Minimum time to fulfil all orders

Geek is organizing a party at his house. For the party, he needs exactly n donuts for the guests. Geek decides to order the donuts from a nearby restaurant, which has l chefs and each chef has a rank r. A chef with rank r can make 1 donut in the first r minutes, 1 more donut in next 2r minutes, 1 more donut in 3r minutes, and so on. find the minimum time in which all the orders will be fulfilled.

Examples:

Input: n = 10, rank[] = [1, 2, 3, 4]
Output: 12
Explanation: Chef with rank 1, can make 4 donuts in time 1 + 2 + 3 + 4 = 10 mins
Chef with rank 2, can make 3 donuts in time 2 + 4 + 6 = 12 mins
Chef with rank 3, can make 2 donuts in time 3 + 6 = 9 mins
Chef with rank 4, can make 1 donuts in time = 4 minutes
Total donuts = 4 + 3 + 2 + 1 = 10 and 
Maximum time taken by all = 12 minutes.

Input: n = 8, rank[] = [1, 1, 1, 1, 1, 1, 1, 1]
Output: 1
Explanation: As all chefs are ranked 1, so each chef can make 1 donuts 1 min.
Total donuts = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 8 and  min time = 1 minute 

[Naive Approach] Using Linear Search

The idea is to use a brute-force linear search from time 1 up to Integer Maximum to check at each time point whether it is possible to make at least n donuts. At each time point, find the maximum number of donuts each chef can make and compare if the sum of donuts is greater than equal to n. The first time for which this is possible is considered the minimum time to fulfill the order.

12

Time Complexity: O(t × l × sqrt(t)) where t is the maximum time iterated, l is the number of cooks. For each time point and each chef, the number of donuts k is found using 1 × r + 2 × r + 3 × r + ... + k × r <= timeAlloted. Maximum iterations will be sqrt(2 × timeAlloted / r)) which is roughly sqrt(timeAlloted).
Auxiliary Space: O(1)

[Better Approach] Using Binary Search and AP sum

The idea is to use binary search on the time to find the minimum time required to produce at least n donuts. For any given time, we calculate how many donuts each chef can make by summing the terms of an arithmetic progression where each chef takes r, 2r, 3r, ... seconds per donut, with r being the chef's rank. The total number of donuts made is the sum of such APs across all chefs. If the total meets or exceeds n, we try to minimize the time further; otherwise, we increase it. This combination of AP sum logic and binary search helps efficiently reach the answer.

Step by step approach:

• Each chef with rank r takes r seconds for the 1st donut, 2r for the 2nd, and so on.
  So, time to make k donuts is r × (1 + 2 + ... + k) = r × k × (k + 1) / 2.
• The minimum time can be 0, and the maximum time (upper bound) can be the time the fastest chef would take to make all n donuts alone.
  That’s given by: rmin × n × (n + 1) / 2 where rmin is the smallest rank.
• Perform binary search over time from 0 to rmin × n × (n + 1) / 2.
• For each mid time:
  compute how many donuts each chef can make using the quadratic formula derived from AP sum:
• k(k+1)/2 <= t/r → k = floor((-1 + sqrt(1 + 8t/r)) / 2)
  Sum these for all chefs.
  If total donuts ≥ n, update answer and try a smaller time (move left).
  Otherwise, try a larger time (move right).
• Return the Minimum Time Found:
  The binary search narrows down the minimum time required to fulfill all orders.

12

Time Complexity: O(l × log(n)), where l is the length of ranks array, and n is the number of donuts to be made.
Auxiliary Space: O(1).