Question

Initially, you are given N pages each of dimension n*m . each of these pages contains some amount of 1-1 squares drawn on it. you have the special power to shift a square from one page to another page and you want to minimise the number of pages used programming algorithm

Answer 1

Answer:

hut yahan se

Explanation:

i dunno why did i said thet  but if you felt bad then srry

Answer 2

Answer:

Here is the Python algorithm for the above problem statement

Code:

def isPossible(arr, n, m, curr_min):

   studentsRequired = 1

   curr_sum = 0

   for i in range(n):

       if (arr[i] > curr_min):

           return False

       if (curr_sum + arr[i] > curr_min):

           studentsRequired += 1

           curr_sum = arr[i]

           if (studentsRequired > m):

               return False

       else:

           curr_sum += arr[i]

   return True

def findPages(arr, n, m):

   sum = 0

   if (n < m):

       return -1

   for i in range(n):

       sum += arr[i]

   start, end = 0, sum

   result = 10**9

   while (start <= end):

       mid = (start + end) // 2

       if (isPossible(arr, n, m, mid)):

           result = mid

           end = mid - 1

       else:

           start = mid + 1

   return result

arr = [12, 34, 67, 90]

n = len(arr)

m = 2   # No. of students

print("Minimum number of pages = ",findPages(arr, n, m))

Explanation:

• The concept is to employ Binary Search. We decide on a number of pages that is in the middle of the present minimum and maximum.
• Minimum and maximum are set to 0 and sum-of-all-pages, respectively. If a present mid can be a solution, we look in the lower half; otherwise, we look in the upper half.

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