We use a Max-Heap to efficiently retrieve the two heaviest stones in each step.

1. Convert all stones to negative values and heapify them to create a Max-Heap.
2. While there is more than 1 stone:
   • Pop two stones s1 and s2.
   • If s1 != s2, calculate the difference s1 - s2 (or abs(s1-s2)) and push it back.
3. If one stone remains, return its absolute value; otherwise, return 0.

• Time Complexity: O(N log N).
• Space Complexity: O(N).

import heapq

def last_stone_weight(stones):
    # Python uses min-heap, so negate values for max-heap
    h = [-s for s in stones]
    heapq.heapify(h)
    
    while len(h) > 1:
        s1 = heapq.heappop(h)
        s2 = heapq.heappop(h)
        # If they are different, push back the positive difference
        if s1 != s2:
            heapq.heappush(h, s1 - s2)
            
    return -h[0] if h else 0