I am very new to programming, so please bear with me...I have been learning Python and I just did an assessment that involved looping through a list using your current value as the next index value to go to while looping. This is roughly what the question was:
You have a zero-indexed array length N of positive and negative integers. Write a function that loops through the list, creates a new list, and returns the length of the new list. While looping through the list, you use your current value as the next index value to go to. It stops looping when A[i] = -1
For example:
A[0] = 1
A[1] = 4
A[2] = -1
A[3] = 3
A[4] = 2
This would create:
newlist = [1, 4, 2, -1]
len(newlist) = 4
It was timed and I was not able to finish, but this is what I came up with. Any criticism is appreciated. Like I said I am new and trying to learn. In the meantime, I will keep looking. Thanks in advance!
def sol(A):
i = 0
newlist = []
for A[i] in range(len(A)):
e = A[i]
newlist.append(e)
i == e
if A[i] == -1:
return len(newlist)
This might be the easiest way to do it if your looking for the least lines of code to write.
A = [1,4,-1,3,2]
B = []
n = 0
while A[n] != -1:
B.append(A[n])
n = A[n]
B.append(-1)
print(len(B))
First of all, note that for A[i] in range(len(A)) is a pattern you certainly want to avoid, as it is an obscure construct that will modify the list A by storing increasing integers into A[i]. To loop over elements of A, use for val in A. To loop over indices into A, use for ind in xrange(len(A)).
The for loop, normally the preferred Python looping construct, is not the right tool for this problem because the problem requires iterating over the sequence in an unpredictable order mandated by the contents of the sequence. For this, you need to use the more general while loop and manage the list index yourself. Here is an example:
def extract(l):
newlist = []
ind = 0
while l[ind] != -1:
newlist.append(l[ind])
ind = l[ind]
newlist.append(-1) # the problem requires the trailing -1
print newlist # for debugging
return len(newlist)
>>> extract([1, 4, -1, 3, 2])
[1, 4, 2, -1]
4
Note that collecting the values into the new list doesn't really make sense in any kind of real-world scenario because the list is not visible outside the function in any way. A more sensible implementation would simply increment a counter in each loop pass and return the value of the counter. But since the problem explicitly requests maintaining the list, code like the above will have to do.
It's simpler to just use a while loop:
data = [1,4,-1,3,2]
ls = []
i = 0
steps = 0
while data[i] != -1:
ls.append(data[i])
i = data[i]
steps += 1
assert steps < len(data), "Infinite loop detected"
ls.append(-1)
print ls, len(ls)
Related
A=[2,3,4,1] B=[1,2,3,4]
I need to find how many elements of list A appear before than the same element of list B. In this case values 2,3,4 and the expected return would be 3.
def count(a, b):
muuttuja = 0
for i in range(0, len(a)-1):
if a[i] != b[i] and a[i] not in b[:i]:
muuttuja += 1
return muuttuja
I have tried this kind of solution but it is very slow to process lists that have great number of values. I would appreciate some suggestions for alternative methods of doing the same thing but more efficiently. Thank you!
If both the lists have unique elements you can make a map of element (as key) and index (as value). This can be achieved using dictionary in python. Since, dictionary uses only O(1) time for lookup. This code will give a time complexity of O(n)
A=[2,3,4,1]
B=[1,2,3,4]
d = {}
count = 0
for i,ele in enumerate(A) :
d[ele] = i
for i,ele in enumerate(B) :
if i > d[ele] :
count+=1
Use a set of already seen B-values.
def count(A, B):
result = 0
seen = set()
for a, b in zip(A, B):
seen.add(b)
if a not in seen:
result += 1
return result
This only works if the values in your lists are immutable.
Your method is slow because it has a time complexity of O(N²): checking if an element exists in a list of length N is O(N), and you do this N times. We can do better by using up some more memory instead of time.
