Allocate Minimum Pages

You are given an array pages, where pages[i] is the number of pages in the i-th book, and an integer students.
Allocate every book to exactly one student so that:
Each student receives at least one book.
Each student receives one contiguous group of books.
The original book order is preserved.
Return the minimum possible value of the maximum pages assigned to any student. Return -1 when there are more students than books.

Function
allocateMinimumPages(pages: int[], students: int) → long

Examples
Example 1
pages = [12,34,67,90]
students = 2
return = 113
Allocate [12,34,67] and [90]. The larger load is 113, and no valid split has a smaller maximum.
Example 2
pages = [10,20,30,40]
students = 2
return = 60
The optimal groups are [10,20,30] and [40].
Example 3
pages = [10,20]
students = 3
return = -1
Three nonempty contiguous groups cannot be formed from two books.

Constraints
1 <= pages.length <= 100000.
1 <= pages[i] <= 10^6.
1 <= students <= 100000.

The trick is binary search on the answer, not on the array. The lower bound is max(pages), since one student must hold the biggest book. The upper bound is sum(pages). For a candidate limit, greedily walk the books, adding to the current student until the next book would exceed the limit, then start a new student. Count students used. If the count is at most students, the limit works, so search lower. Otherwise search higher. Complexity is O(n log(sum)). Pitfalls: forgetting the -1 case when students exceeds the book count, starting the lower bound at 1 or 0 so a single book can't fit, and using 32-bit ints. The sum can reach 10^11, so use long. Also return the smallest feasible limit, not the first one found. If the greedy check slips under pressure, StealthCoder is the hedge during the live OA.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.