Problem Statement
Dilpreet wants to paint his dog's home that has n boards with different lengths. The length of the ith board is given by arr[i] where arr[] is an array of n integers. He hired k painters for this work, and each painter takes 1 unit time to paint 1 unit of the board.
The problem is to find the minimum time to get this job done if all painters start together, with the constraint that any painter will only paint continuous boards (e.g., boards numbered {2,3,4} or only board {1} or nothing but not boards {2,4,5}).
Example 1
- Input:
n = 5, k = 3, arr[] = {5, 10, 30, 20, 15}
- Output:
35
- Explanation:
- Painter 1 allocation: {5, 10}
- Painter 2 allocation: {30}
- Painter 3 allocation: {20, 15}
- Job will be done when all painters finish, i.e., at time =
max(5+10, 30, 20+15) = 35.
Example 2
- Input:
n = 4, k = 2, arr[] = {10, 20, 30, 40}
- Output:
60
- Explanation:
- Painter 1 allocation: {10, 20, 30}
- Painter 2 allocation: {40}
- Job will be complete at time =
60.
Objective
Complete the function minTime() which takes integers n, k, and the array arr[] as input and returns the minimum time required to paint all partitions.
Expected Complexity
- Time Complexity: O(n log m), where
m = sum of all board lengths
- Space Complexity: O(1)
Approaches
Brute Force
Idea
The brute force approach involves checking all possible ways to allocate boards among the painters and finding the configuration that results in the minimum time. However, this is highly inefficient.
Steps
- Generate all possible partitions of the array into
k painters.
- Calculate the time taken for each partition and find the one with the minimum time.
Time Complexity
- O(2^n) in the worst case due to the exponential growth in partition possibilities.
Space Complexity
Example Code
Not feasible to implement due to its inefficiency.
Dry Run
Skipped due to high inefficiency.
Efficient Approach
Idea
We can use binary search to find the optimal time. We check if a given time x is possible using a helper function check(). This function verifies whether it's possible to paint all boards with k painters if the maximum time each painter can take is x.
Steps
- Initialize
l = max(arr) and h = sum(arr).
- Perform binary search:
- Set
mid = (l + h) / 2.
- Use
check(mid) to verify if the job can be done with k painters within mid time.
- Adjust
l and h based on the result.
- Return the minimum possible time.
Time Complexity
- O(n log m), where
m is the sum of all board lengths.
Space Complexity
Example Code
Dry Run
- For
n = 5, k = 3, arr[] = {5, 10, 30, 20, 15}:
- Binary search will progressively narrow down to
35 as the optimal time.
Example Test Cases
-
Test Case 1
- Input:
n = 5, k = 3, arr[] = {5, 10, 30, 20, 15}
- Output:
35
-
Test Case 2
- Input:
n = 4, k = 2, arr[] = {10, 20, 30, 40}
- Output:
60
Complexity Analysis
Brute Force Approach
- Time Complexity: O(2^n)
- Space Complexity: O(n)
Efficient Approach
- Time Complexity: O(n log m)
- Space Complexity: O(1)
Conclusion
The binary search-based approach provides an optimal solution to the problem, reducing the time complexity significantly compared to the brute force approach. This method efficiently finds the minimum time required to paint all the boards with the given constraints.