A returned string that is not a substring of other strings - is this possible in time less than O (n ^ 2)?

Question

You are given an array of strings. you need to return only those lines that are not substrings of other lines in the array. Login - ['abc','abcd','ab','def','efgd']. There should be a way out - 'abcd'and 'efgd' I came up with a solution in python with O (n ^ 2) time complexity. Is there a possible solution that gives less time complexity? My decision:

def sub(l,s):
  l1=l
  for i in range (len(l)):
        l1[i]=''.join(sorted(l1[i]))
  for i in l1:      
         if s in i:
              return True
  return False

def main(l):
      for i in range(len(l)):
            if sub(l[0:i-1]+l[i+1:],l[i])==False:
                  print l[i]


main(['abc','abcd','ab','def','efgd'])                  

Answer 1

Is memory a problem? You can turn to the tried and true ... TRIE!

Create a suffix tree!

Given your input ['abc','abcd','ab','def','efgd']

We would have a tree

              _
            / | \
           a  e  d
          /   |   \
         b*   f    e
        /     |     \
       c*     g      f*
      /       |
     d*       d*

DFS (- ) , abcd, efgd def

, O(n*m). O(n^2), .

.

deg

              _
            / | \
           a  e  d
          /   |   \
         b*   f    e
        /     |   / \
       c*     g  g*   f*
      /       |
     d*       d*

DFS , .

from typing import List


class Trie(object):
    class Leaf(object):
        def __init__(self, data, is_key):
            self.data = data
            self.is_key = is_key
            self.children = []

        def __str__(self):
            return "{}{}".format(self.data, "*" if self.is_key else "")

    def __init__(self, keys):
        self.root = Trie.Leaf('', False)
        for key in keys:
            self.add_key(key)

    def add_key(self, key):
        self._add(key, self.root.children)

    def has_suffix(self, suffix):
        leaf = self._find(suffix, self.root.children)

        if not leaf:
            return False

        # This is only a suffix if the returned leaf has children and itself is not a key
        if not leaf.is_key and leaf.children:
            return True

        return False

    def includes_key(self, key):
        leaf = self._find(key, self.root.children)

        if not leaf:
            return False

        return leaf.is_key

    def delete(self, key):
        """
        If the key is present as a unique key as in it does not have any children nor are any of its nodes comprised of
         we should delete all of the nodes up to the root
        If the key is a prefix of another long key in the trie, umark the leaf node
        if the key is present in the trie and contains no children but contains nodes that are keys we should delete all
         nodes up to the first encountered key
        :param key:
        :return:
        """

        if not key:
            raise KeyError

        self._delete(key, self.root.children, None)

    def _delete(self, key, children: List[Leaf], parents: (List[Leaf], None), key_idx=0, parent_key=False):
        if not parents:
            parents = [self.root]

        if key_idx >= len(key):
            return

        key_end = True if len(key) == key_idx + 1 else False
        suffix = key[key_idx]
        for leaf in children:
            if leaf.data == suffix:
                # we have encountered a leaf node that is a key we can't delete these
                # this means our key shares a common branch
                if leaf.is_key:
                    parent_key = True

                if key_end and leaf.children:
                    # We've encountered another key along the way
                    if parent_key:
                        leaf.is_key = False
                    else:
                        # delete all nodes recursively up to the top of the first node that has multiple children
                        self._clean_parents(key, key_idx, parents)
                elif key_end and not leaf.children:
                    # delete all nodes recursively up to the top of the first node that has multiple children
                    self._clean_parents(key, key_idx, parents)

                # Not at the key end so we need to keep traversing the tree down
                parents.append(leaf)
                self._delete(key, leaf.children, parents, key_idx + 1, key_end)

    def _clean_parents(self, key, key_idx, parents):
        stop = False
        while parents and not stop:
            p = parents.pop()

