Comparing methods to build an entire binayHeap from a list of keys

Question

While reading a post on building a Binary Heap. I got confused in approach1.

The author's approach1(O(n*log n)):

Given a list of keys, build a binary heap by inserting each key one at a time.
Since you are starting with a list of one item, the list is sorted and you could use binary search to find the right position to insert the next key at a cost of approximately O(logn) operations.
However, inserting an item in the middle of the list may require O(n) operations to shift the rest of the list over to make room for the new key.
Therefore, to insert n keys into the heap would require a total of O(nlogn) operations.

`

class BinHeap:
    def __init__(self):
        self.heapList = [0] 
        self.currentSize = 0

def percUp(self,i):
    while i // 2 > 0:
      if self.heapList[i] < self.heapList[i // 2]:
         tmp = self.heapList[i // 2]
         self.heapList[i // 2] = self.heapList[i]
         self.heapList[i] = tmp
      i = i // 2

def insert(self, k):
    self.heapList.append(k)
    self.currentSize = self.currentSize + 1
    self.percUp(self.currentSize)

I didn't understand why he needs to do Binary search to find the right position to insert the next when I can simply use the insert() on each key at a time and percUp() will take care of restoring the heap's property every time.Also, how my approch differs from his O(n*log n) approach:

def method1(list):

    newHeap = BinHeap()
    for key in list:
        newHeap.insert(key)

    return newHeap.heapList

list_keys= [9,6,5,2,3]
print('Method-1', method1(list_keys)[1:])

and get the result

Method-1 [2, 3, 6, 9, 5]

Please suggest where I am going wrong & what I am missing?

score 1 · Accepted Answer · answered Nov 23 '18 at 18:36

Your analysis is correct. The author is confused. He says:

To finish our discussion of binary heaps, we will look at a method to build an entire heap from a list of keys. The first method you might think of may be like the following. Given a list of keys, you could easily build a heap by inserting each key one at a time. Since you are starting with a list of one item, the list is sorted and you could use binary search to find the right position to insert the next key at a cost of approximately O(logn) operations. However, remember that inserting an item in the middle of the list may require O(n) operations to shift the rest of the list over to make room for the new key. Therefore, to insert n keys into the heap would require a total of O(nlogn) operations.

The discussion of binary search in that paragraph is irrelevant. There is no need ever to do binary search when creating a binary heap, and there's no situation when inserting an item into the binary heap that you need to do O(n) operations to make space for the new item. The whole point of the binary heap structure is to avoid that type of expensive insertion.

Slightly rewriting, and editing out the irrelevant portions of what the author wrote:

To finish our discussion of binary heaps, we will look at a method to build an entire heap from a list of keys. The first method you might think of may be like the following. Given a list of keys, you could easily build a heap by inserting each key one at a time, at a cost of O(log n) operations per insertion. Therefore, to insert n keys into the heap would require a total of O(n log n) operations.

Comparing methods to build an entire binayHeap from a list of keys

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