I'm probably missing something major here, but it seems like in most educational contexts online like Leetcode and courses and such, when the examples iterate over an array they do so like:
array = [1,2,3,4,5]
for i in range(len(array)):
print(array[i])rather than just iterating the thing:
for i in array:
print(i)I'm guessing it's to do with time complexity since indexing arrays is quick, but what's the problem with the latter approach?
python - Iterating through a for loop with the size of an array - Stack Overflow
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Don't use a for loop.
for loops in python are different than in C or Java. In those languages, a for loop has an initial condition, a termination condition, and an increment for each time the loop runs. Whereas in python, a for loop is more of a for each loop - you give it an iterable object, and it runs the code for every item in that iterable object.
Modifying the iterable object while you're running through it is a bad idea that can have difficult-to-predict repercussions and will usually break your code.
However, you can always use a while loop:
a = [0,0,1,0,0,1,0]
idx = 0
while(idx < len(a) - 2):
print(idx)
if a[idx + 2] == 0:
idx += 2
elif a[idx + 2] == 1:
idx += 1
print(idx)
which produces the expected output
0 1 3 4 6
Or, if you change the increments to 3 and 2 respectively, rather than 2 and 1,
0 2 5
Your reasoning is pretty confusing, and I don't see ANY application for this, but here is how I understand your problem...
The reason is because you aren't actually returning the values, you're simply returning the index + 3, which is wrong to begin with. What you're trying to do is point to a new index of the array based on its value and return the index if it contains a value greater than 0.
You need to reference the index you want, check its value, then return the index which contains a value.
a = [0, 0, 1, 0, 0, 1, 0]
for i, v in enumerate(a):
if i == 0:
print(i)
next
if v == 0:
next
else
print(i)
But let's be honest, this is extremely ugly and un-pythonic. Let's simply check for whether a[i] contains a value, and if so, return the index...
for i, v in enumerate(a):
if v or i == 0:
print(i)
The purpose of if v or i == 0 is to check if v has a value, if so, print the index. OR if we are looking at the first element of i.
If you want to EXPLICITLY move the index by two, you must set your index at the start and use a while loop, since enumerate can't help you here and doesn't actually let you move the indicies...
a = [0, 0, 1, 0, 0, 1, 0]
i = 0
while i < len(a) - 1: # -1 to avoid out of bounds error
print(i)
if a[i + 2] == 0:
i += 2
elif a[i + 2] == 1:
i += 1
print(i) # Final print statement for the last index at the end of the while loop
I want to impress upon you the fact that this solution does NOT scale with different or larger lists, which is why it isn't recommended. By simply checking for whether a value does or doesn't exist, you guarantee your accuracy.
You should simply return the index based upon whether or not it contains a value. Even for very large lists, this will be extremely fast, and will always scale, even for values greater than 1.
The only other reason I would see you would want to do this differently is if you're doing string-search algorithms.
Your solution seems correct - but it seems u have a couple of syntax errors if you're using python
You also have to re-initialize sum1 & sum2 between test cases. Right now you get the total sum of all arrs rather than arr[x]-hopefully that makes sense. You also have to remove duplicate numbers from the sum. For instance, if you're getting the sum of all numbers up to 20 - you'll end up adding 15 twice as it is divisible by 3 and 5. So the inner for loops will add it twice to the sum. So you'll need to remove 15 to get rid of duplicates.
for x in range(len(arr)): #by default range starts at o. therefore, range(len(arr)) = range(0, len(arr))
sum1 = 0 # you forgot to initialize sum1
sum2 = 0 # you forgot to initialize sum2
duplicates = 0 #you have to remove duplicates from the answer
for b5 in range (0, arr[x], 5): #you have to add the colons here
sum1 = sum1 + b5
for b3 in range (0, arr[x], 3): #you have to add the colons here
sum2 = sum2 + b3
for dup in range(0, arr[x], 3*5): # removes duplicates from the final sum
duplicates = duplicates + dup
sum = sum1 + sum2 - duplicates
print(sum)
This is an O(n^2) solution - you can drop it down to O(n) using a little bit of math.
You'll notice that the inner for loops can be represented using the formula sum(n)=ฮฃd*i=d*ฮฃi-where the summation starts at i = 0, end at โ(n-1)/dโ and d is the divisor (in the case of your question d=3 or 5).
for b5 in range (0, arr[x], 5):
sum1 = sum1 + b5
(https://en.wikipedia.org/wiki/Summation)
There is a very common summation formula that is commonly used to convert summations into a closed-form expression (something with finite steps - which is O(1))
ฮฃi=n*(n+1)/2
In the case of the inner loop - it would be sum(n) = d*(โ(n-1)/dโ)*(โ(n-1)/dโ+1)/2.
let,
f(n,d) = (โ(n-1)/dโ+1)/2
Therefore, the solution to your problem would be f(n,3)+f(n,5)-f(n,3*5)
Which would convert the inner for loops from O(n) to O(1). Which, means your entire solution would be O(n).
I'll let you figure out the code on your own. However, theoretically, there is a better solution; such that as arr grows indefinitely the work scales linearly rather than quadratically.
- Create Array with input number (Eg.
arr = [12, 15, 4]) - Iterate over the created Array
- In
forloop check if number are divisible by 3 and 5
Sample code:
arr = [12, 15, 4]
total = 0
for num in arr:
if num % 3 == 0 and num % 5 == 0:
total = total + num
print(total) # 15