First, iterate over b and create a dictionary mapping the values to the first index that value occurs at:
b_map = {}
for index, value in enumerate(b):
if value not in b_map:
b_map[value] = index
b_map is now {1: 0, 2: 1, 3: 2, 4: 3}
Next, iterate over a, counting how many elements have an index less than that element's value in the dictionary we just created:
result = 0
for index, value in enumerate(a):
if index < b_map.get(value, -1):
result += 1
Which gives the expected result of 3.
b_map.get(value, -1) is used to protect against the situation when a value in a doesn't occur in b, and you don't want to count it towards the total: .get returns the default value of -1, which is guaranteed to be less than any index. If you do want to count it, you can replace the -1 with len(a).
The second snippet can be replaced by a single call to sum:
result = sum(index < b_map.get(value, -1)
for index, value in enumerate(a))
You can make a prefix-count of A, which is an array where for each index you keep track of the number of occurrences of each element before the index.
You can use this to efficiently look-up the prefix-counts when looping over B:
import collections
A=[2,3,4,1]
B=[1,2,3,4]
prefix_count = [collections.defaultdict(int) for _ in range(len(A))]
prefix_count[0][A[0]] += 1
for i, n in enumerate(A[1:], start=1):
prefix_count[i] = collections.defaultdict(int, prefix_count[i-1])
prefix_count[i][n] += 1
prefix_count_b = sum(prefix_count[i][n] for i, n in enumerate(B))
print(prefix_count_b)
This outputs 3.
This still could be O(NN) because of the copy from the previous index when initializing the prefix_count array, if someone knows a better way to do this, please let me know*
I have read in several places that it is bad practice to modify an array/list during iteration. However many common algorithms appear to do this. For example Bubble Sort, Insertion Sort, and the example below for finding the minimum number of swaps needed to sort a list.
Is swapping list items during iteration an exception to the rule? If so why?
Is there a difference between what happens with enumerate and a simple for i in range(len(arr)) loop in this regard?
def minimumSwaps(arr):
ref_arr = sorted(arr)
index_dict = {v: i for i,v in enumerate(arr)}
swaps = 0
for i,v in enumerate(arr):
print("i:", i, "v:", v)
print("arr: ", arr)
correct_value = ref_arr[i]
if v != correct_value:
to_swap_ix = index_dict[correct_value]
print("swapping", arr[to_swap_ix], "with", arr[i])
# Why can you modify list during iteration?
arr[to_swap_ix],arr[i] = arr[i], arr[to_swap_ix]
index_dict[v] = to_swap_ix
index_dict[correct_value] = i
swaps += 1
return swaps
arr = list(map(int, "1 3 5 2 4 6 7".split(" ")))
assert minimumSwaps(arr) == 3
An array should not be modified while iterating through it, because iterators cannot handle the changes. But there are other ways to go through an array, without using iterators.
This is using iterators:
for index, item in enumerate(array):
# don't modify array here
This is without iterators:
for index in range(len(array)):
item = array[index]
# feel free to modify array, but make sure index and len(array) are still OK
If the length & index need to be modified when modifying an array, do it even more "manually":
index = 0
while index < len(array):
item = array[index]
# feel free to modify array and modify index if needed
index += 1
Modifying items in a list could sometimes produce unexpected result but it's perfectly fine to do if you are aware of the effects. It's not unpredictable.
You need to understand it's not a copy of the original list you ar iterating through. The next item is always the item on the next index in the list. So if you alter the item in an index before iterator reaches it the iterator will yield the new value.
That means if you for example intend to move all items one index up by setting item at index+1 to current value yielded from enumerate(). Then you will end up with a list completely filled with the item originally on index 0.
a = ['a','b','c','d']
for i, v in enumerate(a):
next_i = (i + 1) % len(a)
a[next_i] = v
print(a) # prints ['a', 'a', 'a', 'a']
And if you appending and inserting items to the list while iterating you may never reach the end.
In your example, and as you pointed out in a lot of algorithms for e.g. combinatoric and sorting, it's a part of the algorithm to change the forthcoming items.
An iterator over a range as in for i in range(len(arr)) won't adapt to changes in the original list because the range is created before starting and is immutable. So if the list has length 4 in the beginning, the loop will try iterate exactly 4 times regardless of changes of the lists length.
