Easy
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 smallest possible weight of the left 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
class Solution {
public int lastStoneWeight(int[] stones) {
PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b-a); // max heap, Collections.reverseOrder()
for(int stone : stones)
pq.offer(stone);
while(pq.size() > 1) {
int x = pq.poll();
int y = pq.poll();
if(x != y) {
pq.offer(x-y); // abs not rqd as x would always be greater than equal to y
}
}
return pq.isEmpty()?0:pq.peek();
}
}