DevonMeyer212
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It's also important to note that you can get it by taking the modulo of subsequently decreasing parts, as another answer here indicates (- Y):
return ( ( ( n % 20 ) % 9 ) % 6 == 0 )
But keep in mind that the code above is exactly what the hardware does in this case. It subtracts 20 until the number is below 20, then subtracts 9 from that number until that number is below 9, and then subtracts six from that number until that number is below 6. If the remaining number is 0, return True.
Not really important overall, but being able to mention both in an interview would definitely give you some nerd points xD
This problem is the same as the ol' "minimum many quarters, dimes, nickels, and pennies do you need to make up x cents?" only quite a bit easier, because you don't need to keep track of a, b, c: you just need to determine if a, b, and c can exist such that their combination satisfies the conditional statement (6a + 9b + 20c == n)
And here it is in Python
def mcnuggets(n):
d = n
while d >= 20:
d = d - 20
while d >= 9:
d = d-9
while d >= 6:
d = d - 6
return d == 0
print mcnuggets(16)
Python has some ways of doing this very elegantly, but typically interviewers want you to implement it at its base level, as can be seen here:
def get_max_repeated(string):
chars = {}
max = 1
for c in string:
if c not in chars:
chars[c] = 1
else:
current = chars[c]+1
chars[c] = current
if current > max:
max = current
return max
print get_max_repeated("coffee tuffee")
Basically, think of the problem as a graph. For a string of length n, you have n sub-graphs. Each sub-graph then has n-1 sub-graphs. The total number of nodes in all of the sub-graphs is n!
So you start with a root and a remainder. You iteratively add each character from the remainder to the root, and then recurse down to the next level. When you have a remainder with only one character in it, you know you have hit a leaf, so you print the root plus the remainder.
In the case of no character repetitions, it's as simple as:
def get_perms(root, s):
if len(s) == 1:
print root+s
else:
for i in range(len(s)):
get_perms(root+s[i], s[:i]+s[i+1:])
get_perms("", "abcd")
I wrote a solution which also handles the case where the string includes duplicate characters:
def get_perms(root, s, items):
if len(s) == 1:
if root+s not in items:
items.append(root+s)
print root+s
else:
for i in range(len(s)):
get_perms(root+s[i], s[:i]+s[i+1:], items)
get_perms("", "abcdd", [])
It's pretty clear that this is O(n!) in time complexity, and if you have to handle duplicate characters, it becomes O(n!) space complexity as well.
Dynamic programming certainly could offer a solution here as well, but that requires storing all of the data. Therefore, in the case of no character repetitions, recursion will certainly be optimal, as you don't have to store anything other than the string. But in the case of repeated characters, perhaps there's a better way using dynamic programming.
This is my solution as well, but you encounter an issue whenever you have duplicate characters in the string. (i.e. abdcd).
Check my solution for an example of how to avoid this (but it requires maintaining a list of all options, therefore requiring O(n!) space complexity)
This is super easy to implement in python:
s1 = "110"
s2 = "01101"
def max_len(s1, s2)
[a, b] = [s1, s2] if len(s1) >= len(s2) else [s2, s1]
b = "".join(['0']*(len(a)-len(b)))+b
i = len(a) - 1
j = len(b) - 1
c = '0'
result = []
while i >= 0:
s = 0
for x in [a[i], b[j], c]:
if x == '1':
s = s + 1
c = '1' if s >=2 else '0'
result.insert(0, ('0' if s % 2 == 0 else '1'))
i = i - 1
j = j - 1
return "".join(result)
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I see a lot of
but as referenced here:
stackoverflow-dot-com/questions/3501382/checking-whether-a-variable-is-an-integer-or-not
it is better to assume your item is a list, and if it is not, run certain code on what you know, now, to be an integer. As can be seen here:
- DevonMeyer212 September 05, 2013