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# last-stone-weight

We have a collection of stones, 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 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 1:

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 last stone.

Note:

1 <= stones.length <= 30

1 <= stones[i] <= 1000

From problem description, notice that every time smash 2 heaviest stones,

- if two heaviest stones weight not equal, put diff into stones array
- if two heaviest stones are equal, discard both

Use maximum priority queue to store stones, pull top 2 stones to do smash, until priority queue size <= 1, stop

- if priority queue size = 1, return last stone weight
- if no stone left in queue, return 0.

For example:

Move Zeroes

**Time Complexity:**

`O(NlogN)`

**Space Complexity:**

`O(N)`

- N - the length of array stones.

**Java code**

class Solution {

public int lastStoneWeight(int[] stones) {

if (stones.length == 1) return stones[0];

// init max priority queue

PriorityQueue<Integer> pq = new PriorityQueue<>((a,b) -> b - a);

pq.addAll(Arrays.stream(stones).boxed().collect(Collectors.toList()));

// smash 2 heaviest stones each time

while (pq.size() > 1) {

int first = pq.poll();

int second = pq.poll();

if (first > second) pq.offer(first - second);

}

return pq.size() == 1 ? pq.poll() : 0;

}

}

**Python code**

class Solution:

def lastStoneWeight(self, stones: List[int]) -> int:

heap = [-num for num in stones]

# init max heap

heapq.heapify(heap)

# smash two heaviest stones from heap each time

while len(heap) > 1:

diff = heapq.heappop(heap) - heapq.heappop(heap)

if diff != 0:

heapq.heappush(heap, diff)

return -heap[0] if len(heap) == 1 else 0

Last modified 3yr ago