cixak95211:


* push
line 23 is not necessary
as you can return the increased length on line 22.

* unshift
line 42 is not necessary
as you can directly assign this.stack to [value, ...this.stack]

depending on some school of thought, it's best to outline each line seperately,
but you'd be wasting memory declaring variables that are really not necessary
a smart code editor e.g. any of the JetBrains family, would have also warned you about it.

Its amazing we have smart minds here, so why did y'all allow this place to be overran
by some "website for 20k" minds?

const calculateFee= (amount) => {
switch (amount)) {
case 0:
case undefined:
return 0;

case (amount > 0 && amount <= 5000):
return 10;

case amount > 5000 && amount <= 50000:
return 25;

default:
return 50;
}
};

cixak95211:
Ques 3:
What is the problem with this block of code?
What is the fix?

cixak95211:


* push
line 23 is not necessary
as you can return the increased length on line 22.

* unshift
line 42 is not necessary
as you can directly assign this.stack to [value, ...this.stack]

depending on some school of thought, it's best to outline each line seperately,
but you'd be wasting memory declaring variables that are really not necessary
a smart code editor e.g. any of the JetBrains family, would have also warned you about it.

Its amazing we have smart minds here, so why did y'all allow this place to be overran
by some "website for 20k" minds?

lahp:



bro array sizes can never be floats

stop checking if an array is <0

i also think u are to return the new length of array

u should check ur array keys
it says if the key is zero increase the key and add item

have u forgotten an array is indexed from zero?

Karleb:


I've not run the code but from what I see.

Case 0 returns nothing. The switch statement will get stucked there.

Solution, let case 0 return something.

case amount > 5000 && amount <= 50000:

I doubt if that is a valid case statement. Javascript will get confused at this stage. The logic should be wrapped around a parenthesis since it involves comparison.

cixak95211:


Case 0 returns 0. If you look well, you'll see the cases are stacked which is a valid switch expression
Parenthesis are redundant. A smart code editor, once again, will remove the parentheses if you add it.
Prettier will also remove it.
Try again.

cixak95211:
Ques 3:
What is the problem with this block of code?
What is the fix?

Grandlord:
Minds at work

This is my hobby

bunnae:

Grand.Lord, where you enter since? Happy new year

SegFault:
Shouldn't there be a standard algorithm syntax that explains the problem well? If there isn't then this threads name is bogus.

Grandlord:

Do you mean pseudocodes and flowcharts?

Na extra work o

SegFault:

Only pseudocodes are enough, can't be reading code I don't understand.

Karleb:


Please explain this hint more.

It seems you want a deeper level of PQ. as in, a PQ in a PQ in a PQ...

Grandlord:

I think pseudocodes will be too much to ask for here.

Writing comments should be okay.

Brukx:
Exercise
Write a pagination function. This function will convert a 1 dimensional array into a paginated 2 dimensional array. It will take 2 arguments. The first argument is the array to be paginated and the second one is the number of elements per pagination.

Examples
list=[1,2,3,4,5,6,7,8,9,10]

paginate(list, 5) should return
[[1,2,3,4,5], [6,7,8,9,10]]

paginate(list,2) should return
[[1,2], [3,4], [5,6], [7,8], [9,10]]

paginate(list, 3) should return
[[1,2,3], [4,5,6], [7,8,9], [10]]

def page(lst, move):
    length = len(lst)
    output = [ ]
    for i in range(0, length,move):
        if i+ move < length:
            output.append(lst[i: i+move])
        else:
            output.append(lst[i:])

   return output
        
print(page(list,5))

Deicide:
You are given an array and asked to find the element in the array that appears k times.

eg

arr = {1, 2, 2, 3, 3,3,4,4,4,4}

fun = foo(arr, k) k = 2

ans = 2

JFOD:


Hmm, if I'm to solve this

First attempt:
Get the unique numbers
loop through it
get the count of each number
Append it to a list /print it out if the count = 2

Deicide:
You are given an array and asked to find the element in the array that appears k times.

eg

arr = {1, 2, 2, 3, 3,3,4,4,4,4}

fun = foo(arr, k) k = 2

ans = 2

def findK(lst, x):
    unique_num = {lst}
    num_with_x_occurence = [ ]

    for i in unique_num:
        occurrence = lst.count(i)

        if occurrence == x:
            num_with_x_occurence.append(i)

    return num_with_x_occurence
             

JFOD:


I'll make use of python (it's easy to get the unique numbers using set)

I'm assuming, list will be passed

def findK(lst, x):
    unique_num = {lst}
    num_with_x_occurence = [ ]

    for i in unique_num:
        occurrence = lst.count(i)

        if occurrence == x:
            num_with_x_occurence.append(i)

    return num_with_x_occurence
             

Deicide:
did you test this code? am getting error for unique_num = {lst} unhashable type. though the solutions is correct. but then answer wouldn't be unique cause I have to remove {}

cbrass:
Write a function that returns the maximum number N of an array using the following functions POP and PUSH in your code
HINT: Input = 15, Output = 15, input = 10 Output = 10

Modified
Please find the image below for the solution.
I purposely didn't want to give more hint,
From the question you aught to create an array from the given input.

Karleb:


I'm guessing this will require a stack data structure?