Problem B Double Rainbow Time Limit: 1 Second Let 𝑃 be a set of 𝑛 points on the 𝑥-axis and each of the points is colored with one of the colors 1,2, . . . , 𝑘. For each color 𝑖 of the 𝑘 colors, there is at least one point in 𝑃 which is colored with 𝑖. For a set 𝑃 ′ of consecutive points from 𝑃, if both 𝑃 ′ and 𝑃 ∖ 𝑃 ′ contain at least one point of each color, then we say that 𝑃 ′ makes a double rainbow. See the below figure as an example. The set 𝑃 consists of ten points and each of the points is colored by one of the colors 1, 2, 3, and 4. The set 𝑃 ′ of the five consecutive points contained in the rectangle makes a double rainbow. Given a set 𝑃 of points and the number 𝑘 of colors as input, write a program that computes and prints out the minimum size of 𝑃 ′ that makes a double rainbow. Input Your program is to read from standard input. The input starts with a line containing two integers 𝑛 and 𝑘 (1 ≤ 𝑘 ≤ 𝑛 ≤ 10,000), where 𝑛 is the number of the points in 𝑃 and 𝑘 is the number of the colors. Each of the following 𝑛 lines consists of an integer from 1 to 𝑘, inclusively, and the 𝑖-th line corresponds to the color of the 𝑖-th point of 𝑃 from the left. Output Your program is to write to standard output. Print exactly one line. The line should contain the minimum size of 𝑃 ′ that makes a double rainbow. If there is no such 𝑃 ′ , print 0. The following shows sample input and output for two test cases. 具体步骤

首先，我们可以使用两个指针来表示𝑃′的起始和结束位置。初始时，这两个指针都指向𝑃的开头。然后，我们可以使用一个辅助数据结构来记录𝑃′中每种颜色的出现次数。

接下来，我们可以使用一个循环来移动右指针，直到𝑃′中包含了所有颜色。在每次移动右指针之后，我们可以检查当前𝑃′是否满足双彩虹的条件。如果满足条件，则更新最小𝑃′的大小。

接着，我们可以使用另一个循环来移动左指针，并在每次移动之后更新颜色的出现次数。当左指针移动到右指针所在位置时，我们可以检查当前𝑃′是否仍然满足双彩虹的条件。如果满足条件，则继续更新最小𝑃′的大小。

最后，我们可以输出最小𝑃′的大小作为结果。

下面是一个可能的实现（使用Python语言）：

def find_minimum_double_rainbow(n, k, colors):
    color_counts = [0] * k
    left = 0
    min_size = float('inf')

    for right in range(n):
        color_counts[colors[right] - 1] += 1

        while len(set(color_counts)) == k:
            min_size = min(min_size, right - left + 1)
            color_counts[colors[left] - 1] -= 1
            left += 1

    return min_size if min_size != float('inf') else 0

# 读取输入
n, k = map(int, input().split())
colors = [int(input()) for _ in range(n)]

# 寻找最小双彩虹的大小
min_size = find_minimum_double_rainbow(n, k, colors)

# 输出结果
print(min_size)

这个实现的时间复杂度为O(n)，其中n是𝑃的大小。在最坏情况下，我们需要遍历𝑃两次，因此总的时间复杂度为O(2n)，即O(n)。