The following code to find the k largest elements of an array is causing a TLE error. How can I optimize it to make it run faster?
import heapq
for _ in range(int(input())):
n,k=map(int,input().split())
lists=list(map(int,input().split()))
heapq.heapify(lists)
for i in range(k+1):
klargest=heapq.nlargest(i,lists)
print(*klargest)
for i in range(k+1):
klargest=heapq.nlargest(i,lists)
The time complexity of each klargest operation is O(k*log n)) where n is the number of elements in heap. In the above code snippet, this operation is running for k+1 times for the values [0,k].
Calculating the time for the loop:
iteration value (Time)
i == 0 (0*log(n))
i == 1 (1*log(n))
i == 2 (2*log(n))
....
i == k-1 ((k-1)*log(n))
i == k ((k)*log(n))
Total time will be the sum of time taken in each operation = (0.log(n)) + (1*log(n)) + .... + ((k-1)*log(n)) + ((k)*log(n))
Total time = (0+1+2...+(k-1)+k)log(n) = ((k(k+1))/2)*log(n)
Total time ~~ O(k^2*(log(n)))
That's why the above code results in TLE.
OPTIMISED APPROACH:
import heapq
for _ in range(int(input())):
n,k=map(int,input().split())
lists=list(map(int,input().split()))
heapq.heapify(lists)
for i in range(n-k):
heapq.heappop(lists)
klargest = list(lists) # converting heap to list
print(*klargest)
As Inbuilt heap in python is min-heap. So the above code is popping out minimum n-k elements from lists. Popping out each operation will take log(n) time. Thus total time will be ~~ (n-k)*logn. The remaining k elements in the heap are the klargest elements that we want to find.
Thus, the time complexity for the above solution is O((n-k)*log(n)) == O(nlog(n)) which is a optimised time complexity.