CF 1511C - Yet Another Card Deck

Rating: 1100
Tags: brute force, data structures, implementation, trees
Solve time: 2m 13s
Verified: yes

Solution

Problem Understanding

We are given a deck of n cards, each with a color represented by an integer from 1 to 50. The deck is arranged from top to bottom, and we receive q queries. Each query specifies a color, and for each query, we need to locate the first card of that color in the current deck, report its position (1-based index), and then move that card to the top of the deck. The challenge is that the deck dynamically changes with each query, so positions of cards shift over time.

The constraints are significant: both n and q can reach up to 300,000. A naive solution that scans the deck for every query would perform O(n * q) operations, which could reach 9 * 10^10 in the worst case. This is far beyond what fits in a 2-second time limit, so we need a more efficient approach. The limited range of colors (1 to 50) hints that we can use this small universe to manage positions efficiently.

A subtle edge case arises when a color appears multiple times in the deck. Moving the first occurrence to the top changes all subsequent positions. For example, if the deck is [1, 2, 1] and we query 1, we output 1 and move it to the top, leaving [1, 2, 1]. The next query for 1 should now return 1 again because the previously moved card is still on top. Careless implementations that recalculate indices without updating positions dynamically will produce wrong results.

Approaches

The brute-force approach iterates through the deck for each query, searching for the first occurrence of the requested color. Once found, we record its position, remove it from the deck, and insert it at the top. While correct, this approach requires scanning up to n cards for each of the q queries, giving a time complexity of O(n * q). With n and q up to 3 * 10^5, this could perform on the order of 10^10 operations, which is impractical for a 2-second limit.

The key insight is that the number of colors is small. We can maintain a mapping from each color to the first occurrence in the current deck. Instead of shifting cards in an array, we can simulate their relative positions using a list of colors and updating indices. When we process a query for a color, we read its current position from the map, print it, then adjust the positions of all other colors that appear before it in the deck by incrementing their index. Finally, we move the queried color to the top, which is equivalent to setting its position to 1. This approach exploits the bounded color range and avoids full deck scans, reducing the complexity to O(n * c) where c = 50 is the number of colors, which is acceptable.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n * q)	O(n)	Too slow
Optimal	O(q * c)	O(c)	Accepted

Algorithm Walkthrough

Create an array pos of size 51 to track the current position of the first card of each color. Initialize it by scanning the deck once, setting pos[color] to the first index where it appears. This gives the starting position for each color.
Iterate through each query t_j. Use pos[t_j] to find the current position of the first card of that color and print it.
For all colors c in the deck, if pos[c] is smaller than pos[t_j], increment pos[c] by 1. This simulates that these cards have been shifted down by the move-to-top operation.
Set pos[t_j] to 1 because the queried card is now on top.
Repeat steps 2-4 for all queries.

Why it works: The invariant is that pos[c] always represents the current 1-based position of the first card of color c in the deck. Moving a card to the top increments positions of all cards that were above it, preserving correct relative ordering. Since the number of colors is limited, updating pos takes at most 50 operations per query, which is efficient enough.

Python Solution

import sys
input = sys.stdin.readline

n, q = map(int, input().split())
a = list(map(int, input().split()))
t = list(map(int, input().split()))

# initialize position mapping
pos = [0] * 51
for idx, color in enumerate(a):
    if pos[color] == 0:
        pos[color] = idx + 1

result = []

for color in t:
    result.append(pos[color])
    # increment positions of colors above the current one
    for c in range(1, 51):
        if pos[c] < pos[color]:
            pos[c] += 1
    pos[color] = 1

print(*result)

The first loop establishes the initial positions of each color in the deck. The second loop handles each query efficiently using the pos array. By updating all colors above the queried card and moving the queried card to the top, we maintain correct positions without shifting the entire deck. It is important that the pos array uses 1-based indexing to match the problem's output requirements.

Worked Examples

For the sample input:

7 5
2 1 1 4 3 3 1
3 2 1 1 4

Query	pos before	Output	pos after update
3	[2:1,1:2,3:5,4:4]	5	[2:2,1:3,3:1,4:5]
2	[2:2,1:3,3:1,4:5]	2	[2:1,1:4,3:2,4:6]
1	[2:1,1:4,3:2,4:6]	3	[2:2,1:1,3:3,4:7]
1	[2:2,1:1,3:3,4:7]	1	unchanged
4	[2:2,1:1,3:3,4:7]	5	[2:3,1:2,3:4,4:1]

This trace shows the invariant pos[c] is always the first occurrence of color c. Every update reflects the move-to-top operation accurately.

Another small example:

Input:

3 3
1 2 1
1 2 1

Query	pos before	Output	pos after update
1	[1:1,2:2]	1	[1:1,2:3]
2	[1:1,2:3]	3	[1:2,2:1]
1	[1:2,2:1]	2	[1:1,2:2]

The positions adjust correctly even when the same color is queried multiple times.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(q * 50) = O(q)	Each query updates positions of at most 50 colors
Space	O(51) = O(1)	`pos` array stores first occurrence per color

With q up to 3 * 10^5, this gives roughly 15 million operations, which fits comfortably within 2 seconds. Memory usage is minimal, as we only store positions for each color.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    n, q = map(int, input().split())
    a = list(map(int, input().split()))
    t = list(map(int, input().split()))

    pos = [0] * 51
    for idx, color in enumerate(a):
        if pos[color] == 0:
            pos[color] = idx + 1

    result = []
    for color in t:
        result.append(pos[color])
        for c in range(1, 51):
            if pos[c] < pos[color]:
                pos[c] += 1
        pos[color] = 1
    return ' '.join(map(str, result))

# provided sample
assert run("7 5\n2 1 1 4 3 3 1\n3 2 1 1 4\n") == "5 2 3 1 5", "sample 1"

# minimum input
assert run("2 1\n1 2\n1\n") == "1", "minimum size"

# all equal
assert run("4 4\n1 1 1 1\n1 1 1 1\n") == "1 1 1 1", "all equal colors"

# alternating colors
assert run("5 5\n1 2 1 2 1\n2 1 2 1 2\n") == "2 1 3 1 2", "alternating pattern"

# maximum colors
assert run