You are given an array weights
where weights[i]
is the weight of the ith
student, and an infinite number of boats where each boat can carry a maximum weight of maxWeight
. Each boat carries at most two people at the same time, provided the sum of the weight of those people is at most maxWeight
and the absolute difference of their weights shouldn't exceed maxDiff
.
Return the minimum number of boats to carry every given student.
Example 1:
Input: weights = [2, 4, 2, 1], maxWeight = 6, maxDiff = 1 Output: 3 Explanation: 3 boats: (1, 2), (2) and (4)
Example 2:
Input: weights = [81, 37, 32, 88, 55, 93, 45, 72], maxWeight = 100, maxDiff = 10 Output: 6 Explanation: 6 boats (81), (37, 32), (88), (55, 45), (93), (72)
The core challenge of this problem is to minimize the number of boats required to transport all students given the constraints on weight and weight difference. This problem is significant in scenarios like evacuation planning, resource allocation, and load balancing.
Potential pitfalls include not considering the weight difference constraint or not optimizing the pairing of students to minimize the number of boats.
To solve this problem, we can use a greedy algorithm. The idea is to sort the weights and try to pair the lightest and heaviest students that can share a boat. This approach ensures that we minimize the number of boats used.
Initial naive solution might involve trying all possible pairs, but this would be computationally expensive. Instead, sorting the array and using a two-pointer technique is more efficient.
1. **Sort the weights array.**
2. **Use two pointers:** one starting at the beginning (lightest) and one at the end (heaviest).
3. **Try to pair the lightest and heaviest students:** If they can share a boat, move both pointers inward. If not, move only the pointer for the heaviest student.
4. **Count the number of boats used.**
Here is a step-by-step breakdown of the algorithm:
def min_boats(weights, maxWeight, maxDiff):
# Sort the weights array
weights.sort()
# Initialize two pointers
left, right = 0, len(weights) - 1
boats = 0
# Use two-pointer technique to find the minimum number of boats
while left <= right:
# If the lightest and heaviest can share a boat
if weights[left] + weights[right] <= maxWeight and abs(weights[left] - weights[right]) <= maxDiff:
left += 1 # Move the left pointer inward
# Move the right pointer inward in both cases
right -= 1
# Increment the boat counter
boats += 1
return boats
# Example usage
print(min_boats([2, 4, 2, 1], 6, 1)) # Output: 3
print(min_boats([81, 37, 32, 88, 55, 93, 45, 72], 100, 10)) # Output: 6
The time complexity of this approach is O(n log n) due to the sorting step. The space complexity is O(1) as we are using only a few extra variables.
Compared to a naive O(n^2) approach, this is significantly more efficient.
Potential edge cases include:
These cases are handled by the algorithm as it inherently checks the constraints and processes the array accordingly.
To test the solution comprehensively, consider:
Using a testing framework like `unittest` in Python can help automate and validate these tests.
When approaching such problems:
Understanding and solving this problem helps in developing skills in greedy algorithms and two-pointer techniques. These are widely applicable in optimization problems.
Practice and exploration of similar problems can further enhance your problem-solving abilities.
For further reading and practice: