Word Comparison

coastaltiger39 said:I need some help here!

I'm trying to compare two words to see how many matching letters they share.
Here’s what I mean:
apple, egg = 2 (e, p)
apple, butter = 2 (p, e)
potato, bread = 3 (p, t, o)

coastaltiger39 said:I need some help here!

I'm trying to compare two words to see how many matching letters they share.
Here’s what I mean:
apple, egg = 2 (e, p)
apple, butter = 2 (p, e)
potato, bread = 3 (p, t, o)

Michelle Bennett5 said:Basically, you build a boolean matrix for tracking usage, sized to the number of letters in the first word.
Then you run a loop through every letter of the second word.
Inside that, you run another loop through all the letters of the first word.
If the current letter from the second word matches one from the first, and that spot in the first word hasn't been "used" yet...
Mark that position in the first word as used.
Add the matching letter to a list.
Break out of the inner loop.
Once the outer loop finishes, just count up how many—and which ones—were found.
--
This logic should mirror the task requirements and match those example solutions.

bluecyclist64 said:Menščini, I think you’ve overcomplicated this because you missed the core point of the assignment.
There is absolutely no need for some complex boolean matrix here.

Michelle Bennett5 said:An egg produces one "a," while butter results in two "a"s.
I don't think your little suggestion can actually handle that.
Just whip it up in Python and see for yourself—it’s literally just a few lines of code.

bluecyclist64 said:You're right; I completely misread the requirements. I dropped the ball on this one.
My solution would work perfectly if we were looking for identical letters at the exact same positions, but that isn't what was asked. That's where my logic failed.

Michelle Bennett5 said:Basically, you build a boolean matrix for tracking usage, sized to the number of letters in the first word.
Then you run a loop through every letter of the second word.
Inside that, you run another loop through all the letters of the first word.
If the current letter from the second word matches one from the first, and that spot in the first word hasn't been "used" yet...
Mark that position in the first word as used.
Add the matching letter to a list.
Break out of the inner loop.
Once the outer loop finishes, just count up how many—and which ones—were found.
--
This logic should mirror the task requirements and match those example solutions.

coastaltiger39 said:Thanks for the help ... any chance you could write out the whole thing using functions? I'm still pretty much a total newbie at this stuff... 🙂

coastaltiger39 said:And honestly, I just need the count of duplicate letters, not actually knowing which ones they are.

Benjamin Brooks2 said:int c = 0;
string sentence = "Benjamin Brooks2";
string sentence2 = "coastaltiger39";
string istaslova = "";
char[] charArr = sentence.ToCharArray();
char[] charArr2 = sentence2.ToCharArray();
foreach (char ch in charArr)
{
if (sentence2.Contains(ch))
{
if (istaslova.Contains(ch)) { }
else { istaslova = istaslova + ch; ++c; }

}
}
Console.WriteLine(istaslova);
Console.WriteLine(c);

******************************************

// Looks good now. It shows the unique letters and the total count. --> ok,2

Michelle Bennett5 said:anyway, can you try breaking this down in Python?
there's an online runner here https://www.w3schools.com/python/try...e=demo_default

makes it pretty easy to check if the logic holds up and see the results.
just run it against the examples from the assignment definition:
apple, egg = 2 (p, e)
apple, butter = 2 (a, e)
potato, bread = 3 (p, o, t)

Michael Jackson10 said:a=10
print(a)

And then it just goes to hell.

Michael Jackson10 says:
Python C = 0
istaslova=""
sentence=input("Enter first sentence: ")
sentence2=input("Enter second sentence: ")
for i in range(len(sentence)):
--if sentence2.find(sentence)>-1:
----if istaslova.find(sentence)==-1:
------istaslova=istaslova+sentence
------c=c+1

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

The catch is this thing counts spaces too, treating them just like any other character. You could probably just skip those when you're testing...

Michelle Bennett5 said:

Michael Jackson10 says:
Python C = 0
istaslova=""
sentence=input("Enter first sentence: ")
sentence2=input("Enter second sentence: ")
for i in range(len(sentence)):
--if sentence2.find(sentence)>-1:
----if istaslova.find(sentence)==-1:
------istaslova=istaslova+sentence
------c=c+1

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

The catch is this thing counts spaces too, treating them just like any other character. You could probably just skip those when you're testing...

The prompt actually specifies a 'word,' so spaces shouldn't even be in the running.
Check the requirements—if you input something like "apple, butter," the result should be 2, but your code only gives you 1.

Benjamin Brooks2 said:Indices start at 0. len is the letter count.

Benjamin Brooks2 said:Indices start at 0. len is the letter count.