K and Medians (Codeforces 1468H)
Tags: Greedy, constructive algorithms

Let's denote the median of a sequence $$$s$$$ with odd length as the value in the middle of $$$s$$$ if we sort $$$s$$$ in non-decreasing order. For example, let $$$s = [1, 2, 5, 7, 2, 3, 12]$$$. After sorting, we get sequence $$$[1, 2, 2, \underline{3}, 5, 7, 12]$$$, and the median is equal to $$$3$$$.You have a sequence of $$$n$$$ integers $$$[1, 2, \dots, n]$$$ and an odd integer $$$k$$$.In one step, you choose any $$$k$$$ elements from the sequence and erase all chosen elements except their median. These elements do not have to go continuously (gaps are allowed between them).For example, if you have a sequence $$$[1, 2, 3, 4, 5, 6, 7]$$$ (i.e. $$$n=7$$$) and $$$k = 3$$$, then the following options for the first step are possible:  choose $$$[1, \underline{2}, 3]$$$; $$$2$$$ is their median, so it is not erased, and the resulting sequence is $$$[2, 4, 5, 6, 7]$$$;  choose $$$[2, \underline{4}, 6]$$$; $$$4$$$ is their median, so it is not erased, and the resulting sequence is $$$[1, 3, 4, 5, 7]$$$;  choose $$$[1, \underline{6}, 7]$$$; $$$6$$$ is their median, so it is not erased, and the resulting sequence is $$$[2, 3, 4, 5, 6]$$$;  and several others. You can do zero or more steps. Can you get a sequence $$$b_1$$$, $$$b_2$$$, ..., $$$b_m$$$ after several steps?You'll be given $$$t$$$ test cases. Solve each test case independently.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$m$$$ ($$$3 \le n \le 2 \cdot 10^5$$$; $$$3 \le k \le n$$$; $$$k$$$ is odd; $$$1 \le m < n$$$) — the length of the sequence you have, the number of elements you choose in each step and the length of the sequence you'd like to get.The second line of each test case contains $$$m$$$ integers $$$b_1, b_2, \dots, b_m$$$ ($$$1 \le b_1 < b_2 < \dots < b_m \le n$$$) — the sequence you'd like to get, given in the ascending order.It's guaranteed that the total sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print YES if you can obtain the sequence $$$b$$$ or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).

Note: In the first test case, you have sequence $$$[1, 2, 3]$$$. Since $$$k = 3$$$ you have only one way to choose $$$k$$$ elements — it's to choose all elements $$$[1, \underline{2}, 3]$$$ with median $$$2$$$. That's why after erasing all chosen elements except its median you'll get sequence $$$[2]$$$. In other words, there is no way to get sequence $$$b = [1]$$$ as the result. In the second test case, you have sequence $$$[1, 2, 3, 4, 5, 6, 7]$$$ and one of the optimal strategies is following:   choose $$$k = 3$$$ elements $$$[2, \underline{3}, 4]$$$ and erase them except its median; you'll get sequence $$$[1, 3, 5, 6, 7]$$$;  choose $$$3$$$ elements $$$[3, \underline{5}, 6]$$$ and erase them except its median; you'll get desired sequence $$$[1, 5, 7]$$$; In the fourth test case, you have sequence $$$[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]$$$. You can choose $$$k=7$$$ elements $$$[2, 4, 6, \underline{7}, 8, 10, 13]$$$ and erase them except its median to get sequence $$$b$$$.

Sample Inputs:
4
3 3 1
1
7 3 3
1 5 7
10 5 3
4 5 6
13 7 7
1 3 5 7 9 11 12

Sample Outputs:
NO
YES
NO
YES

Solution (Python):
for _ in range(int(input())):
    (n, k, m) = tuple(map(int, input().split(' ')))
    ml = set(map(int, input().split(' ')))
    havitada = [i for i in range(1, n + 1) if i not in ml]
    saab = False
    if len(havitada) % (k - 1) != 0:
        print('no')
        continue
    for i in range(k // 2 - 1, len(havitada) - k // 2):
        if havitada[i + 1] - havitada[i] != 1:
            print('yes')
            break
    else:
        print('no')

Solution (Python):
for _ in range(int(input())):
    (n, k, m) = tuple(map(int, input().split(' ')))
    ml = set(map(int, input().split(' ')))
    havitada = []
    for i in range(1, n + 1):
        if i not in ml:
            havitada.append(i)
    saab = False
    if len(havitada) % (k - 1) != 0:
        print('no')
        continue
    for i in range(k // 2 - 1, len(havitada) - k // 2):
        if havitada[i + 1] - havitada[i] != 1:
            print('yes')
            break
    else:
        print('no')

Solution (Python):
import sys
for _ in range(int(input())):
    (n, k, m) = tuple(map(int, input().split(' ')))
    ml = set(map(int, input().split(' ')))
    havitada = []
    for i in range(1, n + 1):
        if i not in ml:
            havitada.append(i)
    saab = False
    if len(havitada) % (k - 1) != 0:
        print('no')
        continue
    for i in range(k // 2 - 1, len(havitada) - k // 2):
        if havitada[i + 1] - havitada[i] != 1:
            print('yes')
            break
    else:
        print('no')