#1046

Easy

auto_awesomeArray plus Heap (Priority Queue)

LeetCode Problem Workspace

Last Stone Weight

In the 'Last Stone Weight' problem, we smash the two heaviest stones until one remains, using heaps or sorting.

Problem Statement

You are given an array stones[] where each element represents the weight of a stone. On each turn, you must select the two heaviest stones, smash them together, and replace them with their absolute difference. The game ends when at most one stone remains.

For example, with stones = [2,7,4,1,8,1], you simulate smashing stones: the heaviest two, 8 and 7, are smashed to yield 1. This continues until only one stone remains. The task is to find the final stone weight, or return 0 if no stones remain.

Examples

Example 1

Input: stones = [2,7,4,1,8,1]

Output: 1

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.

Example 2

Input: stones = [1]

Output: 1

Example details omitted.

Constraints

1 <= stones.length <= 30
1 <= stones[i] <= 1000

Solution Approach

Use a Max-Heap

Convert the list of stones into a max-heap. This allows efficient retrieval of the two largest stones at each step. By repeatedly extracting and updating the heap, we simulate the stone-smashing process until only one stone remains.

Sort the Stones

Alternatively, you can sort the stones in descending order and simulate the process by picking the first two elements, performing the smash, and inserting the result back into the sorted array. This approach is less efficient than using a heap but still works within the problem's constraints.

Handle Edge Cases

Consider edge cases where the number of stones is very small (like one stone) or when all stones are of the same weight. These cases are simpler but need special handling in the solution to avoid unnecessary operations.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Both heap and sorting approaches involve repeated operations to find and modify the two largest stones. The heap approach typically has O(log n) complexity for insertion and extraction, while the sorting method involves O(n log n) complexity per turn. Thus, the time complexity depends on the chosen method and number of stones.

What Interviewers Usually Probe

Candidate uses efficient heap operations to solve the problem.
Candidate explains the trade-off between heaps and sorting for this problem.
Candidate demonstrates how to handle edge cases such as one stone or identical stones.

Common Pitfalls or Variants

Common pitfalls

Not using a heap efficiently, leading to higher time complexity than necessary.
Over-complicating the solution with unnecessary steps when the problem can be solved with basic sorting or heaps.
Failing to handle the case where only one stone remains.

Follow-up variants

Extend the problem to handle different game variations where the number of stones can be modified after each turn.
Instead of smashing stones, consider keeping track of the two smallest stones.
Adjust the problem to allow multiple rounds of stone smashing, where new stones are added to the array after each turn.

FAQ

What is the best approach for solving the Last Stone Weight problem?

Using a max-heap provides the most efficient solution for the Last Stone Weight problem by allowing quick retrieval and modification of the two largest stones.

Can I solve the problem with sorting?

Yes, sorting works as an alternative approach, though it's less efficient than using a heap. Sorting allows easy access to the two largest stones but involves higher time complexity.

How do I handle edge cases in this problem?

Edge cases like a single stone or stones with identical weights can be handled by checking the length of the array or ensuring no unnecessary operations are performed when only one stone remains.

How does GhostInterview help with Last Stone Weight?

GhostInterview provides step-by-step assistance in understanding heap and sorting methods, helping you implement a clear and efficient solution.

What is the time complexity of the heap solution?

The heap solution has a time complexity of O(n log n), where n is the number of stones, due to repeated heap insertions and extractions.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        h = [-x for x in stones]
        heapify(h)
        while len(h) > 1:
            y, x = -heappop(h), -heappop(h)
            if x != y:
                heappush(h, x - y)
        return 0 if not h else -h[0]

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