I will be dropping some DSA questions for relaxation.

Rule 1
Please drop your pseudo-code and solution

Rule 2
Include time and space complexity

Rule 3
All devs are allowed to review and argue constructively on different solutions.

Rule 4
Working code is always better than wrong code as optimization would be suggested by other devs.

Rule 5
Only for interested devs

We will be progressing from easy to medium to difficult

LABS will be dropped twice a week.

WEEK1

LAB 1.

Given a non-empty string like "Code" return a string like "CCoCodCode".

stringSplosion("Code" → "CCoCodCode"
stringSplosion("abc" → "aababc"
stringSplosion("ab" → "aab"

function (str) {

return str;
}


LAB 2.

A team of developers, have a project to build a new programming language,
they need a developer to build them a syntax reader for braces only to check if braces are properly closed
this determines if the syntax is correct or not.
your task is to build a syntax reader to check if syntax is correct or return the position of the error brace

given a string of syntax of length N containing only braces "[]{}()<>",
check that the syntax is correct or return the position of the wrong braces
if correct return "ok", if wrong, return the position of the first error.

constraint:
0 <= N <= 100000000
N contain only these characters {}[]()<>

For example:
readSyntax('{{}}') => 'ok'
readSyntax('[[>]') => 2
readSyntax('[[>]{}()<>') => 2

ENJOY!!!!!!!!

daddynasa:
LAB 1.

Given a non-empty string like "Code" return a string like "CCoCodCode".

stringSplosion("Code" → "CCoCodCode"
stringSplosion("abc" → "aababc"
stringSplosion("ab" → "aab"

function (str) {

return str;
}

python

def foo(s):
----r = ""
----for i in range(len(s)+1):
--------r += s[:i]
----return r

Altairx440:

python

def foo(s):
----r = ""
----for i in range(len(s)+1):
--------r += s[:i]
----return r

This is O(n) time and O(n) space

daddynasa:

This is O(n) time and O(n) space

Oh my i didn't even include that, thanks.

//JavaScript
function stringSplosion(str){
let result = '';
for (let i=0; i<str.length; i++){
result += str.slice(0, i+1)
}
return result
}

//Time: O(n) , Space: O(n)

DDayve:
//JavaScript
function stringSplosion(str){
let result = '';
for (let i=0; i<str.length; i++){
result += str.slice(0, i+1)
}
return result
}

//Time: O(n) , Space: O(n)

Nice

I have added the second lab for the week

DDayve:
//JavaScript
function stringSplosion(str){
let result = '';
for (let i=0; i<str.length; i++){
result += str.slice(0, i+1)
}
return result
}

//Time: O(n) , Space: O(n)

This is for loop

Following!

This is not the best solution but lemme drop it. i would take correction from my seniors here. Though it passes the 3 test you provided for it with Time Complexity of O(n) and space complexity of O(1).
//Javascript
function readSyntax(str){
let blk =0;
let curl =0;
let circ = 0;
let grt = 0
function checking(char){
switch (char) {
case '[':
blk +=1 ;
break;
case ']':
blk -=1 ;
break;

case '{':
curl +=1 ;
break;
case '}':
curl -=1 ;
break;

case '(':
circ +=1 ;
break;
case ')':
circ -=1 ;
break;

case '<':
grt +=1 ;
break;
case '>':
grt -=1 ;
break;

default:
break;
}
}
for (let i=0; i<str.length; i++){
checking(str[i])
}

if (blk===0 && curl===0 && circ===0 && grt===0){
return 'Ok'
} else if (blk !=0){
return str.lastIndexOf('[') + 1
} else if (curl !=0){
return str.lastIndexOf('{') + 1
} else if (circ !=0){
return str.lastIndexOf('(') + 1
} else{
return str.lastIndexOf('<') + 1
}
}

daddynasa:
WEEK1

LAB 2.

A team of developers, have a project to build a new programming language,
they need a developer to build them a syntax reader for braces only to check if braces are properly closed
this determines if the syntax is correct or not.
your task is to build a syntax reader to check if syntax is correct or return the position of the error brace

given a string of syntax of length N containing only braces "[]{}()<>",
check that the syntax is correct or return the position of the wrong braces
if correct return "ok", if wrong, return the position of the first error.

constraint:
0 <= N <= 100000000
N contain only these characters {}[]()<>

For example:
readSyntax('{{}}') => 'ok'
readSyntax('[[>]') => 2
readSyntax('[[>]{}()<>') => 2

ENJOY!!!!!!!!

daddynasa:
WEEK1
LAB 2

readSyntax('{{}}') => 'ok'
readSyntax('[[>]') => 2
readSyntax('[[>]{}()<>') => 2

ENJOY!!!!!!!!

