People recommend leetCode more.
Software Development and Programming Careers (Official Discussion Thread)
- Thread starter Blackking
- Start date
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- Tags
- careers programming
Doesn't make a difference though, practice is practice. Might be more likely to see a Leet Code problem but all that matters at the end of the day is if you can solve it.
Also @Neo The Resurrected ONE here's one for you: write a function that validates matching parentheses/braces/brackets.
Input: ()
Output: true
Input: {[]}()
Output: true
Input: (})
Output: false
You can assume the input will only contain a combination of ()[]{}.
Input: ()
Output: true
Input: {[]}()
Output: true
Input: (})
Output: false
You can assume the input will only contain a combination of ()[]{}.
Neo. The Only. The One.
THE ONE
Neo. The Only. The One.
THE ONE
I forgot about this until a few min ago and I sat down to solve it, thought I had it done in 5 min but then I see it ain't as easy as it looks
I love it
I know what I need to do tho. I'm working on it now.
Neo. The Only. The One.
THE ONE
Alright, breh, I got it.
This is my JavaScript solution.
Code:
function matchInput (a) {
let b;
let c;
a.toString('');
if (a.includes('(') && a.includes('[') && a.includes('{') ) {
if (a.includes(')') && a.includes(']') && a.includes('}')) {
b = true;
c = true;
} else {
b = false;
c = false;
}
} else if (a.includes('(') && a.includes('[')) {
if (a.includes(')') && a.includes(']')) {
b = true;
c = true;
} else {
b = false;
c = false;
}
} else if (a.includes('(') && a.includes('{')) {
if (a.includes(')') && a.includes('}')) {
b = true;
c = true;
} else {
b = false;
c = false;
}
} else if (a.includes('{') && a.includes('[')) {
if (a.includes('}') && a.includes(']')) {
b = true;
c = true;
} else {
b = false;
c = false;
}
} else if (a.includes(')') && a.includes(']') && a.includes('}') ) {
if (a.includes('(') && a.includes('[') && a.includes('{')) {
c = true;
b = true;
} else {
c = false;
b = false;
}
} else if (a.includes(')') && a.includes(']')) {
if (a.includes('(') && a.includes('[')) {
c = true;
b = true;
} else {
c = false;
b = false;
}
} else if (a.includes(')') && a.includes('}')) {
if (a.includes('(') && a.includes('{')) {
c = true;
b = true;
} else {
c = false;
b = false;
}
} else if (a.includes('}') && a.includes(']')) {
if (a.includes('{') && a.includes('[')) {
c = true;
b = true;
} else {
c = false;
b = false;
}
} else if (a.includes(')')) {
if (a.includes('(')) {
c = true;
b = true;
} else {
c = false;
b = false;
}
} else if(a.includes('}')) {
if (a.includes('{')) {
c = true;
b = true;
} else {
c = false;
b = false;
}
} else if(a.includes(']')) {
if(a.includes('[')) {
c = true;
b = true;
} else {
c = false;
b = false;
}
} else if (a.includes('(')) {
if (a.includes(')')) {
b = true;
c = true;
} else {
b = false;
c = false;
}
} else if(a.includes('{')) {
if (a.includes('}')) {
b = true;
c = true;
} else {
b = false;
c = false;
}
} else if(a.includes('[')) {
if(a.includes(']')) {
b = true;
c = true;
} else {
b = false;
c = false;
}
} else {
return false;
}
if(b && c ) {
return true;
} else {
return false;
}
}
LOL this is wayyyy too much code. If it feels like it's too much code it probably is.
jajahoe
Rookie
Yea, this is way to much code man. When you see problems like this first thing you should do is think if there is any data structure that can help solve this problem. For this problem a stack would make the solution way easier.
Neo. The Only. The One.
THE ONE
I haven't made it to data structures yet
Im still learning
My solution is long but it does work right?
It doesn't look like it properly accounts for ordering. It looks like it would take )( as valid because both halves are there, but it's invalid because of the ordering.I haven't made it to data structures yet
Im still learning
My solution is long but it does work right?
Neo. The Only. The One.
THE ONE
Not sure if this was posted before but Meta has a boot camp style 1 year rotational program for engineers with 2+ years of experience. underrepresented minorities are encouraged to apply. From what i hear it's a leetcode style interview but a bit easier. After successful completion of the program, you'll match into a team at the E4 level.
Rotational Software Engineer
Rotational Software Engineer
it was a nice try. Once you start to learn data structures and see this problem again , a lightbulb should turn on right away.
Heres a coding problem that doesnt require a data structure. Just string manipulation.
Return a version of the given string, where for every star (*) in the string the star and the chars immediately to its left and right are gone.
So "ab*cd" yields "ad"
and "ab**cd" also yields "ad".
example outputs:
starOut("ab*cd") → "ad"
starOut("ab**cd") → "ad"
starOut("sm*eilly") → "silly"
Return a version of the given string, where for every star (*) in the string the star and the chars immediately to its left and right are gone.
So "ab*cd" yields "ad"
and "ab**cd" also yields "ad".
example outputs:
starOut("ab*cd") → "ad"
starOut("ab**cd") → "ad"
starOut("sm*eilly") → "silly"