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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