Last Stone Weight

Problem

You are given an array of integers stones where stones[i] is the weight of the ith stone.

We are playing a game with the stones. On each turn, we choose the heaviest two stones and smash them together. Suppose the heaviest two stones have weights x and y with x <= y. The result of this smash is:

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

At the end of the game, there is at most one stone left.

Return the weight of the last remaining stone. If there are no stones left, return 0.

Example 1:

Input: stones = [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.

Example 2:

Input: stones = [1]
Output: 1

Constraints:

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

Solution

Approach: Max Heap

The key insight is to use a max heap to efficiently get the two heaviest stones each turn.

Implementation

import heapq

class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        # Python's heapq is a min heap, so negate values for max heap
        heap = [-stone for stone in stones]
        heapq.heapify(heap)

        while len(heap) > 1:
            # Get two heaviest stones
            first = -heapq.heappop(heap)
            second = -heapq.heappop(heap)

            # If they're different, add the difference back
            if first != second:
                heapq.heappush(heap, -(first - second))

        # Return last stone weight or 0 if no stones left
        return -heap[0] if heap else 0

Complexity Analysis

Time Complexity: O(n log n), where n is the number of stones. We do n heap operations, each taking O(log n).
Space Complexity: O(n) for the heap.

Key Insights

Max Heap Simulation: We need to repeatedly find the two largest elements, which is perfect for a max heap.
Negative Values Trick: Python’s heapq is a min heap. By negating values, we simulate a max heap.
Destructive Process: The heap shrinks over time as stones are destroyed or combined.
Final State: The loop stops when 0 or 1 stone remains. If 0, we return 0; if 1, we return its weight.
Heap Efficiency: Without a heap, finding the two largest would take O(n) per round, making total complexity O(n²).