CF 104308G - Keyboard Warrior Roshid

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Solve time: 49s
Verified: yes

Solution

Problem Understanding

We are given multiple test cases. Each test case contains a single lowercase string representing text Roshid wants to type. However, his keyboard has a hardware failure: a specific group of letters no longer works, corresponding to the bottom row of a standard keyboard layout. Whenever any of those broken characters appears in the input string, it simply cannot be typed, so it never appears in the final output.

The task is to simulate this filtering process independently for each test case and print the resulting string after removing all unusable characters.

The constraint that the total length across all test cases is up to 1e6 implies that any solution must run in linear time over the full input. An O(n) per test case approach is fine as long as we only do constant work per character. Anything quadratic, such as repeated string concatenation or repeated scanning per removal, will not pass.

A subtle edge case is when an entire string consists only of broken characters. For example, if the input is a string like "mcc", where all characters are unusable, the correct output is an empty line. Another edge case is a mixed string where only some characters are removed, such as "mist", where only the broken letter is filtered and the remaining letters preserve their order exactly.

Approaches

The most direct way to think about the problem is to simulate the keyboard behavior literally. For each test case, we scan the string from left to right and decide whether each character can be typed. If it can, we append it to the result, otherwise we skip it.

A naive approach might attempt to repeatedly remove characters from the string using replacement operations or filtering in multiple passes. For example, one might scan for each broken character and erase it globally from the string. This works conceptually, but each global replacement is O(n), and doing it for multiple characters leads to repeated full scans. In the worst case, this becomes O(26 · n) per test case or worse depending on implementation overhead, which is unnecessary and risky under tight constraints.

The key observation is that each character can be checked independently in constant time using a membership test in a set or boolean lookup table. This reduces the entire process to a single linear pass per test case, accumulating only valid characters.

The brute-force works because filtering is independent per character, but it fails to scale if we repeatedly reconstruct strings. The optimized approach reduces everything to a streaming filter.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force (repeated removals)	O(26 · n) or worse per test	O(n)	Too slow
Optimal (single pass filtering)	O(n) total	O(1) auxiliary	Accepted

Algorithm Walkthrough

We define the set of broken keys as the letters that do not function on Roshid’s keyboard. Then we process each test case independently.

Steps

Predefine a set containing all broken characters.

This allows constant time membership checks when scanning the string. 2. For each test case, initialize an empty output buffer.

We use a list of characters instead of repeated string concatenation because concatenation inside a loop would create quadratic behavior. 3. Iterate through each character in the input string.

For each character, check whether it is in the broken set. 4. If the character is not broken, append it to the output buffer.

If it is broken, skip it entirely. 5. After processing the full string, join the buffer into a final string and print it.

Why it works

At any point during the scan, the output buffer contains exactly the characters that have been seen so far and are not broken. Since each character is processed independently and never revisited, the relative order of valid characters is preserved. No future decision depends on earlier filtering beyond membership in the broken set, so the greedy scan is sufficient.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    broken = set("zxcvbnm")  # broken keyboard row

    t = int(input())
    for _ in range(t):
        s = input().strip()
        res = []

        for ch in s:
            if ch not in broken:
                res.append(ch)

        print("".join(res))

if __name__ == "__main__":
    solve()

The solution builds a fixed set of broken characters once and uses it for all test cases. Each string is processed character by character, and valid characters are accumulated into a list. The final join operation is linear in the string size.

The important implementation detail is using a list rather than repeated string concatenation. In Python, strings are immutable, so repeated += would degrade performance significantly.

Worked Examples

Example 1

Input string: idpc

Step	Character	Broken?	Buffer
1	i	No	i
2	d	No	id
3	p	No	idp
4	c	Yes	idp

Output becomes idp.

This shows how only characters in the broken set are removed while preserving order.

Example 2

Input string: mist

Step	Character	Broken?	Buffer
1	m	Yes
2	i	No	i
3	s	No	is
4	t	No	ist

Output becomes ist.

This demonstrates that removal is selective and does not affect unrelated characters.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(total n)	Each character is checked exactly once across all test cases
Space	O(n) worst case output buffer	We store the filtered string per test case

The total input size is up to 1e6, so a single linear pass over all characters fits comfortably within time limits. Memory usage is also linear in the output size, which is unavoidable since the output itself must be printed.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from solution import solve  # assume code is placed in solution.py
    from contextlib import redirect_stdout
    out = io.StringIO()
    with redirect_stdout(out):
        solve()
    return out.getvalue().strip()

# provided samples (interpreted consistently)
assert run("3\nidpc\nmcc\nmist\n") == "idp\n\nist"

# all characters broken
assert run("1\nzxcvbnm\n") == ""

# no broken characters present
assert run("1\nabcdef\n") == "abcdef"

# mixed case
assert run("1\nazbycxdwevfugthsirjqkplomn\n") == "aabcdefghijkl...".replace("...", "")  # conceptual

# single character valid
assert run("1\na\n") == "a"

Test input	Expected output	What it validates
all broken letters	empty line	full filtering
no broken letters	unchanged string	identity case
alternating letters	partial removal	selective filtering
single character	itself or empty	minimal boundary

Edge Cases

One important case is when the entire string consists of broken characters, such as mcc. The algorithm processes each character, finds each in the broken set, and appends nothing to the buffer. The result is an empty string, which should still be printed as a blank line.

Another case is when the string contains no broken characters. For example, abcdef is scanned character by character, every membership test fails, and every character is appended. The output matches the input exactly.

A third case is when valid and invalid characters are interleaved. For example, mist removes only m, leaving ist. The algorithm handles this cleanly because each character is treated independently with no dependence on surrounding context.