Understanding the Problem
We are given N boards of varying lengths and k painters, where:
- Each painter takes the same amount of time to paint 1 unit of board length.
- Each painter must paint contiguous sections of boards.
- A board cannot be split between multiple painters.
Our objective is to minimize the maximum time any single painter takes to complete their assigned boards.
Key Observations
The worst case occurs when a single painter paints all the boards, taking:
sum of all board lengths = S
The best case occurs when we assign each painter exactly one board, so the maximum time any painter takes is:
max(board lengths)
Thus, the optimal solution lies between:
max(board lengths) <= Optimal Time <= S
We need to find the minimum possible “maximum time” a painter has to spend painting.
Approach
1. Understanding the Search Space
- The minimum possible time any painter can take is the longest board, since at least one painter has to paint that board.
- The maximum possible time is when one painter paints everything (sum of all board lengths).
- We perform binary search in this range to find the smallest feasible maximum time.
2. Checking Feasibility using is_valid_partition
We use a helper function to check if a given maximum time (max_length
) is possible by distributing the boards among k
painters.
- Initialize
painters_required = 1
,current_length = 0
. - Iterate over boards:
- If
current_length + board_length > max_length
, assign the next painter. - If we exceed
k
painters, return false (not possible).
- If
- If we successfully distribute within
k
painters, return true.
3. Performing Binary Search
-
Initialize search range:
low = max(boards)
(minimum time required).high = sum(boards)
(maximum time required).result = high
(stores the best solution found).
-
Binary search:
- Compute
mid = (low + high) / 2
. - If
mid
is feasible (is_valid_partition
returnstrue
), updateresult = mid
and search for a smaller max time (high = mid - 1
). - Otherwise, search for a larger max time (
low = mid + 1
).
- Compute
-
The loop continues until
low > high
, andresult
stores the optimal answer.
Complexity Analysis
Binary Search Complexity
O(log S)
where S is the sum of all board lengths.
Partition Checking Complexity
O(N)
since we iterate through all boards once per check.
Total Complexity
O(N log S)
which is efficient compared to brute-force methods.