1
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Is the complexity of this code \$O(n^3)$? How can I improve the efficiency and reduce the complexity?

A = [3,2,4,2,4,2,4,5,3]
n= len(A)
def ALG2 (A,i,j):  #to check whether all entries between A[i] and A[j] are the same
    if i==j:
        return True     
    for k in range (i,j-1):
        if A[k] != A[k+1]:  
            return False
    return True

def  ALG1(A,n):   #find the max length of the arrays that are found in ALG2 
    l=0
    for i in range (0,n-1):
        for j in range (i+1,n-1):    
            if ALG2(A,i,j) and (j-i)>l:
                l= j-i

    return l

result = ALG1(A,n)



print (result)
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6
  • 2
    \$\begingroup\$ I'm not sure why this has a down vote :( Could you add a little more description on what this code is doing? If this is a programing challenge including the challenge description and a link would be all that's needed! \$\endgroup\$
    – Peilonrayz
    Feb 15, 2019 at 21:17
  • \$\begingroup\$ Yes, I believe if you have a loop in a function being called inside of a loop you must also add that as a nested loop. \$\endgroup\$ Feb 15, 2019 at 21:38
  • \$\begingroup\$ Welcome to Code Review! I changed the title so that it describes what the code does per site goals: "State what your code does in your title, not your main concerns about it.". Feel free to edit and give it a different title if there is something more appropriate. \$\endgroup\$ Feb 15, 2019 at 21:55
  • \$\begingroup\$ It is not a challenge. It is my assignment actually . It is my first day to use stackexchange. really sorry for causing any inconvenience \$\endgroup\$
    – Wing Ho Lo
    Feb 15, 2019 at 23:01
  • \$\begingroup\$ What is the expected result for A = [3,2,4,2,4,2,4,5,3]? \$\endgroup\$
    – vnp
    Feb 16, 2019 at 0:50

2 Answers 2

1
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  1. You should use better function names ALG1 and ALG2 don't really say what they're performing.
  2. You should keep the creation of A, n and result at the bottom of your code.
  3. You should keep your 'main' code in a main guard block:

    if __name__ == '__main__':
        A = [3,2,4,2,4,2,4,5,3]
        n= len(A)
        result = ALG1(A,n)
        print(result)
    
  4. You should follow PEP 8, so that it's easier for others to read your code.

  5. ALG2 can be simplified by slicing the array from i to j, A[i:j], and then checking if the set has a size of one.

    def ALG2(A, i, j):
        return len(set(a[i:j])) == 1
    
  6. Your entire code can be drastically simplified by using itertools.groupby.

    import itertools
    
    
    def ALG1(array):
        sizes = (
            sum(1 for _ in g)
            for _, g in itertools.groupby(array)
        )
        return max(sizes, default=0)
    
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2
  • \$\begingroup\$ wow so is the itertools installed in python? and what is the itertools really for? \$\endgroup\$
    – Wing Ho Lo
    Feb 16, 2019 at 17:05
  • \$\begingroup\$ @WingHoLo Yes it's installed with Python. It's a collection of functions to help with iterators. \$\endgroup\$
    – Peilonrayz
    Feb 16, 2019 at 17:08
-1
\$\begingroup\$

Your algorithm is indeed n^3, but you can do it in linear time. https://ideone.com/Ke3q5o

A = [3,2,5,4,4,4,4,2,4,4]

def findLongestContinuousSection(A):
    if len(A) == 0:
        return
    longestStart = 0
    longestStop = 0
    longestLength = 0
    longestVal = 0
    curStart = 0
    curStop = 0
    curLength = 1
    curVal = A[0]
    for k in range(1,len(A)-1):
        if curVal != A[k]: # record cur as longest
            longestVal = curVal
            longestStart = curStart
            longestStop = curStop
            longestLength = curLength
            curStart = k
            curStop = k
            curLength = 1
            curVal = A[k]
        else: # continue to build current streak
            curStop = k
            curLength += 1
    return (longestStart, longestStop, longestLength, longestVal)

print findLongestContinuousSection(A)
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3
  • \$\begingroup\$ I'm confused where the Code Review is. Also this code looks needlessly long. \$\endgroup\$
    – Peilonrayz
    Feb 16, 2019 at 11:56
  • \$\begingroup\$ I only focused on the complexity. \$\endgroup\$ Feb 16, 2019 at 12:57
  • \$\begingroup\$ so here we return 4 elements in a function. But it cannot be done in python or any programming at all right? \$\endgroup\$
    – Wing Ho Lo
    Feb 16, 2019 at 17:07

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