2
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Problem:

Find the longest sequence consisting only of 0's, and the longest sequence consisting only of 1's?

Solution:

import sys

longestSequenceForOne = None
startIndexForOne = None
longestSequenceForZero = None
startIndexForZero = None

def setLongestSequence(array, longestSequence, startIndex):
    global longestSequenceForOne
    global startIndexForOne
    global longestSequenceForZero
    global startIndexForZero

    if array[startIndex] == 1:
        if longestSequenceForOne == None or longestSequenceForOne < longestSequence:
            longestSequenceForOne = longestSequence
            startIndexForOne = startIndex
    elif array[startIndex] == 0:
        if longestSequenceForZero == None or longestSequenceForZero < longestSequence:        
            longestSequenceForZero = longestSequence
            startIndexForZero = startIndex

def findLongestSequence(array, size):
    """
    Find the longest sequence of 0's and 1's
    """
    longestSequence = 1
    beginIndex = None
    index = 1
    startIndex = 0
    sameElementCount = 1

    while (index < size):
        if array[index] == array[index-1]:
            sameElementCount += 1
        elif sameElementCount > longestSequence:
            longestSequence = sameElementCount
            startIndex = index - sameElementCount
            sameElementCount = 1
            setLongestSequence(array, longestSequence, startIndex)
        else:
            sameElementCount = 1
        index += 1
    if sameElementCount > longestSequence:
        longestSequence = sameElementCount
        startIndex = index - sameElementCount
        setLongestSequence(array, longestSequence, startIndex)

if __name__ == '__main__':
    size = int(input())
    if size < 2 or size > 100:
        sys.exit()
    array = [int(x) for x in input().split() if x == '1' or '0']
    if len(array) != size:
        sys.exit()
    findLongestSequence(array, size)
    print(startIndexForZero, longestSequenceForZero)
    print(startIndexForOne, longestSequenceForOne)

Correctness

Input

12
0 1 0 1 1 1 0 1 0 0 0 0

Output

8 4
3 3

Input
17
1 1 0 1 0 1 0 0 1 0 0 0 0 1 0 1 0

Output

9 4
0 2

Questions:

  1. Does the code require better error handling?

  2. Can this code get more pythonic?

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3
  • 1
    \$\begingroup\$ If you want to just increment by 1 each time, then for index in range(size) is both more pythonic and faster than while (index<size). Also, if your array is large, you can significantly improve the run time by incrementing by the current longest sequence instead of by 1. Check out this question of mine - it's not exactly the same, but the general idea of the algorithm still applies. \$\endgroup\$
    – Zyx
    Jun 29, 2017 at 18:08
  • \$\begingroup\$ @ZyTelevan Yup for index in range(1, len(array)) works \$\endgroup\$ Jun 29, 2017 at 18:30
  • \$\begingroup\$ you could use regex and shorten the code too \$\endgroup\$
    – depperm
    Jun 29, 2017 at 19:38

1 Answer 1

4
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  • Correctness

    Input
    1 1 1 1 0 0 0
    
    Output
    None, None
    0, 4
    

    The reason is obvious: the longestSequence does not distinguish between longest sequence of zeroes and longest sequence of ones, so 0 0 0 is not considered long enough.

  • General

    • Avoid globals; prefer returning values.

    • There is no need to pass size; a list already knows its size. For the same reason, size = int(input()) is redundant and reduces usability.

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1
  • \$\begingroup\$ longestSequence = 1 after setLongestSequence() worked \$\endgroup\$ Jun 29, 2017 at 18:17

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