CF 102697038 - Unshuffle the Cards

The task is simply to restore a partially complete deck to its canonical sorted order. Each card has a rank and a suit. Ranks are ordered numerically, with ace considered higher than every other rank.

codeforcescompetitive-programming

CF 102697038 - Unshuffle the Cards

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Solve time: 49s
Verified: yes

Solution

Problem Understanding

The task is simply to restore a partially complete deck to its canonical sorted order. Each card has a rank and a suit. Ranks are ordered numerically, with ace considered higher than every other rank. When two cards have the same rank, suits are ordered as clubs, diamonds, hearts, then spades. The input gives the number of cards followed by the shuffled cards, and the output must contain exactly those same cards in the required order. Cards that are missing from the standard 52-card deck are simply absent from the output.

The deck contains at most 52 cards, so the input is tiny. Even an O(n²) sorting algorithm performs at most 52 × 51 / 2 = 1326 pair comparisons, which is trivial under the one-second time limit and 256 MB memory limit. The constraints do not require an advanced data structure or a specialized sorting algorithm. A normal comparison sort is comfortably fast, with O(n log n) time.

The main source of mistakes is the custom ordering. Treating ace as the smallest rank produces the wrong result because ace is explicitly the highest rank. For example,

4
AS 2C KD 3S

must produce

2C 3S KD AS

A naive implementation that maps ace to 1 would put AS before the numbered cards.

Another easy mistake is forgetting that suit only breaks ties between equal ranks. For example,

4
5S 5C 5H 5D

must produce

5C 5D 5H 5S

Sorting only by rank would leave the four cards in an arbitrary input order because all four ranks are equal.

A third boundary case is a deck containing only one rank but not every suit. For example,

2
QS QC

must produce

QC QS

The implementation must not assume that all four suits of a rank are present.

Approaches

A direct brute-force interpretation would be to repeatedly look for an inverted pair of cards and swap it, as in bubble sort. The method is correct because whenever two adjacent cards are in the wrong canonical order, swapping them removes that inversion, and after all inversions disappear the entire sequence is sorted. In the worst case it performs n(n - 1) / 2 comparisons, which is 1326 comparisons for n = 52. That is easily fast enough for this particular problem, so calling this approach too slow would be inaccurate.

A more general brute-force method would be to generate every permutation and check which one is sorted. That approach has n! candidates, and for a complete deck it would involve 52! possibilities, approximately 8.07 × 10^67. That is obviously infeasible, but there is no reason to approach the problem this way because the desired ordering is a standard total ordering.

The useful observation is that every card can be represented by a pair of integers, its rank order and its suit order. Once those two components are defined, the entire problem becomes an ordinary sorting problem. Python's sorting algorithm can compare these keys directly, giving O(n log n) time.

The brute-force swapping approach works because every inversion can eventually be removed, but it performs unnecessary repeated comparisons. The observation that each card has a fixed sortable key lets us replace all of that work with one standard sort.

Approach Time Complexity Space Complexity Verdict
Brute Force, repeated swaps O(n²) O(1) extra Accepted for n ≤ 52
Optimal, key-based sort O(n log n) O(n) Accepted

Algorithm Walkthrough

  1. Read all cards from the input. Each card is represented by a two-character string, where the first character is the rank and the second character is the suit.
  2. Convert each rank into its position in the required rank ordering. The characters 2 through 9 have their natural numeric order, T would normally represent ten but cannot occur in this input, J, Q, K follow, and A receives the largest rank value.
  3. Convert the suit into an integer using the required order C < D < H < S. The exact numeric values do not matter, only their relative order.
  4. Use the pair (rank_order, suit_order) as the sorting key. Python first compares the rank component, and only when two ranks are equal does it compare the suit component. This exactly matches the definition of sorted order.
  5. Print the sorted cards separated by spaces. Since sorting rearranges existing elements rather than creating new cards, missing cards remain missing automatically.

Why it works

For every card, the sorting key contains exactly the two properties that determine its position: rank first and suit second. Consider any two cards. If their ranks differ, the first component of their keys places the lower rank first. If their ranks are equal, the first components are equal and the suit component places clubs before diamonds, diamonds before hearts, and hearts before spades. Thus comparing two cards by their keys is equivalent to comparing them according to the problem's required order. A comparison sort therefore produces exactly the desired deck ordering.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n = int(input())
    cards = input().split()

    rank_order = {
        '2': 2,
        '3': 3,
        '4': 4,
        '5': 5,
        '6': 6,
        '7': 7,
        '8': 8,
        '9': 9,
        'J': 11,
        'Q': 12,
        'K': 13,
        'A': 14,
    }

    suit_order = {
        'C': 0,
        'D': 1,
        'H': 2,
        'S': 3,
    }

    cards.sort(key=lambda card: (
        rank_order[card[0]],
        suit_order[card[1]]
    ))

    print(*cards)

if __name__ == "__main__":
    solve()

The rank_order dictionary explicitly encodes the unusual part of the ordering. In particular, ace receives 14, so it comes after king rather than before the numbered cards.

