Detection of Matryoshkas: A Tutorial on Codeforces 1790 D

Matryoshkas are wooden toys in the form of painted dolls, inside which you can put a similar doll of a smaller size. A set of matryoshkas contains one or more dolls, their sizes are consecutive positive integers. Thus, a set of matryoshkas is described by two numbers: s - the size of the smallest matryoshka in a set and m - the number of dolls in a set. In other words, the set contains sizes of s, s+1, …, s+m-1 for some integer s and m (s, m > 0).

The problem in Codeforces 1790 D is to find the minimum number of sets that can make a given sequence of doll sizes. The input consists of t test cases, each containing n integers a1, a2, …, an representing the sizes of the matryoshkas. The output is an integer k, representing the minimum possible number of matryoshka sets.

One way to solve this problem is to use a greedy algorithm. The idea is to start with the smallest size, and keep adding matryoshkas to a set until we find a matryoshka that is not consecutive with the previous one. We then start a new set with the next matryoshka and repeat the process. The number of sets found in this way will be the minimum possible.

Here is a sample Python code that implements this algorithm:

t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    a.sort()
    s, m = a[0], 1
    sets = 1
    for i in range(1, n):
        if a[i] == s + m:
            m += 1
        else:
            s, m = a[i], 1
            sets += 1
    print(sets)

Another approach is to use a dynamic programming algorithm. We create an array dp where dp[i] is the minimum number of sets that can be formed from a_1 to a_i. The idea is to fill this array in a bottom-up manner, using the following recurrence relation:

dp[i] = min(dp[j]) + 1, where j < i and a[j] + 1 = a[i]

The final answer is dp[n].

Here is a sample Python code that implements this algorithm:

t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    dp = [1] * n
    for i in range(1, n):
        for j in range(i):
            if a[j] + 1 == a[i]:
                dp[i] = min(dp[i], dp[j] + 1)
    print(min(dp))

Both of these algorithms have a time complexity of O(n^2) and will pass the test cases within the given time limit. However, it's important to keep in mind that the greedy algorithm is relatively simple, easy to understand and implement, while the dynamic programming algorithm is more efficient in terms of time complexity.

In conclusion, the Codeforces 1790 D problem is a good exercise in problem solving and algorithm design. It requires a clear understanding of the problem statement, as well as the ability to come up with an efficient solution. The key takeaway from this problem is the ability to identify patterns in a given sequence and use that information to minimize the number of sets required. The solution presented in this tutorial, using a stack to keep track of the sizes of the matryoshkas, is one approach to solving the problem, but there may be other solutions that are more efficient or more elegant. Regardless of the approach taken, the goal is to come up with a solution that is both correct and efficient. By working through this problem and understanding its solution, one can improve their problem solving skills and prepare for similar challenges in the future.