99 problems in Julia programming Language (1-10)

Prolog and Haskell got 99 problems to understand basic logic constructs .This is an attempt to create Julia versions of solutions to those problems.Here first 10 problems are solved by me.Remaining will be looked in further articles.If you like to contribute to my work mail me at narenarya@live.com.

Problem 1:

Q) Find the last element of a list
Ex: julia>myLast([1,2,3,4,5])
5
Solution:
julia>myLast(l)= l[end]

Prblem 2:

Q) Find last but one element from list
Ex: julia>myButOne([1,2,3,4,5])
4
Solution:
julia>myButOne(l)=l[end-1]

Problem 3:

Q) Find the kth element of list
Ex: julia>kthElement([12,3,4,16,43,9],4)
16
since 16 is 4th element in the list.
Solution: 
julia>kthElement(l,k)=l[k]
Note:expression is so simple because Julia indexes from 1 not 0.

Problem 4:
Q)Find the no of elements in list
Ex:julia>ElementNo([1,2,3,4])
4
Solution:
julia>ElementNo(l)=length(l)

Problem 5:
Q)Reverse a list
Ex:julia>myReversedList([1,2,3,4])
[4,3,2,1]
Solution:
julia>myReversedList(l)=reverse(l)

Problem 6:
Q)Find out whether a list is a palindrome. A palindrome can be read forward or backward; e.g. (x a m a x).
Ex:julia>Palindrome([1,2,3,2,1])
true
Solution:
julia>Palindrome(l)= return l==reverse(l)

Problem 7:
Q)Flatten a nested list structure.
Ex:julia>myFlatten([1,[2,3,],16,[45,76,84],0])
1
2
3
16
45
76
84
0
Flattening means removing recursive lists i.e lists inside lists.So [1,[2,3]] list is broken into [1,2,3]
Solution:
julia>myFlatten(l)=l
This simple,yes it is utterly simple Julia automatically flats your nested structure.

Problem 8:
Q) Eliminate consecutive duplicates of list elements.
Ex:julia>deleteConsDuplicates([1,1,1,2,2,3,3,3,3,4,4,5])
1
2
3
4
5
Solution:
julia> function deleteConsDuplicates(l)
            myset=Set()
            for i=1:endof(l)
               push!(myset,l[i])
            end
           sort([x for x in myset])
        end
we can simply cheat using below oneliner
julia>deleteConsDuplicates(l)=unique(l)

Problem 9:
Q)Pack consecutive duplicates of list elements into sublists. If a list contains repeated elements they should be placed in separate sublists.
Ex: julia>packDuplicates([1,1,1,2,2,3,3,3,4,4,4,4,4])
[(1,1,1),(2,2),(3,3,3),(4,4,4,4)]
Solution:
julia>function packDuplicates(x)
               lest=[]
               for c in deleteConsDuplicates(x)
                     tup=()
                     for i=1:counter(x,c)
                             tup=tuple(tup...,(c))
                     end
                    lest=[lest,tup]
              end
              lest
       end
julia> function counter(l,x)
              temp=0
              for i=1:endof(l)
                   if l[i]==x
                   temp=temp+1
                   end
             end
             temp
        end

Problem 10:
Q)Run-length encoding of a list. Use the result of problem P09 to implement the so-called run-length encoding data compression method. Consecutive duplicates of elements are encoded as lists (N E) where N is the number of duplicates of the element E.
Ex: julia>encodeList([a,a,a,b,b,c,c,c,d,d,d,d])
[(3,a),(2,b),(3,c),(4,d)]
Solution:
 julia> function encodeList(x)
               finlist=[]
               for c in deleteConsDuplicates(x)
                       finlist=[finlist,(counter(x,c),c)]
               end
              finlist
         end

We cleverly used comprehensions for list and tuple to generate new lists,tuples respectively.From now each week new 10 problems will be solved until 99 project is completed,so please be updated with learnjulia.