To solve this problem, we need to determine the minimum number of flips required to turn all coins heads up, where each flip operation must flip exactly k coins. If it's impossible, we return -1.

Approach

1. Count Tails: First, count the number of tails (cnt) in the initial state. If there are no tails, return 0 immediately.
2. Case when k equals n: If we have to flip all coins each time, then it's only possible to turn all coins heads up if all coins are tails (return 1) or no coins are tails (return 0). Otherwise, return -1.
3. Case when k < n:
   • Parity Check: If the parity (even/odd) of the number of tails (cnt) is different from the parity of k:
     • If k is even, it's impossible to turn all coins heads up (return -1).
     • If k is odd, we need to add 1 to the minimum number of flips (since flipping k coins once more will adjust the parity).
   • Same Parity: If the parity of cnt and k is the same, compute the minimum number of flips as the ceiling of cnt/k and adjust if necessary to ensure the difference between the total flips and cnt is even.

Solution Code

def min_flips(coins, k):
    cnt = coins.count('T')
    n = len(coins)
    if cnt == 0:
        return 0
    if k == n:
        return 1 if cnt == n else -1
    # k < n
    if (cnt % 2) != (k % 2):
        if k % 2 == 0:
            return -1
        else:
            # k is odd, add 1 to minimal m to adjust parity
            m = (cnt + k - 1) // k + 1
            return m
    else:
        # same parity, find minimal m where m*k >= cnt and (m*k - cnt) even
        m = (cnt + k - 1) // k
        while (m * k - cnt) % 2 != 0:
            m += 1
        return m

Explanation

• Count Tails: This gives us the number of coins we need to turn heads up.
• k equals n: Flipping all coins once will either turn all tails to heads (if all are tails) or create a loop (if there are mixed coins).
• k < n:
  • Parity Check: The parity of the total number of flips (m*k) must match the parity of the number of tails (since each tail needs to be flipped an odd number of times and each head even).
  • Adjusting Flips: For same parity, we start with the minimal number of flips (ceiling of cnt/k) and adjust until the difference between total flips and cnt is even, ensuring we can flip the necessary coins to turn all heads up.

This approach efficiently handles all cases and provides the minimal number of flips or indicates impossibility. The complexity is O(n) for counting tails and O(1) for the rest, making it optimal for large inputs.

作者声明:本文包含人工智能生成内容。