I have two strings: one a word and one a scramble of letters. I want to see if this scramble of letters holds enough letters to spell the word. I have come up with an algorithm to do this but it isn't efficient enough and I was hoping I could get some help making it faster.
Here's what I have so far:
s1 = 'hypochondriac'
s2 = 'yqhpwoewnqlchpijcdrxpoa'
temp = list(s1)
for X in s2:
for Y in temp:
if X == Y:
X = '@'
if temp == :
print('Found ', s1)
Your code is not efficient, no, because you iterate in a double loop. For each letter in
s1, in the worst-case scenario (no matches) you loop over all of
Counter object instead; these act as multi-sets, where you can both test if a character is present in O(1) time and manage remaining counts:
from collections import Counter def contains(s1, s2): s2set = Counter(s2) for c in s1: count = s2set[c] if not c: return False if count == 1: del s2set[c] else: s2set[c] = count - 1 return True
You can also turn
s1 into a multi-set, and check if the multi-set for
s2 contains enough letters for each entry:
def contains(s1, s2): s1set = Counter(s1) s2set = Counter(s2) for c, count in s1set.items(): if count > s2set[c]: return False return True
The latter can be reduced further with the
all() function, which returns
False early if any of the results it is passed is
def contains(s1, s2): s2set = Counter(s2) return all(count <= s2set[c] for c, count in Counter(s1).items())
In all these you only have to iterate over both
s2 once (either directly or to produce the multi-set).
Demo of the latter:
>>> from collections import Counter >>> def contains(s1, s2): ... s2set = Counter(s2) ... return all(count <= s2set[c] for c, count in Counter(s1).items()) ... >>> s1 = 'hypochondriac' >>> s2 = 'yqhpwoewnqlchpijcdrxpoa' >>> contains(s1, s2) True >>> contains(s1 + 'b', s2) False