CF 2094A - Trippi Troppi

Each test case describes a “country name” that is stored in an older format consisting of three separate words.

codeforcescompetitive-programmingstrings

CF 2094A - Trippi Troppi

Rating: 800
Tags: strings
Solve time: 1m 27s
Verified: yes

Solution

Problem Understanding

Each test case describes a “country name” that is stored in an older format consisting of three separate words. The modern representation is formed by compressing those three words into a single abbreviation: we take the first character of the first word, the first character of the second word, and the first character of the third word, then concatenate them in order.

So the task is purely about string parsing. For every line, we split it into exactly three words and extract one character from each.

The constraints are extremely small. There are at most 100 test cases, and each word has length at most 10. This means the total input size is tiny, and any approach that reads and processes each line once is already optimal. Even repeated scanning or rebuilding strings would be fast enough, but the structure of the problem strongly suggests a direct extraction approach.

There are no tricky numerical edge cases, but there is one subtle failure mode that shows up in incorrect implementations: treating the entire line as a single string and taking fixed positions like s[0], s[1], s[2]. That breaks immediately because spaces separate words, and the second and third letters of the full line are often spaces rather than characters. For example, in the input oh my god, the string starts with 'o', then a space, so indexing blindly gives wrong results. The correct behavior always depends on splitting by whitespace first.

Approaches

The brute-force view is to take each line and for every character, try to determine whether it is the first character of a word. One could scan the string and maintain a boolean flag that indicates whether we are at the start of a word (either at index 0 or after a space). Every time the flag is true, we append the character. This works, but it does unnecessary work because we are iterating over every character in every line, even though we only ever need three characters per test case.

The key observation is that the structure is fixed. Each line always contains exactly three words, already separated cleanly by spaces. Once we split the line into tokens, the answer is determined in constant time per test case: just access the first character of each token. This reduces the problem to simple string splitting and indexing.

Approach Time Complexity Space Complexity Verdict
Brute Force Scan Characters O(total characters) O(1) Accepted but unnecessary
Split + Direct Access O(total characters for split) O(1) Accepted

The split-based method is strictly simpler and avoids manual state tracking.

Algorithm Walkthrough

  1. Read the number of test cases t, which determines how many independent lines we will process.
  2. For each test case, read the entire line and split it into three words using whitespace separation. This guarantees we isolate each word cleanly regardless of spacing.
  3. Extract the first character of each of the three words. Since each word is guaranteed to be non-empty and consists of lowercase letters, indexing at position 0 is always valid.
  4. Concatenate these three characters in order and output the result immediately.

Why it works

Each input line is structurally fixed into exactly three tokens, and the modern name is defined purely as a deterministic projection of those tokens. The operation of taking the first character is independent for each word, so no interaction exists between positions. As long as tokenization is correct, the mapping from input line to output string is lossless and unique.

Python Solution

import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    a, b, c = input().split()
    print(a[0] + b[0] + c[0])

The solution relies on Python’s built-in split() which automatically handles arbitrary spacing and ensures we correctly isolate the three words. We then destructure directly into three variables, which makes the intent explicit and avoids indexing errors on a list.

Each word is guaranteed to exist, so unpacking into a, b, c is safe under the problem constraints.

The final concatenation uses simple string addition, which is efficient given the fixed size of 3 characters.

Worked Examples

We trace two inputs to see how the transformation behaves.

Example 1

Input line: binary indexed tree

Step Token 1 Token 2 Token 3 Output
Split binary indexed tree
First letters b i t bit
Final bit

This shows how each word independently contributes exactly one character, with no dependence on position or length.

Example 2

Input line: skibidi slay sigma

Step Token 1 Token 2 Token 3 Output
Split skibidi slay sigma
First letters s s s sss
Final sss

This case demonstrates that repeated initials are handled naturally; no special logic is needed for duplicates or identical letters.

Complexity Analysis

Measure Complexity Explanation
Time O(total input size) Each line is split once and only three characters are accessed
Space O(1) Only a constant number of variables are stored per test case

The input bounds are tiny, so even a character-by-character scan would be safe. The chosen approach is optimal and comfortably within limits.

Test Cases

import sys, io

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

    t = int(input())
    out = []
    for _ in range(t):
        a, b, c = sys.stdin.readline().split()
        out.append(a[0] + b[0] + c[0])
    return "\n".join(out)

# provided sample
assert run("""7
united states america
oh my god
i cant lie
binary indexed tree
believe in yourself
skibidi slay sigma
god bless america
""") == """usa
omg
icl
bit
biy
sss
gba"""

# custom 1: minimal repetition
assert run("""1
a b c
""") == "abc"

# custom 2: identical words
assert run("""1
same same same
""") == "sss"

# custom 3: mixed words
assert run("""1
code forces round
""") == "cfr"

# custom 4: multiple cases
assert run("""2
hello world again
alpha beta gamma
""") == """hwa
abg"""
Test input Expected output What it validates
a b c abc minimal case
same same same sss repeated initials
code forces round cfr normal mixed words
two lines hwa, abg multi-test correctness

Edge Cases

A potential pitfall is assuming fixed character positions in the raw string instead of tokenizing. For example, input oh my god cannot be handled by taking s[0], s[1], s[2], because s[1] is a space. The correct processing splits first:

Input: oh my god

After splitting: ["oh", "my", "god"]

Then:

  • first word gives 'o'
  • second word gives 'm'
  • third word gives 'g'

Result: omg

Another edge case is repeated letters across words. Input skibidi slay sigma produces sss, which confirms we do not need any deduplication or special handling.