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EWSTechSupport:
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for i in range(len(lst) + 1):
     for j in range(i):
          sub = lst[j: i]

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 {}

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
             

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

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))