Word Comparison

Michael Jackson10 said:Nah, he's actually right. I totally misread the prompt—I thought we weren't supposed to count duplicate letters...

Whatever, just comment out that second if statement

Michelle Bennett5 said:Just a heads up—if you follow the instructions (like with "apple" or "egg"), the answer should be 2. If you just comment that part out, you're gonna end up with 3.
You really need an array to track which letters from the first word have already been used, otherwise, there's no way to pull this off.

Michelle Bennett5 said:Just a heads up—if you follow the instructions (like with "apple" or "egg"), the answer should be 2. If you just comment that part out, you're gonna end up with 3.
You really need an array to track which letters from the first word have already been used, otherwise, there's no way to pull this off.

Benjamin Brooks2 says:
I've never touched Python before, but I figure it would go something like this...

c=0
istaslova=""
sentence=input("Enter the first sentence: ")
sentence2=input("Enter the second sentence: ")
for i in range(len(sentence)):
for j in range(len(sentence2)):
for c1 in range(len(istaslova))
if sentence2.find(sentence)********* if it found the same letter in both words
if istaslova.NOTfind istaslova = istaslova + ......c=c+1****** if it DIDN'T find the letter in "istaslova" already, then add the letter

print("Duplicate letters:",istaslova)
print("Total count:",c)

Michelle Bennett5 said:

Benjamin Brooks2 says:
I've never touched Python before, but I figure it would go something like this...

c=0
istaslova=""
sentence=input("Enter the first sentence: ")
sentence2=input("Enter the second sentence: ")
for i in range(len(sentence)):
for j in range(len(sentence2)):
for c1 in range(len(istaslova))
if sentence2.find(sentence)********* if it found the same letter in both words
if istaslova.NOTfind istaslova = istaslova + ......c=c+1****** if it DIDN'T find the letter in "istaslova" already, then add the letter

print("Duplicate letters:",istaslova)
print("Total count:",c)

It’s not gonna work like that. Check post #3; I already went over this there.
You need some kind of status matrix for every position in that first word—something to track what's been used up, right?

Benjamin Brooks2 said:c = 0
a = "Benjamin Brooks2"
b = "Michelle Bennett5 Kok"
istaslova = ""
for x in a:
for y in b:
if x == y:
if x not in istaslova:
istaslova = istaslova + x
c = c + 1
print(c)
print(istaslova)

Michelle Bennett5 said:It's fine, I guess, but if you actually want to solve the problem correctly, you have to match those specific examples from the very first post. That's where the real trick is hidden.

Michael Jackson10 says:
I haven't had a spare second until now, but here it is. At least, this is my version of a shorter way:

c=0
istaslova=""
word=input("Enter first word: ")
word2=input("Enter second word: ")
for i in range(len(word)):
--if istaslova.count(word)<min(word.count(word), word2.count(word)):
----istaslova=istaslova+word
----c=c+1

print("Common letters are:",istaslova)
print("Total count:",c)

That previous example using find() was a total disaster, because if you use find(), you have to come up with some kind of pointer logic to track where you are moving through the words...

Anyway, isn't this decent for someone who’s only had about four hours of Python classes while helping out a middle schooler nephew??
You should see me trying to mess around with Turtle 🤣
The Syntax trips me up sometimes. I've got one tab glued to a W3Schools reference and another on Replit, but Python is actually pretty sweet for learning logical structure. You can interpret things immediately or just drop in some code to check variable states. It's a bit of a pain with these indentation-only structures, and I'm constantly checking my casting since nothing is declared—I'm never quite sure what a function expects or if I just botched the Syntax... man, my last "masterpiece" was back when I was messing with Cleveland... 🤦

Besides, you didn't really define the task right in your first post: it's not "how many identical letters they have" (since that doesn't account for repeats: "aaaa" and "aaa" only share one unique letter), it's a "common subset" (where you treat every single letter as its own entity, regardless of whether the shape exists elsewhere in the set)

Michael Jackson10 says:
Haven't had the time until now, but here it is. At least, this is my shorter version:

c=0
istaslova=""
word=input("Enter first word: ")
word2=input("Enter second word: ")
for i in range(len(word)):
--if istaslova.count(word)<min(word.count(word), word2.count(word)):
----istaslova=istaslova+word
----c=c+1

print("Common letters for the words are:",istaslova)
print("Total count:",c)

My previous attempt using find() was a complete bust. You'd have to manage some kind of pointer to track where you are in the words...

Anyway, isn't this decent for someone who's only had about 4 hours of Python classes with a nephew in middle school?
Wait until you see me struggle with Turtle 🤣
The Syntax trips me up. I've got one tab open with a W3Schools reference and another on Replit, but Python is actually pretty good for learning logic. You can interpret it instantly or drop in code to check variable states. The indentation-only structure is a pain, and I'm constantly checking casting since nothing is declared—I'm never sure what a function accepts or if I just messed up the Syntax... damn, my last "masterpiece" was in Cleveland... 🤦

Besides, you didn't even define the task right in your first post. It wasn't "how many identical letters they have" (since that ignores repeats: "aaaa" and "aaa" only share one unique letter), it's a "common subset" (where you treat every single letter as its own entity regardless of shape)

Benjamin Brooks2 said:

Michael Jackson10 says:
Haven't had the time until now, but here it is. At least, this is my shorter version:

c=0
istaslova=""
word=input("Enter first word: ")
word2=input("Enter second word: ")
for i in range(len(word)):
--if istaslova.count(word)<min(word.count(word), word2.count(word)):
----istaslova=istaslova+word
----c=c+1

print("Common letters for the words are:",istaslova)
print("Total count:",c)

My previous attempt using find() was a complete bust. You'd have to manage some kind of pointer to track where you are in the words...

Anyway, isn't this decent for someone who's only had about 4 hours of Python classes with a nephew in middle school?
Wait until you see me struggle with Turtle 🤣
The Syntax trips me up. I've got one tab open with a W3Schools reference and another on Replit, but Python is actually pretty good for learning logic. You can interpret it instantly or drop in code to check variable states. The indentation-only structure is a pain, and I'm constantly checking casting since nothing is declared—I'm never sure what a function accepts or if I just messed up the Syntax... damn, my last "masterpiece" was in Cleveland... 🤦

Besides, you didn't even define the task right in your first post. It wasn't "how many identical letters they have" (since that ignores repeats: "aaaa" and "aaa" only share one unique letter), it's a "common subset" (where you treat every single letter as its own entity regardless of shape)

Man, we should start a new thread for coding challenges. We could post problems and try to hunt down solutions online, though people would probably accuse us of being "intellectually dishonest" or whatever. Is there anyone here doing actual "coding" who isn't intellectually dishonest, or are we just going to play around with Syntax errors?

Is there a thread for problem-solving? Not just Syntax, but actual discussion on how things should be done?