            # Need to stop processing a removal at a branch
            if len(p.children) > 1:
                stop = True

            # Locate our branch and kill its children
            for i in range(len(p.children)):
                if p.children[i].data == key[key_idx]:
                    p.children.pop(i)
                    break
            key_idx -= 1

    def _find(self, key, children: List[Leaf]):
        if not key:
            raise KeyError

        match = False
        if len(key) == 1:
            match = True

        suffix = key[0]
        for leaf in children:
            if leaf.data == suffix and not match:
                return self._find(key[1:], leaf.children)
            elif leaf.data == suffix and match:
                return leaf
        return None

    def _add(self, key, children: List[Leaf]):
        if not key:
            return

        is_key = False
        if len(key) == 1:
            is_key = True

        suffix = key[0]
        for leaf in children:
            if leaf.data == suffix:
                self._add(key[1:], leaf.children)
                break
        else:
            children.append(Trie.Leaf(suffix, is_key))
            self._add(key[1:], children[-1].children)

        return

    @staticmethod
    def _has_children(leaf):
        return bool(leaf.children)


def main():
    keys = ['ba', 'bag', 'a', 'abc', 'abcd', 'abd', 'xyz']
    trie = Trie(keys)
    print(trie.includes_key('ba'))  # True
    print(trie.includes_key('b'))  # False
    print(trie.includes_key('dog'))  # False
    print(trie.has_suffix('b'))  # True
    print(trie.has_suffix('ab'))  # True
    print(trie.has_suffix('abd'))   # False

    trie.delete('abd')  # Should only remove the d
    trie.delete('a')    # should unmark a as a key
    trie.delete('ba')   # should remove the ba trie
    trie.delete('xyz')  # Should remove the entire branch
    trie.delete('bag')  # should only remove the g

    print(trie)

if __name__ == "__main__":
    main()

, trie DFS; , .

Answer 2

Aho-Corasick O(n) ( " ", ). , , ( , , ..).

pyahocorasick, , , :

import ahocorasick

# This code assumes no duplicates in mystrings (which would make them mutually
# substrings). Easy to handle if needed, but simpler to avoid for demonstration

mystrings = ['abc','abcd','ab','def','efgd']

# Build Aho-Corasick automaton, involves O(n) (in combined length of mystrings) work
# Allows us to do single pass scans of a string for all strings in mystrings
# at once
aut = ahocorasick.Automaton()
for s in mystrings:
    # mapping string to itself means we're informed directly of which substring
    # we hit as we scan
    aut.add_word(s, s)
aut.make_automaton()

# Initially, assume all strings are non-substrings
nonsubstrings = set(mystrings)

# Scan each of mystrings for substrings from other mystrings
# This only involves a single pass of each s in mystrings thanks to Aho-Corasick,
# so it only O(n+m) work, where n is again combined length of mystrings, and
# m is the number of substrings found during the search
for s in mystrings:
    for _, substr in aut.iter(s):
        if substr != s:
           nonsubstrings.discard(substr)

# A slightly more optimized version of the above loop, but admittedly less readable:
# from operator import itemgetter
# getsubstr = itemgetter(1)
# for s in mystrings:
#     nonsubstrings.difference_update(filter(s.__ne__, map(getsubstr, aut.iter(s))))

for nonsub in nonsubstrings:
    print(nonsub)

. , , pyahocorasick , , , , .

Answer 3

. , . . O (n log n)

:

def not_substrings(l):
    mask = [True]*len(l)
    for i in range(len(l)):
        if not mask[i]:
            continue
        for j in range(i+1, len(l)):
            if len(l[i]) > len(l[j]):
                if l[j] in l[i]:
                    mask[j] = False
            elif l[j] == l[i]:
                mask[j] = False
                mask[i] = False
            else:
                if l[i] in l[j]:
                    mask[i] = False
        if mask[i]:
            print l[i]

, . , , [True]*len(l). , n log n, , .

Answer 4

Use an object setto save all substrings. It's faster, but use a lot of memory, if each line is short, you can try this.

import string
import random
from itertools import combinations

def get_substrings(w):
    return (w[s:e] for s, e in combinations(range(len(w)+1), 2))

def get_not_substrings(words):
    words = sorted(set(words), key=len, reverse=True)
    substrings = set()

    for w in words:
        if w not in substrings:
            yield w
            substrings.update(get_substrings(w))

words = ["".join(random.choice(string.ascii_lowercase) 
    for _ in range(random.randint(1, 12))) for _ in range(10000)]
res = list(get_not_substrings(words))