# What's the time complexity of functions in heapq library

## Question:

My question is from the solution in leetcode below, I can’t understand why it is `O(k+(n-k)log(k))`

.

Supplement: Maybe the complexity isn’t that, in fact I don’t know the time complexity of `heappush()`

and `heappop()`

```
# O(k+(n-k)lgk) time, min-heap
def findKthLargest(self, nums, k):
heap = []
for num in nums:
heapq.heappush(heap, num)
for _ in xrange(len(nums)-k):
heapq.heappop(heap)
return heapq.heappop(heap)
```

## Answers:

`heapq`

is a binary heap, with O(log n) `push`

and O(log n) `pop`

. See the heapq source code.

The algorithm you show takes O(n log n) to push all the items onto the heap, and then O((n-k) log n) to find the kth largest element. So the complexity would be O(n log n). It also requires O(n) extra space.

You can do this in O(n log k), using O(k) extra space by modifying the algorithm slightly. I’m not a Python programmer, so you’ll have to translate the pseudocode:

```
# create a new min-heap
# push the first k nums onto the heap
for the rest of the nums:
if num > heap.peek()
heap.pop()
heap.push(num)
# at this point, the k largest items are on the heap.
# The kth largest is the root:
return heap.pop()
```

The key here is that the heap contains just the largest items seen so far. If an item is smaller than the kth largest seen so far, it’s never put onto the heap. The worst case is O(n log k).

Actually, `heapq`

has a `heapreplace`

method, so you could replace this:

```
if num > heap.peek()
heap.pop()
heap.push(num)
```

with

```
if num > heap.peek()
heap.replace(num)
```

Also, an alternative to pushing the first `k`

items is to create a list of the first `k`

items and call `heapify`

. A more optimized (but still O(n log k)) algorithm is:

```
# create array of first `k` items
heap = heapify(array)
for remaining nums
if (num > heap.peek())
heap.replace(num)
return heap.pop()
```

You could also call `heapify`

on the entire array, then pop the first `n-k`

items, and then take the top:

```
heapify(nums)
for i = 0 to n-k
heapq.heappop(nums)
return heapq.heappop(nums)
```

That’s simpler. Not sure if it’s faster than my previous suggestion, but it modifies the original array. The complexity is O(n) to build the heap, then O((n-k) log n) for the pops. So it’s be O((n-k) log n). Worst case O(n log n).

heapify() actually takes linear time because the approach is different than calling heapq.push() N times.

heapq.push()/heapq.pop() takes log n time because it adjust all the nodes at a given hight/level.

when you pass an array in heapify() it makes sure that the left and right children of the node are already maintaining the heap property whether it is a min heap or max heap.

you can see this video:

https://www.youtube.com/watch?v=HqPJF2L5h9U

https://www.youtube.com/watch?v=B7hVxCmfPtM

Hope this would help.

Summarize from @Shivam purbia ‘s post:

- Using
`heaps.heapify()`

can reduce both**time**and**space**complexity because`heaps.heapify()`

is an in-place heapify and costs linear time to run it. - both
`heapq.heappush()`

and`heapq.heappop()`

cost**O(logN)**time complexity

Final code will be like this …

```
import heapq
def findKthLargest(self, nums, k):
heapq.heapify(nums) # in-place heapify -> cost O(N) time
for _ in range(len(nums)-k): # run (N-k) times
heapq.heappop(heap) # cost O(logN) time
return heapq.heappop(heap)
```

- Total time complexity is
**O((N – k)logN)** - Total space complexity is
**O(1)**

for just creating and heapify the elements, it’s O(nlogn).

But for just heapify the elements, it’s o(n).

In the question, popout the smallest from heap is not the best answer

lets say ur input has 1 million items, then u need to pop 1m – k time

instead, in python , we can use maxheap, you will only require to have O(k) on pop, instead O(n-k), when n is super large

```
def findKthLargest(self, nums: List[int], k: int) -> int:
_heapify_max(nums)
while k > 0:
val = _heappop_max(nums)
k-=1
if k == 0 :
return val
```