CF 102697062 - Delete Characters

CF 102697062 - Delete Characters

Rating: -
Tags: -
Solve time: 1m 16s
Verified: yes

Solution

Problem Understanding

The task is a direct string filtering problem. We have a string S, followed by a collection C containing the characters that must disappear. Every occurrence in S of a character listed in C has to be removed, while every other character must stay in exactly the same relative order. The output is the resulting string after all such deletions. The official problem gives the example howdy, with w and y forbidden, producing hod.

The published problem page gives a 1 second time limit and 256 MB of memory, but it does not expose explicit bounds for the length of S or for N. That means the safe interpretation is to avoid an algorithm whose running time multiplies the string length by the number of characters to delete. A linear scan of the input is the natural target, since every character has to be inspected at least once to determine whether it survives.

The main edge cases come from the fact that the deletion rule applies to every occurrence, not just the first occurrence. For example, with input

banana
1
a

the correct output is

bnn

A careless implementation that removes only one occurrence of a would incorrectly leave either bnana or banan.

Another boundary case is when no characters need to be deleted. With input

abc
0

the correct output is

abc

There are no deletion characters, so the original string must be preserved exactly. An implementation that assumes at least one character follows the integer N can fail here.

The opposite case is when every character belongs to the deletion set. With input

abc
3
a
b
c

the correct output is the empty string. A solution that builds the result correctly but forgets to print the terminating newline can still be accepted by most judges, but a solution that assumes the result is nonempty can fail or accidentally retain a character.

Repeated deletion characters are another possible implementation trap if the input is not trusted to enforce distinctness. For example,

aaa
2
a
a

should still produce an empty string. Membership in the deletion set should be treated as a boolean property, so inserting the same character twice has no additional effect.

Approaches

A straightforward brute-force solution can process every character of S and, for each one, scan the entire deletion list until it finds a match. This is correct because a character survives exactly when it does not occur in C. If |S| = n and N characters are supplied, the worst case performs nN membership comparisons. With both values around 10^5, that is about 10^10 comparisons, far beyond what a 1 second limit can support.

The brute force works because it directly checks the definition of deletion, but the repeated search is unnecessary. The key observation is that membership is the only question we need to ask about C. We do not care where a character occurs in the deletion list or how many times it occurs. We can put all deletion characters into a set, making membership testing expected O(1). Then one left-to-right scan of S is enough.

For each character in S, we check whether it belongs to the deletion set. If it does, we skip it. Otherwise, we append it to the answer. Because characters are processed from left to right and only unwanted characters are skipped, the surviving characters automatically retain their original order.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(	S	N)
Optimal	O(	S	+ N) expected

Algorithm Walkthrough

1. Read the original string S, then read N and the next N characters. Store those characters in a set called deleted. A set gives us direct membership testing instead of repeatedly scanning the deletion list.
2. Create an empty list answer. Using a list is preferable to repeatedly concatenating Python strings because repeated string concatenation can cause unnecessary copying.
3. Scan every character ch of S from left to right. If ch is in deleted, ignore it. Otherwise, append ch to answer. The decision depends only on whether the character is forbidden, so no information about earlier positions is needed.
4. Join the surviving characters and print the resulting string. If every character was deleted, the list is empty and ''.join(answer) correctly produces the empty string.

Why it works

The invariant during the scan is that answer contains exactly the characters from the already processed prefix of S that are not in deleted, in their original order. When the next character belongs to deleted, omitting it is exactly the required operation. When it does not belong to deleted, keeping it is mandatory because only characters in the deletion set may be removed. Thus the invariant remains true after every character, and after the entire string has been processed, answer is exactly the required final string.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    s = input().rstrip('\n')
    n = int(input())
    deleted = {input().strip() for _ in range(n)}

    answer = [ch for ch in s if ch not in deleted]
    print(''.join(answer))

if __name__ == "__main__":
    solve()

The first line reads the string that will be filtered. rstrip('\n') removes only the line ending, so other characters in the string are not accidentally discarded.

The set comprehension reads all N deletion characters and stores them in deleted. The problem describes these as different characters, but using a set also handles accidental repetitions naturally.

The list comprehension implements the scan from the algorithm walkthrough. Every character not present in deleted is retained. Since Python iterates through the string in order, the resulting list already has the correct order.

Finally, join constructs the output string in one operation. There are no index calculations, so there are no off-by-one boundaries to manage. Python integers also have arbitrary precision, although this problem never requires arithmetic large enough for overflow to matter.

