Soyo

Day 12 of 30 days Leetcode challenge

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


Solution

Use Priority queue (heap) to simulate the process.

class Solution {
    func lastStoneWeight(_ stones: [Int]) -> Int {
        var heap = Heap<Int>(elements: stones, priorityFunction: >) // O(n)
        
        while heap.count > 1 {
            let first = heap.dequeue()! // O(logn) time 
            let second = heap.dequeue()! // O(logn) time
            let newVal = abs(first - second)
            if newVal > 0 {
                heap.enqueue(newVal)
            }
        }
        
        if heap.count == 0 { return 0 }
        
        return heap.peek()!
    }
}

Detail about how to implement Priority Queue can be found here

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