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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Lisa's Workbook
  5. Discussions

Lisa's Workbook

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937 Discussions

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  • alban_tyrex
     + 0 comments

    Here is my c++ solution , you can watch the explanation here : https://youtu.be/3L-haDbsjAg

    int workbook(int n, int k, vector<int> arr) {
    int page = 1, chapter = 1, fe = 1, res = 0;
    while(chapter <= arr.size()){
        int le = min(fe + k - 1, arr[chapter-1]);
        if(fe <= page && page <= le) res++;
        fe = le + 1;
        if(fe > arr[chapter-1]) {
            chapter++;
            fe = 1;
        }
        page ++;
    }   
    return res;
}

  • illogicalsleepi1
     + 0 comments
    def workbook(n, k, arr):
    book=[]
    ans=0
    for i in range(n):
        for h in range(arr[i]//k):
            book.append(list(range(h*k+1,h*k+k+1)))
        if arr[i]%k!=0:
            book.append(list(range((arr[i]//k)*k+1,arr[i]+1)))
    print(book)
    for d in range(len(book)):
        if d+1 in book[d]:
            ans+=1
    return ans

  • Karolis_katinas1
     + 0 comments
    book = []
page = -1
ans = 0

for i in range(n):
    for j in range(arr[i]):
        if j % k == 0:
            page += 1
            book.append([])
        book[page].append(j+1)
        if page == j:
            ans += 1
return ans

  • pratyushkumar274
     + 0 comments

    Instead of iterating over each problem, we can check if current_page falls within the range of problem_number to last_problem_on_page.

    This approach reduces the time complexity for each page from O(k) to O(1) , where k is the maximum number of problems per page. Because it performs a single comparison per page rather than multiple comparisons for each problem on the page.

    def work_book(n, k, arr):
    special_problems = 0
    current_page = 1

    for chapter in range(n):
        total_problems = arr[chapter]
        problem_number = 1

        while problem_number <= total_problems:
            # Calculate the last problem on the current page
            last_problem_on_page = min(problem_number + k - 1, total_problems)

            # Check if the current page contains a special problem
            if problem_number <= current_page <= last_problem_on_page:
                special_problems += 1

            # Move to the next page and update the problem number
            current_page += 1
            problem_number = last_problem_on_page + 1

    return special_problems

  • jmcparland
     + 0 comments

    Haskell

    module Main where

import Data.List.Split (chunksOf)

solve :: Int -> [Int] -> Int
solve k arr =
    length -- count the True
        . filter id -- only the Trues
        . zipWith elem [1 ..] -- is the page number in the page?
        $ concatMap (chunksOf k . flip take [1 ..]) arr -- list of pages' contents

main :: IO ()
main = do
    [n, k] <- map read . words <$> getLine :: IO [Int]
    arr <- map read . words <$> getLine :: IO [Int]
    print $ solve k arr