Closing the Tweets Problem’s Solution with Approach – CodeChef

Problem Statement

Little kids, Jack and Evan like playing their favorite game Glass-and-Stone. Today they want to play something new and came across Twitter on their father’s laptop.

They saw it for the first time but were already getting bored to see a bunch of sentences having at most 140 characters each. The only thing they liked to play with it is, closing and opening tweets.

There are N tweets on the page and each tweet can be opened by clicking on it, to see some statistics related to that tweet. Initially all the tweets are closed. Clicking on an open tweet closes it and clicking on a closed tweet opens it. There is also a button to close all the open tweets. Given a sequence of K clicks by Jack, Evan has to guess the total number of open tweets just after each click. Please help Evan in this game.

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Input

The first line contains two integers N K, the number of tweets (numbered 1 to N) and the number of clicks respectively (1 ≤ N, K ≤ 1000). Each of the following K lines has one of the following.

• CLICK X , where X is the tweet number (1 ≤ X ≤ N)
• CLOSEALL

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Output

Output K lines, where the i^th line should contain the number of open tweets just after the i^th click.

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Example

Input:
3 6
CLICK 1
CLICK 2
CLICK 3
CLICK 2
CLOSEALL
CLICK 1

Output:
1
2
3
2
0
1

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Solution 

Approach

1. Firstly you have to create a list to store the status of the tweets
2. Then initialize the list for the N number of zeroes
3. take the input for K times 
4. If the input says ‘CLOSEALL‘, again initialize the list to N zeroes
5. If not then check is the tweet is open or not 
6. If it is active change the value at its position to 0
7. Else make it active
8. Lastly print the number of open tweets

Code

# Taking the inputs
N,K = [int(x) for x in input().split()]

# intializing the main list to store the status of tweet
arr = [0]*N
for i in range(K):
    e = input()
    if e != 'CLOSEALL':
        com, pos = [x for x in e.split()]
        if arr[int(pos) - 1] == 1:
            arr[int(pos)-1] = 0
        else:
            arr[int(pos)-1] = 1
    else:
        arr = [0]*N
    print(arr.count(1))

Question Link – link