The “Allocate Minimum Pages” problem is a classic binary search question that is often asked in coding interviews. <\/p>\n\n\n\n
It teaches you how to optimize allocation problems using binary search. In this blog, we\u2019ll explain the problem, walk you through both brute force and optimal solutions, and make everything easy to understand with code comments, dry runs, and clear explanations.<\/p>\n\n\n\n
Here’s the [Problem Link<\/span><\/mark><\/a><\/strong>] to begin with.<\/p>\n\n\n\n You are given an array If it is not possible to allocate, return\u00a0 Constraints:<\/strong> Let\u2019s solve the problem in the most straightforward way:<\/p>\n\n\n\n This approach is easy to understand but not efficient for large inputs.<\/p>\n\n\n\n
\n\n\n\nProblem Statement<\/strong><\/h2>\n\n\n\n
arr<\/code> where each element represents the number of pages in a book. There are k<\/code> students, and the books are arranged in a line.
Your task is to allocate books to students such that:<\/p>\n\n\n\n\n
-1<\/code>.<\/p>\n\n\n\nInput: <\/strong>arr[] = [12, 34, 67, 90], k = 2\nOutput: <\/strong>113\nExplanation: <\/strong>Allocation can be done in following ways:\n[12] and [34, 67, 90] Maximum Pages = 191\n[12, 34] and [67, 90] Maximum Pages = 157\n[12, 34, 67] and [90] Maximum Pages = 113.\nTherefore, the minimum of these cases is 113, which is selected as the output.<\/pre>\n\n\n\n
Input: <\/strong>arr[] = [15, 17, 20], k = 5
Output: <\/strong>-1
Explanation: <\/strong>Allocation can not be done.<\/pre>\n\n\n\n
1 <=\u00a0 arr.size() <= 106<\/sup>
1 <= arr[i] <= 103
<\/sup>1 <= k <= 103\u00a0<\/sup><\/p>\n\n\n
\n\n\n\nBrute Force Solution (Try Every Possible Maximum)<\/h2>\n\n\n\n
Intuition and Approach<\/h3>\n\n\n\n
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Try every possible maximum pages value from the largest single book to the sum of all pages.<\/li>\n\n\n\n
For each value, check if it is possible to allocate books so that no student gets more than that many pages.<\/li>\n\n\n\n
The first value for which allocation is possible is our answer.<\/li>\n\n\n\n
We check every possible value, so we\u2019re sure to find the minimum one that works.<\/li>\n<\/ol>\n\n\n\nCode Implementation<\/h3>\n\n\n\n