D

Sample 2

Input lengths: A=3, B=5, C=4, D=6

Choice	Length	Smaller than twice others?	Larger than twice others?	Great?
A	3	False	False	False
B	5	False	False	False
C	4	False	False	False
D	6	False	False	False

No unique great choice. Output: C

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Four options and constant comparisons
Space	O(1)	Store four lengths and a small list of great choices

The solution is extremely lightweight, well within the 1-second time limit and 256 MB memory limit.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    choices = ['A', 'B', 'C', 'D']
    lengths = []
    for _ in range(4):
        line = input().strip()
        lengths.append(len(line[2:]))
    great = []
    for i in range(4):
        smaller = all(lengths[i] * 2 <= lengths[j] for j in range(4) if j != i)
        larger = all(lengths[i] >= 2 * lengths[j] for j in range(4) if j != i)
        if smaller or larger:
            great.append(choices[i])
    if len(great) == 1:
        return great[0]
    return 'C'

# provided samples
assert run("A.VFleaKing_is_the_author_of_this_problem\nB.Picks_is_the_author_of_this_problem\nC.Picking_is_the_author_of_this_problem\nD.Ftiasch_is_cute\n") == "D"
assert run("A.Ab\nB.Abc\nC.Abcd\nD.Abcde\n") == "C"

# custom cases
assert run("A.A\nB.AAA\nC.AAA\nD.AAA\n") == "A"  # A is twice shorter than all
assert run("A.AAA\nB.A\nC.A\nD.A\n") == "A"       # A is twice longer than all
assert run("A.AAA\nB.AA\nC.AA\nD.AA\n") == "C"    # No unique great
assert run("A.A\nB.A\nC.A\nD.A\n") == "C"        # All equal

Test input	Expected output	What it validates
A.A\nB.AAA\nC.AAA\nD.AAA	A	Unique shortest choice
A.AAA\nB.A\nC.A\nD.A	A	Unique longest choice
A.AAA\nB.AA\nC.AA\nD.AA	C	Multiple near-great choices, should default
A.A\nB.A\nC.A\nD.A	C	All equal lengths, no great choice

Edge Cases

When all choices have equal length, the algorithm evaluates both the smaller and larger checks as false for every choice. No great choices are found, so it prints C, exactly as required. For example, with A=1, B=1, C=1, D=1, the output is C.

If one option is exactly twice as short as others, like lengths [2,5,5,5], the smaller condition is satisfied for A because 2*2 <= 5 for all others, marking A as great. The remaining checks fail, so A is correctly chosen.

If two options both satisfy the great condition, like lengths [2,4,1,2], the list of great choices has length greater than one. The algorithm then prints C, matching the tie-breaking rule.