CF 102697135 - Anti-sort

We are given a collection of lowercase words and must print the same words in reverse alphabetical order. In other words, the word that would appear last in ordinary dictionary order must be printed first, and the word that would appear first must be printed last.

codeforcescompetitive-programming

CF 102697135 - Anti-sort

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Solve time: 2m 23s
Verified: yes

Solution

Problem Understanding

We are given a collection of lowercase words and must print the same words in reverse alphabetical order. In other words, the word that would appear last in ordinary dictionary order must be printed first, and the word that would appear first must be printed last.

The input contains a positive integer n, followed by n words. Every word contains only lowercase English letters. The output contains exactly those same n words, one per line, but arranged in descending lexicographical order. Equal words may appear next to each other in any order because they are indistinguishable.

The published statement does not give an explicit upper bound for n or for the word lengths, but the time limit is 1 second and the memory limit is 256 MB. That makes an obviously quadratic sorting procedure undesirable for large inputs. A standard comparison sort is the natural choice because it reduces the number of word comparisons to O(n log n). Python's built-in sort is implemented in optimized native code, so it is also preferable to implementing a sorting algorithm manually.

There are a few small cases where an implementation can silently go wrong. With a single word, for example,

1
apple

the correct output is

apple

A solution that assumes there must be at least two words could accidentally access a nonexistent second element.

Duplicate words must also be preserved. For

3
cat
cat
apple

the output is

cat
cat
apple

A solution that removes duplicates by converting the input to a set would incorrectly print only two words.

The direction of the sort is another common mistake. For

3
apple
banana
cat

the correct output is

cat
banana
apple

Using ordinary ascending sorting without reversing it would produce the exact opposite order.

Lexicographical comparison is performed on the complete words, not by length. For example,

3
aa
b
ab

must become

b
ab
aa

because 'b' is greater than 'a', while ab is greater than aa. Sorting by length would produce a different and incorrect result.

Approaches

The most direct brute-force approach is to repeatedly find the largest remaining word, print it, and remove it from consideration. On the first iteration we inspect all n words, on the second we inspect n - 1, and so on. The number of comparisons is exactly

n + (n - 1) + ... + 2 + 1 = n(n - 1) / 2.

The procedure is correct because every iteration explicitly selects the largest word that has not yet been printed. However, its quadratic number of comparisons becomes expensive when n is large. For n = 100000, the exact comparison count is 4,999,950,000, which is far beyond what a 1 second solution should attempt. Repeatedly removing elements from the middle of a Python list can make the situation even worse because removal itself can require linear time.

The key observation is that the problem has no special structure beyond ordinary lexicographical ordering. We do not need to invent a custom greedy rule or exploit the letters individually. The desired result is simply the reverse of the normal sorted order. A comparison sort can arrange all words in O(n log n) comparisons, after which requesting descending order gives exactly the required sequence.

The brute-force method works because it discovers the next maximum directly, but it rediscovers information that a sorting algorithm can organize more efficiently. The observation that the entire task is just descending lexicographical sorting lets us replace the repeated maximum search with one standard sort.

Approach Time Complexity Space Complexity Verdict
Brute Force O(n²) word comparisons O(n) Too slow for large n
Optimal O(n log n) comparisons O(n) auxiliary/reference space Accepted

If word comparisons themselves are considered, their cost depends on the length of the common prefix. With maximum word length L, the practical upper bound for comparison sorting is O(n log n * L). Python stores the input words themselves, so the memory required for the strings is proportional to their total input size.

Algorithm Walkthrough

  1. Read n and then read the n words into a list. We keep every occurrence because the output must contain exactly the input words, including duplicates.
  2. Sort the list in descending lexicographical order. Python's sorted(..., reverse=True) performs an ordinary lexicographical sort and reverses the requested ordering, which is exactly the required dictionary order from largest to smallest.
  3. Print the sorted words, one per line. Joining them with newline characters avoids repeated output calls and produces the required format directly.

Why it works

After the sorting step, the list is ordered so that for every adjacent pair a[i] and a[i + 1], a[i] is lexicographically greater than or equal to a[i + 1]. Thus the first element is the largest word, the second is the largest word among those remaining, and so on. Since sorting does not remove or duplicate elements, the final list contains exactly the original words in the required reverse alphabetical order.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n = int(input())
    words = [input().strip() for _ in range(n)]

    words.sort(reverse=True)

    sys.stdout.write("\n".join(words))

if __name__ == "__main__":
    solve()

The first line reads the number of words. The value of n is needed only to know how many subsequent lines belong to the input.

The list comprehension reads exactly those n words. Calling strip() removes the newline inserted by readline() without changing the lowercase letters that form the word.

words.sort(reverse=True) is the central operation. Python compares strings lexicographically, character by character, which matches alphabetical ordering for lowercase English words. reverse=True changes the direction from ascending to descending.

The final join produces one newline-separated string. There is no need to add a special newline after the final word because the output format accepts the final line without one.

No integer arithmetic involving the size of the input is performed, so integer overflow is irrelevant in Python. The main implementation detail to avoid is accidentally using sorted(words) without reverse=True, which would produce ascending rather than descending order.

Worked Examples

For the first sample, the input contains two words.

Step Words before sorting Action Result
1 ["coderams", "club"] Read both words ["coderams", "club"]
2 ["coderams", "club"] Descending lexicographical sort ["coderams", "club"]
3 ["coderams", "club"] Print in order coderams, club

coderams comes after club in ascending lexicographical order because both begin with c, but the second characters are o and l, with o > l. Reversing the normal order consequently puts coderams first.

For the second sample, the words are arranged as follows.

Step Current state Action
1 hello, world, we, hope, you, enjoy, the, contest, today, good, luck Read all words
2 hello, world, we, hope, you, enjoy, the, contest, today, good, luck Sort descending
3 you, world, we, today, the, luck, hope, hello, good, enjoy, contest Print

The resulting order illustrates why complete lexicographical comparison matters. you comes before world because y > w, while world comes before we because their common prefix is w and then o > e. The same comparison rule is applied consistently to every pair.

Complexity Analysis

Measure Complexity Explanation
Time O(n log n · L) Sorting performs O(n log n) comparisons, with each comparison taking up to O(L) for words of maximum length L
Space O(n · L) The input list stores all words, while Python's sort also uses auxiliary memory

The intended algorithm replaces the quadratic repeated maximum search with a standard O(n log n) comparison sort. With a large number of words, this difference is substantial, and the implementation uses Python's optimized built-in sorting routine to stay comfortably within the stated 1 second time limit for the intended input size. The memory usage is dominated by storing the input words and is linear in the input size.

Test Cases

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

def solve():
    input = sys.stdin.readline
    n = int(input())
    words = [input().strip() for _ in range(n)]
    words.sort(reverse=True)
    sys.stdout.write("\n".join(words))

def run(inp: str) -> str:
    old_stdin = sys.stdin
    old_stdout = sys.stdout

    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()

    try:
        solve()
        return sys.stdout.getvalue()
    finally:
        sys.stdin = old_stdin
        sys.stdout = old_stdout

# Provided sample 1
assert run("""2
coderams
club
""") == """coderams
club""", "sample 1"

# Provided sample 2
assert run("""11
hello
world
we
hope
you
enjoy
the
contest
today
good
luck
""") == """you
world
we
today
the
luck
hope
hello
good
enjoy
contest""", "sample 2"

# Minimum-size input
assert run("""1
apple
""") == """apple""", "single word"

# All words equal
assert run("""4
same
same
same
same
""") == """same
same
same
same""", "duplicates"

# Lexicographical order differs from length order
assert run("""5
aa
b
ab
a
ba
""") == """ba
b
ab
aa
a""", "lexicographical ordering"

# Large input
large_words = [f"{i:06d}" for i in range(100000)]
large_input = "100000\n" + "\n".join(large_words) + "\n"
large_expected = "\n".join(reversed(large_words))
assert run(large_input) == large_expected, "large input"
Test input Expected output What it validates
1 / apple apple Minimum-size input and single-element handling
Four copies of same Four copies of same Duplicate preservation
aa, b, ab, a, ba ba, b, ab, aa, a True lexicographical ordering rather than length sorting
100000 numeric strings Reverse lexicographical sequence Large-input performance and sorting direction

Edge Cases

The single-word case is handled without any special branch. For the input

1
apple

the list initially contains only apple. Sorting a one-element list leaves it unchanged, so the algorithm prints apple. A careless implementation that assumes two or more elements could fail here, but the built-in sort naturally handles the boundary.

Duplicate values remain present because the algorithm sorts the original list instead of constructing a set. For

3
cat
cat
apple

the sorted list is cat, cat, apple. Both occurrences of cat participate in the sort and both are printed. The invariant that the output contains exactly the original elements is preserved.

The direction of ordering is handled by reverse=True. For

3
apple
banana
cat

ascending sorting would produce apple, banana, cat, but descending sorting produces cat, banana, apple, which is the required output. The implementation specifies the direction directly rather than sorting and manually reversing an index range, avoiding an unnecessary source of off-by-one errors.

Words of different lengths are compared lexicographically rather than by their lengths. For

3
aa
a
ab

the algorithm first compares the common prefix a. Since a ends after that prefix, it is smaller than both aa and ab. Between aa and ab, the second characters determine the order, so the final result is

ab
aa
a

This follows the same comparison rule used by Python strings and matches alphabetical ordering.

The large-input case does not require a different algorithm. With 100000 words, repeatedly finding the maximum would require

100000 * 99999 / 2 = 4,999,950,000

pairwise comparisons in the worst case. The sorting solution instead performs on the order of 100000 log₂(100000), roughly 1.7 million comparison positions before accounting for the cost of comparing the characters inside each word. That difference is why using a standard O(n log n) sort is the essential part of the solution.