The Painter’s Partition Problem is all about assigning tasks (boards) to workers (painters) in a way that balances the load as efficiently as possible. Each painter must paint a continuous sequence of boards, and our goal is to minimize the time the most-burdened painter spends.
Understanding the Problem
Let’s say we have a list of board lengths like [10, 20, 30, 40] and two painters. One approach is to divide the boards evenly by count—but that won’t always give the best result. The key is to find a way to divide the array so that the painter who ends up with the most work has as little as possible.
This is where Binary Search helps. Instead of checking all ways to split the array (which is too slow), we search for the minimum possible time using a helper function that checks whether a given maximum time is achievable with the available painters.
Different Scenarios Explained
- Normal Case: When the number of painters is less than the number of boards, we need to group boards cleverly so that all painters are used and the work is balanced. Binary search helps us test mid-values as the upper limit of work a painter can do, narrowing down to the optimal time.
- One Painter: The only painter must paint all the boards. So the answer is just the sum of all board lengths.
- Same Number of Painters and Boards: Each painter gets exactly one board. The result is the maximum board length.
- More Painters Than Boards: Since we can’t split a board among painters, some painters won’t be used. The result is still the maximum single board length.
- Empty Input: If the board list is empty, there’s nothing to paint—so we return 0.
- Zero Painters: If no painter is available, the task can’t begin. We return 0 as a safe default.
By combining Binary Search with this understanding of constraints, we can efficiently find the best answer without testing every possible way to split the work.
Efficiency
This binary search method runs in O(N log S), where N is the number of boards and S is the range between the largest board and total sum. This makes it feasible for large arrays.