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Minimum Number of Boats - Time Complexity: O(n log n) - Java Solution


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)

Understanding the Problem

The core challenge of this problem is to pair students in such a way that the number of boats used is minimized while adhering to the constraints of maximum weight and maximum weight difference. This problem is significant in scenarios like rescue operations where optimizing the number of trips is crucial.

Potential pitfalls include not considering the weight difference constraint or not optimizing the pairing process, leading to suboptimal solutions.

Approach

To solve this problem, we can use a greedy algorithm. The idea is to sort the weights and then use a two-pointer technique to try and pair the lightest and heaviest students who can share a boat. This approach ensures that we are always making the optimal choice at each step.

Naive Solution

A naive solution would involve checking all possible pairs, which would be computationally expensive (O(n^2)) and not feasible for large inputs.

Optimized Solution

The optimized solution involves sorting the array and using two pointers to find the optimal pairs. This reduces the complexity to O(n log n) due to the sorting step.

Algorithm

  1. Sort the array of weights.
  2. Initialize two pointers: one at the start (lightest) and one at the end (heaviest) of the sorted array.
  3. Iterate while the start pointer is less than or equal to the end pointer:
    • If the sum of the weights at the two pointers is less than or equal to maxWeight and their difference is less than or equal to maxDiff, pair them and move both pointers inward.
    • Otherwise, move only the end pointer inward (the heavier student gets a boat alone).
  4. Increment the boat count for each iteration.

Code Implementation

import java.util.Arrays;

public class MinimumBoats {
    public static int minNumberOfBoats(int[] weights, int maxWeight, int maxDiff) {
        // Sort the weights array
        Arrays.sort(weights);
        
        int start = 0;
        int end = weights.length - 1;
        int boats = 0;
        
        while (start <= end) {
            // If the lightest and heaviest can share a boat
            if (weights[start] + weights[end] <= maxWeight && Math.abs(weights[start] - weights[end]) <= maxDiff) {
                start++;
                end--;
            } else {
                // Otherwise, the heaviest gets a boat alone
                end--;
            }
            boats++;
        }
        
        return boats;
    }

    public static void main(String[] args) {
        int[] weights1 = {2, 4, 2, 1};
        int maxWeight1 = 6;
        int maxDiff1 = 1;
        System.out.println(minNumberOfBoats(weights1, maxWeight1, maxDiff1)); // Output: 3

        int[] weights2 = {81, 37, 32, 88, 55, 93, 45, 72};
        int maxWeight2 = 100;
        int maxDiff2 = 10;
        System.out.println(minNumberOfBoats(weights2, maxWeight2, maxDiff2)); // Output: 6
    }
}

Complexity Analysis

The time complexity of the optimized solution is O(n log n) due to the sorting step. The space complexity is O(1) as we are using a constant amount of extra space.

Edge Cases

Potential edge cases include:

Each of these cases should be tested to ensure the algorithm handles them correctly.

Testing

To test the solution comprehensively, consider a variety of test cases:

Using a testing framework like JUnit can help automate and manage these tests.

Thinking and Problem-Solving Tips

When approaching such problems:

Conclusion

Understanding and solving optimization problems like this one is crucial for developing efficient algorithms. Practice and exploration of different approaches can significantly enhance problem-solving skills.

Additional Resources

For further reading and practice: