leetcode_gym/_problems/lastStoneWeight.md

layout, title, last_solved, category, leetcode_url, status

layout	title	last_solved	category	leetcode_url	status
page	Last Stone Weight		heap	https://leetcode.com/problems/last-stone-weight	Attempted

Problem

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.

Solution

{% highlight python %}

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

while len(stones)>=2:

    # negate values for max heap
    stones = list(map(lambda x: -x if x>0 else x, stones))
    # stones = [-x for x in stones if x>0 else x]
    print(stones)

    # get top 2 heaviest stones
    heapq.heapify(stones)
    print(stones)
    heavy1 = heapq.heappop(stones)
    heavy2 = heapq.heappop(stones)

    # smash the heaviest stones
    diff = abs(heavy1-heavy2)
    heapq.heappush(stones, diff)
    print(stones)

print(stones)
return stones[0]

{% endhighlight %}