# python
# time: O(n)
# space: O(n/2)

def read_syntax(s):
~~stack = []
~~for i in range(len(s)):
~~~~ch = s[i]
~~~~if ch in '<{[(':
~~~~~~stack.append(ch)
~~~~else:
~~~~~~if len(stack) == 0:
~~~~~~~~return i
~~~~~~ch2 = stack.pop()
~~~~~~if not ((ch2 == '{' and ch == '}') or
~~~~~~~~~(ch2 == '<' and ch == '>') or
~~~~~~~~~(ch2 == '(' and ch == ')') or
~~~~~~~~~(ch2 == '[' and ch == ']')):
~~return i
~~if len(stack) == 0:
~~~~return 'ok'
~~else:
~~~~return 0

@daddynasa

LAB 2 is not well written, specifically, languages with static typing do not allow the return of more than one type.

Altairx440:
@daddynasa

LAB 2 is not well written, specifically, languages with static typing do not allow the return of more than one type.

I get you. May a return of -1 for “ok”
Although I have not ran your solution but
Using a stack data structure is a key to an easy solution
Well done.

Space complexity for your solution is O(n) constants are ignored.

No comment on mine? Also, why are people not participating? Would have loved to see different solutions so I can learn different ways of doing things.

daddynasa:


I get you. May a return of -1 for “ok”
Although I have not ran your solution but
Using a stack data structure is a key to an easy solution
Well done.

Space complexity for your solution is O(n) constants are ignored.

daddynasa:


I get you. May a return of -1 for “ok”
Although I have not ran your solution but
Using a stack data structure is a key to an easy solution
Well done.

Space complexity for your solution is O(n) constants are ignored.

Yeah, true.
It's such a pain to indent with "~" or "-", though IMO is more readable than zero indents.

DDayve:
No comment on mine? Also, why are people not participating? Would have loved to see different solutions so I can learn different ways of doing things.

Most of these people have zero interest for the process, they just want the money.

Omo if I know Dsa I would have participated, na only the first task I understand

DDayve:
No comment on mine? Also, why are people not participating? Would have loved to see different solutions so I can learn different ways of doing things.

Apologies. I will go through yours and other devs can comment too.

The intention was to get most devs to participate so that everyone can learn how things are done differently and why they chose the approach. But most NL devs na copy and paste or those kind of “SELECT * FROM users” developers. NA JOKE O

Anyways let us be patient as more devs will see the post

DDayve:
This is not the best solution but lemme drop it. i would take correction from my seniors here. Though it passes the 3 test you provided for it with Time Complexity of O(n) and space complexity of O(1).
//Javascript
function readSyntax(str){
let blk =0;
let curl =0;
let circ = 0;
let grt = 0
function checking(char){
switch (char) {
case '[':
blk +=1 ;
break;
case ']':
blk -=1 ;
break;

case '{':
curl +=1 ;
break;
case '}':
curl -=1 ;
break;

case '(':
circ +=1 ;
break;
case ')':
circ -=1 ;
break;

case '<':
grt +=1 ;
break;
case '>':
grt -=1 ;
break;

default:
break;
}
}
for (let i=0; i<str.length; i++){
checking(str[i])
}

if (blk===0 && curl===0 && circ===0 && grt===0){
return 'Ok'
} else if (blk !=0){
return str.lastIndexOf('[') + 1
} else if (curl !=0){
return str.lastIndexOf('{') + 1
} else if (circ !=0){
return str.lastIndexOf('(') + 1
} else{
return str.lastIndexOf('<') + 1
}
}

Your code takes 7+ seconds to run and only passed the three test cases. After running your code against all possible cases,
From your code, what happens in the following cases?
1. readSyntax("{<[(" => 0
2. readSyntax("{" => 0

Thanks. I would use some conditionals to correct that. The code does not run in 7secs rather it runs in 7milliSeconds, which is 0.007sec

daddynasa:

Your code takes 7+ seconds to run and only passed the three test cases. After running your code against all possible cases,
From your code, what happens in the following cases?
1. readSyntax("{<[(" => 0
2. readSyntax("{" => 0

function slice(array $data, int $start, int $length, array $result = array ()):array
{

if($start == $length -1)
{
return $result;
}
$result[] = $data[$start];
slice($data, $start +=1, $length, $result);
}

function copy (string $casestudy): string
{
$length = 0;
$result = null;
$newcasestudy = str_split($casestudy);
for($i = 1; $i < count($newcasestudy); $i++)
{
$result = slice($casestudy, 0, $i);
}
return implode(',', $result);
}



Here are my codes for lab1


Still practicing recursion

Not a guru

Please can you explain str.slice(0,i+1)

DDayve:
//JavaScript
function stringSplosion(str){
let result = '';
for (let i=0; i<str.length; i++){
result += str.slice(0, i+1)
}
return result
}

//Time: O(n) , Space: O(n)