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Last Stone Weight - Heap - Easy - LeetCode

Last Stone Weight - Heap - Easy - LeetCode

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

 

public class Solution {
    int[] heap;
    int index = 1;
    public int LastStoneWeight(int[] stones)
    {
        if(stones.Length==0){
            return 0;
        }
        
        if(stones.Length==1){
            return stones[0];
        }
        
        heap = new int[stones.Length + 1];
        for (int i = 0; i < stones.Length; i++)
        {
            Insert(stones[i]);
        }

        int diff = 0;
        while (index-1 > 1)
        {
            int max1 = ExtractMax();
            int max2 = ExtractMax();
            diff = max1 - max2;
            Insert(diff);
        }

        return diff;
    }

    private int ExtractMax()
    {
        int max = heap[1];
        index--;
        heap[1] = heap[index];            
        HeapifyDown(1);
        return max;
    }

    private void HeapifyDown(int idx)
    {
        int left = 2 * idx;
        int right = 2 * idx + 1;
        if (left > index)
        {
            return;
        }
        if (right > index)
        {
            if (heap[left] > heap[idx])
            {
                int temp = heap[left];
                heap[left] = heap[idx];
                heap[idx] = temp;
                HeapifyDown(left);                    
            }
            return;
        }

        if (heap[left] > heap[right] && heap[left] > heap[idx])
        {
            int temp = heap[left];
            heap[left] = heap[idx];
            heap[idx] = temp;
            HeapifyDown(left);
        }
        else if (heap[right] > heap[idx])
        {
            int temp = heap[right];
            heap[right] = heap[idx];
            heap[idx] = temp;
            HeapifyDown(right);
        }
    }

    private void Insert(int val)
    {
        heap[index] = val;
        HeapifyUp(index);
        index++;
    }

    private void HeapifyUp(int idx)
    {
        int parent = idx / 2;
        if (parent < 1)
        {
            return;
        }

        if (heap[parent] < heap[idx])
        {
            int temp = heap[parent];
            heap[parent] = heap[idx];
            heap[idx] = temp;

            HeapifyUp(parent);
        }
    }
}

Time Complexity: O(nlogn)

Space Complexity: O(n)