# This is probably a bad idea
for i in range(len(arr)):
item = arr[i]
if item == 0:
arr.pop()
# This will work (don't ask for a use case)
for i, item in enumerate(arr):
if item == 0:
arr.pop()
I should not use advance function, as this is a logical test during interview.
Trying to remove all digits which appear more than once in array.
testcase:
a=[1,1,2,3,2,4,5,6,7]
code:
def dup(a):
i=0
arraySize = len(a)
print(arraySize)
while i < arraySize:
#print("1 = ",arraySize)
k=i+1
for k in range(k,arraySize):
if a[i] == a[k]:
a.remove(a[k])
arraySize -= 1
#print("2 = ",arraySize)
i += 1
print(a)
result should be : 1,2,3,4,5,6,7
But i keep getting index out of range. i know that it is because the array list inside the loop changed, so the "while" initial index is different with the new index.
The question is : any way to sync the new index length (array inside the loop) with the parent loop (index in "while" loop) ?
The only thing i can think of is to use function inside the loop.
any hint?
Re-Calculating Array Size Per Iteration
It looks like we have a couple issues here. The first issue is that you can't update the "stop" value in your inner loop (the range function). So first off, let's remove that and use another while loop to give us the ability to re-calculate our array size every iteration.
Re-Checking Values Shifted Into Removed List Spot
Next, after you fix that you will run into a larger issue. When you use remove it moves a value from the end of the list or shifts the entire list to the left to use the removed spot, and you are not re-checking the value that got moved into the old values removed spot. To resolve this, we need to decrement i whenever we remove an element, this makes sure we are checking the value that gets placed into the removed elements spot.
remove vs del
You should use del over remove in this case. remove iterates over the list and removes the first occurrence of the value and it looks like we already know the exact index of the value we want to remove. remove might work, but it's usage here over complicates things a bit.
Functional Code with Minimal Changeset
def dup(a):
i = 0
arraySize = len(a)
print(arraySize)
while i < arraySize:
k = i + 1
while k < arraySize: # CHANGE: use a while loop to have greater control over the array size.
if a[i] == a[k]:
print("Duplicate found at indexes %d and %d." % (i, k))
del a[i] # CHANGE: used del instead of remove.
i -= 1 # CHANGE: you need to recheck the new value that got placed into the old removed spot.
arraySize -= 1
break
k += 1
i += 1
return a
Now, I'd like to note that we have some readability and maintainability issues with the code above. Iterating through an array and manipulating the iterator in the way we are doing is a bit messy and could be prone to simple mistakes. Below are a couple ways I'd implement this problem in a more readable and maintainable manner.
Simple Readable Alternative
def remove_duplicates(old_numbers):
""" Simple/naive implementation to remove duplicate numbers from a list of numbers. """
new_numbers = []
for old_number in old_numbers:
is_duplicate = False
for new_number in new_numbers:
if old_number == new_number:
is_duplicate = True
if is_duplicate == False:
new_numbers.append(old_number)
return new_numbers
Optimized Low Level Alternative
def remove_duplicates(numbers):
""" Removes all duplicates in the list of numbers in place. """
for i in range(len(numbers) - 1, -1, -1):
for k in range(i, -1, -1):
if i != k and numbers[i] == numbers[k]:
print("Duplicate found. Removing number at index: %d" % i)
del numbers[i]
break
return numbers
You could copy contents in another list and remove duplicates from that and return the list. For example:
duplicate = a.copy()
f = 0
for j in range(len(a)):
for i in range(len(duplicate)):
if i < len(duplicate):
if a[j] == duplicate[i]:
f = f+1
if f > 1:
f = 0
duplicate.remove(duplicate[i])
f=0
print(duplicate)
I'm trying to write a Python function that counts the number of entries in a list that occur exactly once.
For example, given the list [17], this function would return 1. Or given [3,3,-22,1,-22,1,3,0], it would return 1.
