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

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  • anarod_val25
     + 0 comments

    Too many variables? Maybe. Do I care? Absolutely not!

    int workbook(int n, int k, vector<int> arr) {
    int problems = 0;
    int bookMark = 1;
    
    for(int i = 0; i<arr.size(); i++){
        int pages = (arr[i] + k - 1) / k;
        int firs_exe = 1;
        int last_exe = min(firs_exe + k - 1, arr[i]);
        
        while (pages-->0) {
            
            if(firs_exe <= bookMark && bookMark <= last_exe) problems++;    
            
            firs_exe = last_exe + 1;
            last_exe = min(firs_exe + k - 1, arr[i]);
            bookMark++;
        }
    }
    
    return problems;
}

  • krushnaingale201
     + 0 comments

    JAVA

    public static int workbook(int n, int k, List arr)

    {
int page = 1;
int count = 0;
for(int i=0;i<n;i++){
    int noq = arr.get(i);
    int j =1;
    while(j<=noq){
        int ques = j;
        int qcount = 1;
        while(qcount<=k){
            if(ques==page){
                count++;
            }
            ques++;
            if(ques>noq){
                break;
            }   
            qcount++;             
        }
        page++;
        j+=k;
    }
    }
    return count;
}

  • afzal_saiyed931
     + 0 comments
    def workbook(n, k, arr):
    # Write your code here
    
    current_page_no = 0
    special_problem_count = 0
    for chapter in range(1, n+1):
        pages = math.ceil(arr[chapter-1] / k)
        current_problems = 0
        while pages > 0:
            current_page_no += 1
            problems_on_current_page = k if (arr[chapter-1] - current_problems) > k else arr[chapter-1] - current_problems
            if current_page_no > current_problems and current_page_no <= (current_problems+problems_on_current_page):
                special_problem_count += 1
            current_problems += problems_on_current_page
            pages -= 1            
    return special_problem_count

  • lsloan
     + 0 comments

    Python: Semi-short looping solution…

    def workbook(_, k, arr):
  p = s = 0
  for i in arr:
    for j in range(0, i, k):
      p, s = ((p + 1, s + 1)
        if j < p + 1 <= min(i, j + k)
        else (p + 1, s)) 
  return s

    Here's a semi-short, non-looping functional solution…

    def workbook(_, k, arr):
    pg = lambda m: m[0]
    prob1 = lambda m: m[1][0]
    probN = lambda m: m[1][1]
    return sum(map(
        lambda pgProbs: int(prob1(pgProbs) < pg(pgProbs) <= probN(pgProbs)), 
        enumerate(sum(map(
            lambda chNumProb: list(map(
                lambda pgProb1: (pgProb1, min(chNumProb, pgProb1 + k)), 
                range(0, chNumProb, k))), 
            arr), []), start=1)))

    It could be shorter, but I've seen people complain about the readability of functional code. So, I tried to make this descriptive. I used verbose names for lambda arguments. I also added lambdas for accessing the members of tuples returned from the inner map levels.

    Here's a terse version of the same code…

    def workbook(_, k, arr):
    return sum(map(lambda m: int(m[1][0] < m[0] <= m[1][1]),
            enumerate(sum(map(lambda i:
                list(map(lambda j: (j, min(i, j + k)), 
                range(0, i, k))), 
            arr), []), 1)))

  • yashdeora98294
     + 0 comments

    Here is problem solution in Python, Java, C++, C and Javascript - https://programmingoneonone.com/hackerrank-lisas-workbook-problem-solution.html