Worked Examples

Sample 1

The official sample is howdy with w and y marked for deletion.

Character	Deleted set	Action	Answer so far
h	{w, y}	keep	h
o	{w, y}	keep	ho
w	{w, y}	delete	ho
d	{w, y}	keep	hod
y	{w, y}	delete	hod

The final answer is hod. This demonstrates the central invariant: after each character, the partial answer contains exactly the allowed characters from the processed prefix.

Sample 2

Consider the custom input

banana
1
a

Character	Deleted set	Action	Answer so far
b	{a}	keep	b
a	{a}	delete	b
n	{a}	keep	bn
a	{a}	delete	bn
n	{a}	keep	bnn
a	{a}	delete	bnn

The result is bnn. This trace exercises repeated occurrences of the same deleted character. Every occurrence is checked independently, so none can accidentally survive.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(	S
Space	O(	S

The algorithm performs a constant amount of expected work per input character and does not contain nested loops over the string and deletion list. Since the official limit is only 1 second, this linear approach is the appropriate choice, especially because the published statement does not provide a small explicit bound that would justify quadratic processing.

Test Cases

# helper: run solution on input string, return output string
import sys
import io

def solve():
    s = input().rstrip('\n')
    n = int(input())
    deleted = {input().strip() for _ in range(n)}
    answer = [ch for ch in s if ch not in deleted]
    print(''.join(answer))

def run(inp: str) -> str:
    global input

    old_stdin = sys.stdin
    old_input = input

    sys.stdin = io.StringIO(inp)
    input = sys.stdin.readline

    out = io.StringIO()
    old_stdout = sys.stdout
    sys.stdout = out

    try:
        solve()
    finally:
        sys.stdin = old_stdin
        sys.stdout = old_stdout
        input = old_input

    return out.getvalue()

# Provided sample
assert run("""howdy
2
w
y
""") == "hod\n", "sample 1"

# Minimum-size string, one character deleted
assert run("""a
1
a
""") == "\n", "single character deleted"

# No characters are deleted
assert run("""abc
0
""") == "abc\n", "empty deletion set"

# All characters are deleted
assert run("""abcabc
3
a
b
c
""") == "\n", "all characters deleted"

# Repeated occurrences and alternating survivors
assert run("""banana
1
a
""") == "bnn\n", "repeated deleted characters"

# Boundary case where only the first and last characters survive
assert run("""xabcy
3
a
b
c
""") == "xy\n", "boundary characters"

# Large input to exercise linear behavior
large_s = "ab" * 50000
large_input = large_s + "\n1\nb\n"
assert run(large_input) == ("a" * 50000) + "\n", "large input"

Test input	Expected output	What it validates
a, delete a	empty string	Minimum-size input and empty result
abc, delete nothing	abc	Boundary case N = 0
abcabc, delete a,b,c	empty string	Every character is removed
banana, delete a	bnn	Repeated deleted occurrences
xabcy, delete a,b,c	xy	Characters at both boundaries survive
ab repeated 50,000 times, delete b	50,000 a characters	Large input and linear-time behavior

Edge Cases

For an empty deletion set, the exact input is

abc
0

The set deleted is empty, so every membership test returns false. The scan appends a, b, and c, producing abc. The algorithm does not need a special case for N = 0, which avoids introducing unnecessary boundary logic.

For repeated occurrences of a deleted character, consider

banana
1
a

The set is {a}. The scan removes the second, fourth, and sixth characters while keeping b, n, and n, producing bnn. The decision is made for every position, so the algorithm cannot accidentally stop after the first matching occurrence.

For complete deletion, consider

abc
3
a
b
c

Every membership test succeeds, so answer remains empty throughout the scan. Joining an empty list gives the empty string, which is exactly the required output. No separate empty-output branch is needed.

For characters surviving at the boundaries, consider

xabcy
3
a
b
c

The first character x is retained, the middle characters a, b, and c are removed, and the final y is retained. The result is xy. Since the algorithm processes the first and last positions exactly like every other position, there is no special first-character or last-character off-by-one case.

For a large input, consider a string consisting of ab repeated 50,000 times with only b deleted. Every one of the 100,000 characters is examined once, and every set lookup is expected constant time. The result contains exactly 50,000 a characters, demonstrating why the algorithm scales linearly rather than performing a search through the deletion list for every character.