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1046. Last Stone Weight

Description

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 weight of the last remaining 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

Solutions

• Java
• C++
• Python
• Go
• TypeScript
• Javascript

• class Solution {
      public int lastStoneWeight(int[] stones) {
          PriorityQueue<Integer> q = new PriorityQueue<>((a, b) -> b - a);
          for (int x : stones) {
              q.offer(x);
          }
          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();
      }
  }
  

• class Solution {
  public:
      int lastStoneWeight(vector<int>& stones) {
          priority_queue<int> pq;
          for (int x : stones) {
              pq.push(x);
          }
          while (pq.size() > 1) {
              int y = pq.top();
              pq.pop();
              int x = pq.top();
              pq.pop();
              if (x != y) {
                  pq.push(y - x);
              }
          }
          return pq.empty() ? 0 : pq.top();
      }
  };
  

• class Solution:
      def lastStoneWeight(self, stones: List[int]) -> int:
          h = [-x for x in stones]
          heapify(h)
          while len(h) > 1:
              y, x = -heappop(h), -heappop(h)
              if x != y:
                  heappush(h, x - y)
          return 0 if not h else -h[0]
  
  

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

• 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;
  }
  
  

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

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