##### Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/1046.html

# 1046. Last Stone Weight (Easy)

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. 1 <= stones.length <= 30
2. 1 <= stones[i] <= 1000

Related Topics:
Heap, Greedy

## Solution 1. Max-root Heap

// OJ: https://leetcode.com/problems/last-stone-weight/
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
int lastStoneWeight(vector<int>& A) {
priority_queue<int> pq(begin(A), end(A));
while (pq.size() > 1) {
int a = pq.top();
pq.pop();
int b = pq.top();
pq.pop();
pq.push(a - b);
}
return pq.top();
}
};

• class Solution {
public int lastStoneWeight(int[] stones) {
PriorityQueue<Integer> priorityQueue = new PriorityQueue<Integer>(new Comparator<Integer>() {
public int compare(Integer num1, Integer num2) {
return num1 == null || num2 == null ? 0 : num2 - num1;
}
});
for (int stone : stones)
priorityQueue.offer(stone);
while (priorityQueue.size() > 1) {
int stone1 = priorityQueue.poll();
int stone2 = priorityQueue.poll();
if (stone1 > stone2)
priorityQueue.offer(stone1 - stone2);
}
return priorityQueue.size() == 0 ? 0 : priorityQueue.poll();
}
}

############

class Solution {
public int lastStoneWeight(int[] stones) {
PriorityQueue<Integer> q = new PriorityQueue<>((a, b) -> b - a);
for (int v : stones) {
q.offer(v);
}
while (q.size() > 1) {
int y = q.poll();
int x = q.poll();
if (x != y) {
q.offer(y - x);
}
}
return q.isEmpty() ? 0 : q.poll();
}
}

• // OJ: https://leetcode.com/problems/last-stone-weight/
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
int lastStoneWeight(vector<int>& A) {
priority_queue<int> pq(begin(A), end(A));
while (pq.size() > 1) {
int a = pq.top(); pq.pop();
int b = pq.top(); pq.pop();
int r = abs(a - b);
if (r) pq.push(r);
}
return pq.size() ? pq.top() : 0;
}
};

• class Solution:
def lastStoneWeight(self, stones: List[int]) -> int:
h = [-v for v in stones]
heapify(h)
while len(h) > 1:
x = heappop(h)
y = heappop(h)
if x != y:
heappush(h, x - y)
return 0 if len(h) == 0 else -h[0]

############

class Solution(object):
def lastStoneWeight(self, stones):
"""
:type stones: List[int]
:rtype: int
"""
stones = map(lambda x : -x, stones)
heapq.heapify(stones)
while len(stones) > 1:
x = heapq.heappop(stones)
if stones:
y = heapq.heappop(stones)
if x != y:
heapq.heappush(stones, -abs(x - y))
return 0 if not stones else -stones[0]

• func lastStoneWeight(stones []int) int {
q := &hp{stones}
heap.Init(q)
for q.Len() > 1 {
x, y := q.pop(), q.pop()
if x != y {
q.push(x - y)
}
}
if q.Len() > 0 {
return q.IntSlice[0]
}
return 0
}

type hp struct{ sort.IntSlice }

func (h hp) Less(i, j int) bool  { return h.IntSlice[i] > h.IntSlice[j] }
func (h *hp) Push(v interface{}) { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() interface{} {
a := h.IntSlice
v := a[len(a)-1]
h.IntSlice = a[:len(a)-1]
return v
}
func (h *hp) push(v int) { heap.Push(h, v) }
func (h *hp) pop() int   { return heap.Pop(h).(int) }

• /**
* @param {number[]} stones
* @return {number}
*/
var lastStoneWeight = function (stones) {
const pq = new MaxPriorityQueue();
for (const v of stones) {
pq.enqueue(v);
}
while (pq.size() > 1) {
const x = pq.dequeue()['priority'];
const y = pq.dequeue()['priority'];
if (x != y) {
pq.enqueue(x - y);
}
}
return pq.isEmpty() ? 0 : pq.dequeue()['priority'];
};


• function lastStoneWeight(stones: number[]): number {
const pq = new MaxPriorityQueue();
for (const x of stones) {
pq.enqueue(x);
}
while (pq.size() > 1) {
const y = pq.dequeue().element;
const x = pq.dequeue().element;
if (x !== y) {
pq.enqueue(y - x);
}
}
return pq.isEmpty() ? 0 : pq.dequeue().element;
}