Last Stone Weight in C++

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 efficiently finding the two heaviest stones, which can lead to suboptimal performance.

Approach

To solve this problem, we need to efficiently find and remove the two heaviest stones. A naive approach would involve sorting the list of stones repeatedly, but this is not optimal. Instead, we can use a max-heap (priority queue) to always access the heaviest stones efficiently.

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

Using a max-heap allows us to efficiently find and remove the heaviest stones. The max-heap operations (insertion and extraction) have a time complexity of O(log n), making this approach much more efficient.

Algorithm

1. Insert all stone weights into a max-heap.

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

#include <iostream>
#include <vector>
#include <queue>

using namespace std;

int lastStoneWeight(vector<int>& stones) {
    // Max-heap to store the stones
    priority_queue<int> maxHeap(stones.begin(), stones.end());

    // Process the stones until one or none is left
    while (maxHeap.size() > 1) {
        int stone1 = maxHeap.top(); // Heaviest stone
        maxHeap.pop();
        int stone2 = maxHeap.top(); // Second heaviest stone
        maxHeap.pop();

        if (stone1 != stone2) {
            maxHeap.push(stone1 - stone2); // Push the difference back into the heap
        }
    }

    // If there's one stone left, return its weight, otherwise return 0
    return maxHeap.empty() ? 0 : maxHeap.top();
}

int main() {
    vector<int> stones = {2, 7, 4, 1, 8, 1};
    cout << "Last stone weight: " << lastStoneWeight(stones) << endl;
    return 0;
}

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 in the max-heap takes O(log n) time, and we perform these operations n-1 times in the worst case.

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 a large number of stones to test performance.

Using a testing framework like Google Test can help automate and manage these tests effectively.

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 for the task.
Consider edge cases and how your solution handles them.
Practice similar problems to improve your 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, considered edge cases, and provided tips for problem-solving.

Understanding and solving such problems is crucial for developing efficient algorithms and improving coding skills. Keep practicing and exploring further to enhance your problem-solving abilities.

Additional Resources

For further reading and practice, consider the following resources:

Last Stone Weight - C++ Solution and Time Complexity Analysis