The suit dictionary starts clubs at zero and increases through spades. Any increasing sequence would work, because only relative order matters.

The sorting key is a two-element tuple. Python compares tuples lexicographically, so the rank is always considered first. The suit is consulted only when the ranks match, which is exactly the required rule.

There is no need to handle 10 separately because the statement guarantees that no ten appears. The remaining ranks fit naturally into the two-character representation used by the input.

The value of n is not needed after reading the input, because split() already gives exactly the supplied cards. It is still read because it is part of the input format.

Worked Examples

The official sample is:

10
8C 7D QH QD QC KD 2H AD 3D 7H

The rank and suit keys evolve as follows.

Card Rank key Suit key Position after sorting
8C 8 0 5
7D 7 1 3
QH 12 2 8
QD 12 1 7
QC 12 0 6
KD 13 1 9
2H 2 2 1
AD 14 1 10
3D 3 1 2
7H 7 2 4

Sorting by the pair (rank key, suit key) gives:

2H 3D 7D 7H 8C QC QD QH KD AD

This example demonstrates both levels of ordering. The two sevens are adjacent because rank is considered before suit, and within the queens the suits appear as clubs, diamonds, hearts.

A second example exercises the suit ordering directly:

4
5S 5C 5H 5D

The trace is:

Card Rank key Suit key
5S 5 3
5C 5 0
5H 5 2
5D 5 1

All cards have the same rank key, so the suit key completely determines their order. The result is:

5C 5D 5H 5S

This confirms that the second component of the tuple is used only for equal ranks, while still handling the case where every input card has the same rank.

Complexity Analysis

Measure Complexity Explanation
Time O(n log n) Python sorts n cards using their two-component keys
Space O(n) The sorting operation requires auxiliary space proportional to the input size

With n ≤ 52, the largest possible input contains only 52 cards. Even the O(n²) brute-force approach would perform only 1326 pair comparisons, while the chosen O(n log n) solution is smaller still. The memory usage is negligible compared with the 256 MB limit.

Test Cases

import sys
import io

def solve():
    input = sys.stdin.readline

    n = int(input())
    cards = input().split()

    rank_order = {
        '2': 2,
        '3': 3,
        '4': 4,
        '5': 5,
        '6': 6,
        '7': 7,
        '8': 8,
        '9': 9,
        'J': 11,
        'Q': 12,
        'K': 13,
        'A': 14,
    }

    suit_order = {
        'C': 0,
        'D': 1,
        'H': 2,
        'S': 3,
    }

    cards.sort(key=lambda card: (
        rank_order[card[0]],
        suit_order[card[1]]
    ))

    print(*cards)

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
assert run(
    "10\n"
    "8C 7D QH QD QC KD 2H AD 3D 7H\n"
) == "2H 3D 7D 7H 8C QC QD QH KD AD\n", "sample 1"

# Minimum-size valid input
assert run(
    "2\n"
    "AS 2C\n"
) == "2C AS\n", "minimum size and ace boundary"

# All four suits of one rank
assert run(
    "4\n"
    "5S 5C 5H 5D\n"
) == "5C 5D 5H 5S\n", "suit ordering"

# Same rank, only some suits present
assert run(
    "2\n"
    "QS QC\n"
) == "QC QS\n", "partial suit set"

# Maximum-size input, complete deck in reverse order
ranks = "23456789JQKA"
suits = "CDHS"
deck = [r + s for r in ranks for s in suits]
reverse_deck = list(reversed(deck))

assert run(
    "52\n" + " ".join(reverse_deck) + "\n"
) == " ".join(deck) + "\n", "maximum size"

print("All tests passed.")
Test input Expected output What it validates
10 / 8C 7D QH QD QC KD 2H AD 3D 7H 2H 3D 7D 7H 8C QC QD QH KD AD Official sample and combined rank/suit ordering
2 / AS 2C 2C AS Minimum size and ace being the highest rank
4 / 5S 5C 5H 5D 5C 5D 5H 5S Complete tie on rank, so suits determine everything
2 / QS QC QC QS Partial set of suits and same-rank ordering
Reversed complete 52-card deck Canonical 52-card order Maximum input size and many simultaneous comparisons

Edge Cases

The ace boundary is handled by assigning ace the largest rank key. For

2
AS 2C

the keys are (14, 3) and (2, 0), so sorting produces

2C AS

A careless implementation that uses ace as rank 1 would reverse the intended relationship.

The suit tie-breaker is handled independently of rank. For

4
5S 5C 5H 5D

all rank keys are 5, so Python compares the suit keys 3, 0, 2, 1. The sorted result is

5C 5D 5H 5S

This catches implementations that sort by rank alone or accidentally use the conventional suit order from a different card representation.

Missing cards require no special handling. With

2
QS QC

the algorithm sees only those two cards and sorts them to

QC QS

It never assumes that the four suits of a rank must exist.

The smallest legal input contains two cards, so there is no need for a special case for an empty deck or a single card. The sorting operation already handles the general case directly. A maximum-size input contains all 52 distinct cards, and the same key function orders it without any additional logic.