** Restriction: I cannot import anything into my program.
The incorrect code that I've written so far: I'm going the double-loop route, but the index math is getting over-complicated.
def count_unique(x):
if len(x) == 1:
return 1
i = 0
j = 1
for i in range(len(x)):
for j in range(j,len(x)):
if x[i] == x[j]:
del x[j]
j+1
j = 0
return len(x)
Since you can't use collections.Counter or sorted/itertools.groupby apparently (one of which would usually be my go to solution, depending on whether the inputs are hashable or sortable), just simulate roughly the same behavior as a Counter, counting all elements and then counting the number of elements that appeared only once at the end:
def count_unique(x):
if len(x) <= 1:
return len(x)
counts = {}
for val in x:
counts[val] = counts.get(val, 0) + 1
return sum(1 for count in counts.values() if count == 1)
lst = [3,3,-22,1,-22,1,3,0]
len(filter(lambda z : z[0] == 1,
map(lambda x : (len(filter(lambda y : y == x, lst)), x), lst)))
sorry :)
Your solution doesn't work because you are doing something weird. Deleting things from a list while iterating through it, j+1 makes no sense etc. Try adding elements that are found to be unique to a new list and then counting the number of things in it. Then figure out what my solution does.
Here is the O(n) solution btw:
lst = [3,3,-22,1,-22,1,3,0,37]
cnts = {}
for n in lst:
if n in cnts:
cnts[n] = cnts[n] + 1
else:
cnts[n] = 1
count = 0
for k, v in cnts.iteritems():
if v == 1:
count += 1
print count
A more simple and understandable solution:
l = [3, 3, -22, 1, -22, 1, 3, 0]
counter = 0
for el in l:
if l.count(el) == 1:
counter += 1
It's pretty simple. You iterate over the items of the list. Then you look if the element is exactly one time in the list and then you add +1. You can improve the code (make liste comprehensions, use lambda expressions and so on), but this is the idea behind it all and the most understandable, imo.
you are making this overly complicated. try using a dictionary where the key is the element in your list. that way if it exists it will be unique
to add to this. it is probably the best method when looking at complexity. an in lookup on a dictionary is considered O(1), the for loop is O(n) so total your time complexity is O(n) which is desirable... using count() on a list element does a search on the whole list for every element which is basically O(n^2)... thats bad
from collections import defaultdict
count_hash_table = defaultdict(int) # i am making a regular dictionary but its data type is an integer
elements = [3,3,-22,1,-22,1,3,0]
for element in elements:
count_hash_table[element] += 1 # here i am using that default datatype to count + 1 for each type
print sum(c for c in count_hash_table.values() if c == 1):
There is method on lists called count.... from this you can go further i guess.
for example:
for el in l:
if l.count(el) > 1:
continue
else:
print("found {0}".format(el))
lst = [3, 4, 3, 5, 3]
def lstmaker(lst):
lst1 = []
x = 0
while x < len(lst):
if lst[x] == lst[x+1]:
lst1.append(lst[x])
else:
pass
x+=1
print lst1
lstmaker(lst)
I tried to make this simple program to find the the mode of list, but it just throws the index of out range error.
Thanks
For the last value of x, lst[x+1] x+1 is out of range. The while loop should be while x < len(lst) -1.
As a side note, to calculate the mode, you can simply do: max(set(lst), key=lst.count) as shown here.
The logic is incorrect.
For starters, the reason you're getting an index out of range issue is because of the line
if lst[x] == lst[x+1]
x is correctly incremented throughout your loop, but when x is at the last index, that +1 bit accesses an index that isn't in the list (e.g. index 5 of a list of size 5).
Additionally, what you were actually doing within the loop doesn't appear to be getting you towards a value for the mode. The mode is the element(s) in a list that occurs the most. One approach to tackling this problem could be using a dictionary (dict()) where the "keys" are the elements in your list, and the "values" are the amount of times each element occurs.
Try something like this:
# lst defined up here
occurrences = dict()
for x in lst:
if x in occurrences:
occurrences[x] += 1
else:
occurrences[x] = 1
mode = occurrences.keys()[0]
for k in occurrences:
if occurrences[k] >= mode:
mode = k
print(mode) # or return, etc.
This is perhaps not the most "Pythonic" solution, though it is an intuitive break-down of the logic involved in finding the mode, at least as if you were to do it by hand on paper.