Write a program, with comments, to do the following. Define a list L1 consisting

Question

Write a program, with comments, to do the following. Define a list L1 consisting of at least 10 integers (e.g L1 = [5, -34, 0, 98, -11, 244, 193, 28, -10, -20, 45, 67]) b. Ask the user to enter a value between 0 and n-1, where n is the number of items in the list L1. You may assume the user will enter a positive integer, but it may not be in the specified range. c. Check the value x entered by the user is in the proper range (i.e. 0 lessthanorequalto x lessthanorequalto n-1) d. If x is in the proper range: i. Print the xth element of L1 with a suitable message. ii. If the integer x occurs in L1, print a message indicating the position of x in L1; otherwise print a message indicating x does not occur in L1 iii. If x is an even number and greater than 0, print the first x elements of L1. Of x is an odd number and greater than 0, print the last x elements of L1. If x is 0, print an empty list. e. Otherwise (i.e. x is not in the proper range), print the message 'You did not enter a valid input! Some sample sessions are shown below.

Explanation / Answer

my_list = [5,-34,0,98,-11,244,193,28,-10,-20,45,67]

a=int(input("Enter your number: "))
print "length",len(my_list)
print "a", my_list[a]

if (a<len(my_list)-1):         //if the entered number is less than the length of the list if prints the element at that position
    print "The Xth element is", my_list[a]
if(a in my_list):                  //if the entered number is the element of the list it prints the position of that one int the list
    {
for i in range(len(my_list)):
        {
        if (my_list[i]==a):
            k=i
        }
print "entered number exist in the list at position",i
    }

if(x%2==0 & x>0) //if the entered num is even
{
for(i=0;i<len(x);i++)
{
print "Number less than even::",my_list[i]
}
}

if(x%2 !=0 & x>0)   //if the entered num is odd
{
for(i=len(x)-1;i>0;i--)
{
print "Number less than even::",my_list[i]
}
}
else
{
print "you did not enter a valid number"
}

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