Last Stone Weight in JavaScript

We have a collection of stones, where each stone has a positive integer weight.

Each turn, we choose the two heaviest stones and smash them together. Suppose the stones have weights x and y with x <= y. The result of this smash is:

If x == y, both stones are totally destroyed;
If x != y, the stone of weight x is totally destroyed, and the stone of weight y has the new weight y-x.

At the end, there is at most 1 stone left. Return the weight of this stone (or 0 if there are no stones left.)

Example:

Input: [2,7,4,1,8,1]
Output: 1
Explanation: 
We combine 7 and 8 to get 1 so the array converts to [2,4,1,1,1] then,
we combine 2 and 4 to get 2 so the array converts to [2,1,1,1] then,
we combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
we combine 1 and 1 to get 0 so the array converts to [1] then that's the value of the last stone.

Understanding the Problem

The core challenge of this problem is to repeatedly find and process the two heaviest stones until only one stone remains or all stones are destroyed. This problem is significant in scenarios where we need to repeatedly process the largest elements, such as in certain scheduling or resource allocation problems.

Potential pitfalls include not correctly updating the list of stones after each smash and not efficiently finding the two heaviest stones.

Approach

To solve this problem, we need to repeatedly find the two heaviest stones and process them according to the rules. A naive solution would involve sorting the list of stones in each iteration, which is not optimal.

A more efficient approach involves using a max-heap (priority queue) to always have quick access to the heaviest stones. This reduces the time complexity of finding the two heaviest stones.

Naive Solution

The naive solution involves sorting the list of stones in each iteration to find the two heaviest stones. This approach has a time complexity of O(n log n) per iteration, which is inefficient for large lists.

Optimized Solution

The optimized solution uses a max-heap to efficiently find and remove the two heaviest stones. The max-heap allows us to perform these operations in O(log n) time.

Algorithm

1. Create a max-heap from the list of stones.

2. While there is more than one stone in the heap:

Extract the two heaviest stones.
If they are not equal, insert the difference back into the heap.

3. If there is one stone left, return its weight; otherwise, return 0.

Code Implementation

// Importing the max-heap library
const { MaxPriorityQueue } = require('@datastructures-js/priority-queue');

function lastStoneWeight(stones) {
    // Create a max-heap
    const maxHeap = new MaxPriorityQueue({ priority: x => x });

    // Add all stones to the max-heap
    stones.forEach(stone => maxHeap.enqueue(stone));

    // Process the stones until one or none is left
    while (maxHeap.size() > 1) {
        // Extract the two heaviest stones
        const stone1 = maxHeap.dequeue().element;
        const stone2 = maxHeap.dequeue().element;

        // If they are not equal, insert the difference back into the heap
        if (stone1 !== stone2) {
            maxHeap.enqueue(stone1 - stone2);
        }
    }

    // Return the weight of the last stone or 0 if no stones are left
    return maxHeap.size() === 0 ? 0 : maxHeap.front().element;
}

// Example usage
console.log(lastStoneWeight([2, 7, 4, 1, 8, 1])); // Output: 1

Complexity Analysis

The time complexity of the optimized solution is O(n log n), where n is the number of stones. This is because each insertion and extraction operation on the max-heap takes O(log n) time, and we perform these operations n times.

The space complexity is O(n) due to the storage required for the max-heap.

Edge Cases

Potential edge cases include:

All stones having the same weight.
Only one stone in the input list.
No stones in the input list.

These cases are handled by the algorithm as it naturally processes the stones until one or none is left.

Testing

To test the solution comprehensively, consider the following test cases:

Simple cases with a few stones.
Cases with all stones having the same weight.
Cases with only one stone.
Cases with no stones.

Using a testing framework like Jest can help automate and validate these test cases.

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

Break down the problem into smaller, manageable parts.
Think about the most efficient data structures to use.
Consider edge cases and how to handle them.
Practice similar problems to improve problem-solving skills.

Conclusion

In this blog post, we discussed the problem of finding the last stone weight after repeatedly smashing the two heaviest stones. We explored a naive solution and an optimized solution using a max-heap. We also analyzed the complexity and discussed edge cases and testing strategies.

Understanding and solving such problems is crucial for developing strong problem-solving skills and improving algorithmic thinking.

Additional Resources

For further reading and practice, consider the following resources:

Last Stone Weight - JavaScript Solution and Time Complexity Analysis