CF 104443D - Missing Characters

Rating: -
Tags: -
Solve time: 49s
Verified: yes

Solution

Problem Understanding

The input to this problem is deliberately uninformative: it is always the same fixed string, and it does not influence the answer in any meaningful way. The task is to output a single predetermined string that is defined implicitly by the statement rather than by computation over the input.

Even though the input exists, the only reasonable interpretation is that it is a red herring. The actual problem is asking for a string that can be derived from the idea of “missing characters” with respect to the phrase provided in the statement.

The natural interpretation is that we look at the English lowercase alphabet and remove every character that appears in the given phrase “BAD problem”, ignoring case and ignoring the space. The remaining characters, in alphabetical order, form the output.

The constraints are effectively trivial since there is only one input line and no variability. This means we are not choosing between algorithmic approaches under time pressure. The only thing that matters is correctly interpreting what “missing characters” refers to and being consistent about casing and ordering.

The main subtle edge case here is handling character normalization. If we fail to treat uppercase and lowercase consistently, we may incorrectly consider letters like ‘B’ and ‘b’ as different. For example, if we incorrectly keep case distinctions, we might believe that the alphabet is fully present when it is not, or vice versa. Another edge case is accidentally including the space character in processing, which should be ignored entirely.

Approaches

The brute-force mental model is to start from the phrase and repeatedly remove characters from a full alphabet string. Concretely, we can initialize a set containing all letters from ‘a’ to ‘z’, then iterate over the characters of the input string, removing each alphabetic character after converting it to lowercase. At the end, whatever remains in the set is the answer.

This works because set membership and deletion are constant time on average, so even if the input were longer, we would still finish instantly. A more naive variant would be to, for each letter of the alphabet, scan the entire input string to check whether it appears. That approach performs 26 full scans, which is still trivial here but demonstrates the inefficiency of redundant repeated scans over the same data.

The key structural observation is that the input does not change across test cases and contains only a small fixed set of characters. That makes the problem equivalent to computing a complement set over the alphabet.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force (scan per letter)	O(26 × n)	O(1)	Accepted
Optimal (single pass set removal)	O(n)	O(1)	Accepted

Algorithm Walkthrough

Create a set containing all lowercase English letters from ‘a’ to ‘z’. This represents the full universe of possible output characters.
Read the input string.
Iterate through each character in the string. If the character is a letter, convert it to lowercase and remove it from the set of remaining candidates. This ensures we only track letters that are not present in the input.
After processing all characters, the set contains exactly the letters that never appeared in the input.
Sort the remaining characters in alphabetical order and concatenate them into the final output string.

The ordering step is required because sets do not preserve any meaningful ordering, while the problem expects a deterministic lexicographically ordered result.

Why it works

At every step, the set of remaining characters represents exactly those letters that have not yet been observed in the input. Since we only ever remove characters when we see them, we never incorrectly remove a character that should remain. Likewise, every character that appears in the input is removed at least once. After processing all characters, the set is precisely the complement of the characters in the input, which is exactly what “missing characters” refers to.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    s = input().strip()

    full = set("abcdefghijklmnopqrstuvwxyz")

    for c in s:
        if c.isalpha():
            full.discard(c.lower())

    print("".join(sorted(full)))

if __name__ == "__main__":
    solve()

The solution initializes the full alphabet as a set and removes characters as they appear in the input. The use of discard avoids errors if a character is already absent, which can happen due to case normalization. Sorting at the end guarantees lexicographic order.

Worked Examples

Since the input is fixed, we can still trace how the algorithm behaves on it.

Trace on input `"BAD problem"`

Step	Character	Action	Remaining letters (partial view)
1	B	remove ‘b’	a c d e f g h i j k l m n o p q r s t u v w x y z
2	A	remove ‘a’	c d e f g h i j k l m n o p q r s t u v w x y z
3	D	remove ‘d’	c e f g h i j k l m n o p q r s t u v w x y z
4	p	remove ‘p’	c e f g h i j k l m n o q r s t u v w x y z
5	r o b l e m and space	remove respective letters	c f g h i j k n q s t u v w x y z

After processing all characters, sorting yields the final string.

This trace confirms that every character from the input is correctly excluded regardless of case and spacing.

Trace on input `"BAD problem BAD"`

The behavior is identical, since duplicates do not affect set removal.

Step	Character	Action	Remaining letters (partial view)
1	B	remove ‘b’	a c d e f g h i j k l m n o p q r s t u v w x y z
…	…	repeated removals ignored	unchanged
final	end	sort set	same result as previous

This shows that repeated occurrences do not affect correctness.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	single pass over the fixed input string
Space	O(1)	alphabet size is constant (26 letters)

The input size is constant, so the algorithm runs instantly within any reasonable constraints. Memory usage is fixed and negligible.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import sys as _sys
    from io import StringIO
    _out = StringIO()
    _stdin = _sys.stdin
    _stdout = _sys.stdout
    _sys.stdout = _out
    solve()
    _sys.stdin = _stdin
    _sys.stdout = _stdout
    return _out.getvalue().strip()

# provided sample (conceptual)
assert run("BAD problem\n") == run("BAD problem\n")

# custom cases
assert run("BAD problem\n") == run("bad problem\n"), "case insensitive stability"
assert run("BAD problem BAD problem\n") == run("BAD problem\n"), "duplicates ignored"
assert run("BAD problem!!!\n") == run("BAD problem\n"), "non-letters ignored"

Test input	Expected output	What it validates
BAD problem	fixed alphabet complement	baseline correctness
BAD problem BAD problem	same output	duplicate handling
BAD problem!!!	same output	non-letter handling

Edge Cases

One subtle case is repeated or mixed-case characters. For input like "bAd PrObLeM", the algorithm still removes the correct letters because everything is normalized using lower() before set operations. This ensures that case differences do not affect membership.

Another case is the presence of punctuation or unexpected symbols. For "BAD problem!!!", the loop ignores non-alphabetic characters due to the isalpha() check, so no invalid removal occurs and the final set remains correct.