Last Stone Weight

Problem Overview

The "Last Stone Weight" problem is an interesting problem that involves breaking down a collection of elements until a specific condition is met. This problem is categorized under the Heaps data structure on LeetCode. Here’s a brief overview of the problem statement:

• You're given an array of integers that represent the weights of stones.
• You need to smash two stones together: the stone with the maximum weight and the stone with the next maximum weight.
• The result of this smash operation is either:
  • If they are of equal weight, both stones will be completely crushed and removed.
  • If they are of different weights, the stone with a larger weight will be reduced by the weight of the smaller stone.
• Continue this process until only one stone is left or no stones are left.
• The task is to return the weight of the last remaining stone (if any). If no stones are left, return 0.

Understanding the Problem

To solve this problem, we need to simulate the process of smashing stones until no further operations are possible. For each operation, we need to determine the two largest stones, which makes this problem a great fit for a data structure like a max heap (or priority queue). The operations involve removing elements, which is efficiently doable using heaps.

Key DSA Concepts and Theory

Heaps (Priority Queues)

A heap is a specialized tree-based data structure that satisfies the heap property. It is often implemented as:

• A max heap, where each parent node is greater than or equal to its children.
• A min heap, where each parent node is less than or equal to its children.

For this specific problem, a max heap is particularly useful since we need to efficiently retrieve the largest stones. Heaps support operations like insert, delete_max/delete_min, and find_max/find_min in logarithmic time complexity.

Why Heaps for This Problem?

1. Efficient Maximum Retrieval: The heap structure allows for retrieving and removing the largest element (stone) in O(log n) time.
2. Dynamic Size: As stones are smashed, the collection size decreases, and heaps efficiently handle dynamic collections.
3. Inherent Priority: The heap manages priority (i.e., larger weights) inherently, ensuring that we always handle the two heaviest stones first.

Solution Approach

Let's step through the detailed solution using a max heap.

Solution Steps

1. Max Heap Initialization:

   • Instead of a built-in max heap, LeetCode's default priority queue (heap) is a min heap. We store negative weights to simulate max heap functionality.

2. Insert Elements:

   • Insert each stone weight into a priority queue with its value negated.

3. Smash Stones:

   • Repeatedly extract the two largest stones, simulate the smash:
     • If the stones have different weights, push the result back into the heap.

4. Result Calculation:

   • Once all possible smash operations are done, check if there's any stone left in the heap and return its weight or zero if empty.

C++ Implementation

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

int lastStoneWeight(std::vector<int>& stones) {
    std::priority_queue<int> maxHeap;
    
    for (int stone : stones) {
        maxHeap.push(stone);
    }
    
    while (maxHeap.size() > 1) {
        int stone1 = maxHeap.top();
        maxHeap.pop();
        int stone2 = maxHeap.top();
        maxHeap.pop();
        
        if (stone1 != stone2) {
            maxHeap.push(stone1 - stone2);
        }
    }
    
    return maxHeap.empty() ? 0 : maxHeap.top();
}

Java Implementation

import java.util.Collections;
import java.util.PriorityQueue;

public class Solution {
    public int lastStoneWeight(int[] stones) {
        PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());
        
        for (int stone : stones) {
            maxHeap.add(stone);
        }
        
        while (maxHeap.size() > 1) {
            int stone1 = maxHeap.poll();
            int stone2 = maxHeap.poll();
            
            if (stone1 != stone2) {
                maxHeap.add(stone1 - stone2);
            }
        }
        
        return maxHeap.isEmpty() ? 0 : maxHeap.poll();
    }
}

Time and Space Complexity Analysis

• Time Complexity:

  • Building the initial heap from n elements takes O(n log n).
  • Each smash operation involves extracting the top two elements and possibly inserting a new one, which is O(log n) per operation.
  • In total, for n operations, this gives us O(n log n).

• Space Complexity:

  • The space complexity is O(n) for storing the stones in the heap.

Common Mistakes to Avoid

1. Wrong Heap Initialization: Confusing max heap with min heap and not correctly simulating by using negative weights for max heap logic.
2. Integer Overflow: While not a concern in most cases for this problem, ensure handling integer operations safely if customizing or expanding the solution.

Similar Problems on LeetCode

Understanding heaps can help solve similar problems such as:

• Kth Largest Element in an Array (LeetCode #215)
• Top K Frequent Elements (LeetCode #347)
• Find Median from Data Stream (LeetCode #295)

Additional Resources and References

• Priority Queues and Heaps
• LeetCode Heap Topic
• CLRS Book: Introduction to Algorithms - Chapter on Heaps

This detailed breakdown should equip you with the insights required to tackle the "Last Stone Weight" problem efficiently using heaps, along with a solid understanding of the underlying data structure concepts.