CF 2080C - Card Flip

CF 2080C - Card Flip

Rating: 2300
Tags: *special
Solve time: 54s
Verified: yes

Solution

Problem Understanding

In this problem, we are given a sequence of n cards laid out in a row. Each card has a face showing either 0 or 1. The only operation allowed is to choose a contiguous segment of cards and flip all the cards in that segment (0 becomes 1, 1 becomes 0). Our goal is to transform the original sequence into a sequence where all cards show 1. We need to determine the minimum number of operations required to achieve this.

The input consists of an integer n followed by a string of length n representing the card faces. The output is a single integer, the minimal number of flips.

Given the constraint that n can be up to 10^5 and the time limit is standard (usually 1-2 seconds), we can expect to perform roughly 10^8 operations in the worst case. This rules out any algorithm that tries all possible contiguous segments explicitly, because the number of segments grows quadratically as n(n+1)/2.

The non-obvious edge cases include sequences where there are alternating 0s and 1s, sequences where all cards are already 1, and sequences where all cards are 0. For example, for 1010, a naive approach that flips one 0 at a time might require four operations, but there exists a sequence of flips that uses only two operations: flipping segments 1-1 and 3-3. If all cards are already 1, the answer is 0. If all cards are 0, one contiguous flip of the entire array suffices, giving 1 operation. Careless implementations often overcount operations by flipping each 0 individually.

Approaches

The brute-force approach is simple: iterate over every possible contiguous segment and try flipping it, recursively checking the resulting sequences until we reach all ones. This works because flipping any segment is allowed and eventually will transform the sequence to all ones. However, the number of contiguous segments is n(n+1)/2, and checking all possible combinations is exponential in n, which is completely infeasible for n=10^5.

The key observation is that we do not need to track individual flips explicitly. Each flip only matters in terms of how it changes 0s to 1s. We can scan the sequence from left to right and count the number of contiguous segments of 0s. Each such segment requires at least one flip, and flipping it in one operation is optimal because flipping a larger segment that includes 1s unnecessarily increases complexity without reducing the number of operations. The minimal number of flips is therefore equal to the number of contiguous 0-segments.

The brute-force works because it will eventually reach the all-ones sequence, but fails due to exponential complexity. The observation that contiguous 0s can be flipped together reduces the problem to a single pass over the array, yielding an O(n) solution.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(2^n)	O(n)	Too slow
Optimal	O(n)	O(1)	Accepted

Algorithm Walkthrough

1. Initialize a counter flips to 0. This will store the number of required operations.
2. Initialize an index i to 0 to start scanning from the leftmost card.
3. While i is less than n, check the current card:

a. If it is 1, increment i and continue, because no flip is needed.

b. If it is 0, increment flips by 1, then skip all contiguous 0s by moving i forward until a 1 or the end of the sequence is reached. This ensures each segment of 0s is counted exactly once. 4. Once the end of the sequence is reached, output flips.

Why it works: The invariant is that every contiguous segment of 0s is counted exactly once. Flipping that segment turns all its 0s to 1s in one operation, and flipping beyond the segment would unnecessarily affect already-correct 1s. By scanning left to right and counting segments, we guarantee the minimal number of flips.

Python Solution

import sys
input = sys.stdin.readline

def min_flips_to_all_ones(cards: str) -> int:
    n = len(cards)
    flips = 0
    i = 0
    while i < n:
        if cards[i] == '1':
            i += 1
        else:
            flips += 1
            while i < n and cards[i] == '0':
                i += 1
    return flips

def main():
    n = int(input())
    cards = input().strip()
    print(min_flips_to_all_ones(cards))

if __name__ == "__main__":
    main()

The function min_flips_to_all_ones performs a single linear scan of the sequence. The inner while-loop ensures contiguous 0s are treated as a single segment. Using strip() on input prevents hidden newline characters from interfering with indexing.

Worked Examples

Example 1: cards = "1010"

i	cards[i]	flips	Action
0	1	0	skip
1	0	1	start segment, skip 0s
2	1	1	skip
3	0	2	start segment, skip 0s
end	-	2	finished

We counted 2 segments of 0s. Flipping each segment once results in 1111.

Example 2: cards = "111"

i	cards[i]	flips	Action
0	1	0	skip
1	1	0	skip
2	1	0	skip
end	-	0	finished

No flips needed, as expected.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Single linear scan over the sequence
Space	O(1)	Only a few integer variables needed

Given n <= 10^5, this scan will complete comfortably within standard time limits.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from __main__ import main
    sys.stdout = io.StringIO()
    main()
    return sys.stdout.getvalue().strip()

# Provided samples
assert run("4\n1010\n") == "2", "sample 1"
assert run("3\n111\n") == "0", "sample 2"

# Custom cases
assert run("5\n00000\n") == "1", "all zeros"
assert run("6\n010101\n") == "3", "alternating zeros"
assert run("1\n0\n") == "1", "single zero"
assert run("1\n1\n") == "0", "single one"
assert run("10\n1111100000\n") == "1", "zeros at end"

Test input	Expected output	What it validates
00000	1	Full segment of zeros
010101	3	Alternating zeros
0	1	Single zero
1	0	Single one
1111100000	1	Contiguous zeros at end

Edge Cases

For the alternating pattern 010101, the algorithm correctly counts each isolated zero as a separate segment. Starting at index 0, it skips the 1, counts a flip for 0 at index 1, skips to index 2 (1), counts flip at index 3 (0), and flip at index 5 (0), yielding exactly 3 flips. This matches the minimal possible operations.

For the all-ones sequence, the algorithm skips every card and never increments flips, producing 0, which is correct. For a single-card sequence, it handles both 0 and 1 correctly, flipping only when necessary.