1046. 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. 1 <= stones.length <= 30
2. 1 <= stones[i] <= 1000

# @lc code=start
using LeetCode

using DataStructures

function last_stone_weight(stones::Vector{Int64})

Create a PriorityQueue with unique identifiers for each stone

    heap = PriorityQueue{Tuple{Int64, Int64}, Int64}()
    for (i, stone) in enumerate(stones)
        enqueue!(heap, (stone, i), -stone)  # Use negative stone value for max-heap
    end

    while length(heap) > 1

Extract the two largest stones

        largest, _ = dequeue!(heap)
        second_largest, _ = dequeue!(heap)

If they are not the same, calculate the difference and enqueue it

        if largest != second_largest
            enqueue!(heap, (largest - second_largest, length(heap) + 1), -(largest - second_largest))
        end
    end

Return the last stone or 0 if the heap is empty

    return isempty(heap) ? 0 : first(first(collect(keys(heap))))
end

last_stone_weight([2, 7, 4, 1, 8, 2])

# add your code here:
# @lc code=end

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