Last Stone Weight, an easy LeetCode problem
Happy Sunday, folks!
In this problem, there is a very simple data structure that can be used. Here is the problem: https://leetcode.com/problems/last-stone-weight/
The simplest data structure to solve this problem is a Heap, which I had implemented before as a Priority Queue. Enqueue all the stones, and start the process of dequeuing following the rules outlined in the problem statement. Straightforward. Cheers, ACC:
In this problem, there is a very simple data structure that can be used. Here is the problem: https://leetcode.com/problems/last-stone-weight/
We have a collection of rocks, each rock has a positive integer weight.
Each turn, we choose the two heaviest rocks 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 weightxis totally destroyed, and the stone of weightyhas new weighty-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 <= 301 <= stones[i] <= 1000
The simplest data structure to solve this problem is a Heap, which I had implemented before as a Priority Queue. Enqueue all the stones, and start the process of dequeuing following the rules outlined in the problem statement. Straightforward. Cheers, ACC:
public class Solution
{
public int LastStoneWeight(int[] stones)
{
PriorityQueue pq = new PriorityQueue(2000, true);
foreach (int s in stones) pq.Enqueue(s, s);
while (pq.Count> 0)
{
int first = (int)pq.Dequeue();
if (pq.Count == 0) return first;
int second = (int)pq.Dequeue();
if (first> second)
{
pq.Enqueue(first - second, first - second);
}
}
return 0;
}
}
/*By-the-book PriorityQueue*/
public class PriorityQueue
{
public struct HeapEntry
{
private object item;
private long priority;
public HeapEntry(object item, long priority)
{
this.item = item;
this.priority = priority;
}
public object Item
{
get
{
return item;
}
}
public long Priority
{
get
{
return priority;
}
}
}
private long count;
private long capacity;
private bool descending; //Means that the max element at the top
private HeapEntry[] heap;
public long Count
{
get
{
return this.count;
}
}
public PriorityQueue(long capacity, bool descending)
{
this.capacity = capacity;
this.descending = descending;
heap = new HeapEntry[this.capacity];
}
public object Dequeue()
{
object result = heap[0].Item;
count--;
trickleDown(0, heap[count]);
return result;
}
public void Enqueue(object item, long priority)
{
count++;
bubbleUp(count - 1, new HeapEntry(item, priority));
}
private void bubbleUp(long index, HeapEntry he)
{
long parent = (index - 1) / 2;
// note: (index> 0) means there is a parent
while (
(index> 0) &&
((this.descending && heap[parent].Priority < he.Priority) || (!this.descending && heap[parent].Priority> he.Priority))
)
{
heap[index] = heap[parent];
index = parent;
parent = (index - 1) / 2;
}
heap[index] = he;
}
private void trickleDown(long index, HeapEntry he)
{
long child = (index * 2) + 1;
while (child < count) { if ( ((child + 1) < count) && ((this.descending && heap[child].Priority < heap[child + 1].Priority) || (!this.descending && heap[child].Priority> heap[child + 1].Priority))
)
{
child++;
}
heap[index] = heap[child];
index = child;
child = (index * 2) + 1;
}
bubbleUp(index, he);
}
}
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