What will be the time complexity of the following program

DoomLord Published at Dev

DoomLord

Can anyone find the time complexity of the following python 3.6 program :

I tried to find and my answer is O(N*n), where N is the range of the very first loop and n is the range of the second for loop which is inside the first for loop

This program is for https://www.codechef.com/FEB20B/problems/SNUG_FIT problem on codechef.

N=int(input())
for _ in range(N):
    S=0
    n=int(input())
    A=list(map(int, input().split()))[:n] #Assume amount of inputs to be strictly of n numbers
    B=list(map(int, input().split()))[:n]
    A.sort();A.reverse()
    B.sort();B.reverse()

    for i in range(0,len(A)): # Here range is n (same as len(A))
        if A[i]>=B[i]:
            S=S+B[i]
        else:
            S=S+A[i]
    print(S)

Faruk Hossain

@Pritam Banerjee wrote -

O(N( nlogn + n)) which effectively is O(Nlog(n))

Actually it is not true.

I am describing it more clearly along with the code.

N=int(input())
for _ in range(N):    # O(N)
    S=0
    n=int(input())
    # The following line is O(n)
    A=list(map(int, input().split()))[:n] #Assume amount of inputs to be strictly of n numbers
    # This line is also O(n)
    B=list(map(int, input().split()))[:n]
    A.sort();A.reverse()    # this is O(nlog(n))
    B.sort();B.reverse()    # this is also O(nlog(n))

    for i in range(0,len(A)): # Here range is n (same as len(A))  # this is O(n)
        if A[i]>=B[i]:
            S=S+B[i]
        else:
            S=S+A[i]
    print(S)

So overall O(N( n + n + nlog(n) + nlog(n) + n)

We can calculate in the following way -

  O(N( n + n + nlog(n) + nlog(n) + n)
= O(N( 3n + 2nlog(n))
= O(N (n + nlog(n)) (we can eliminate const number when determining complexity)
= O(N(nlog(n)) (as n > nlog(n), so we can eliminate the small part)
= O(N*nlog(n))

So overall complexity is O(N*nlog(n), not O(Nlog(n).

There are more difference between O(N*nlog(n) and O(Nlog(n).

Hope this post will help you a lot.

Thanks.

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