Question

PYTHON CODE

- Write the body of a function second_instance(s, c) which consumes a string s and a length 1 string c that is contained at least twice in s and returns the index of the second location of c.

second_instance: Str Str -> Nat
Requires:
len(c) == 1
c occurs at least twice in s
  
Examples:
second_instance("banana", "a") => 3
second_instance("bb", "b") => 1

- Write the body of a function make_list(n) which consumes a natural number n and returns a list of strings where in position , the number  is repeated as a string  times. You must use loops in your solution. Do not use recursion or abstract list functions.

Returns the list of strings formed by in position i,
repeating i a total of i times
  
make_list: Nat -> (listof Str)
  
Examples:
make_list(0) => ['']
make_list(3) => ['', '1', '22', '333']

- Write the body of a function niven(n) using recursion that consumes a natural number n and returns True if and only if n is a Niven number and False otherwise. A Niven number is a number which is divisible by the sum of its digits.

Returns True if and only if n is a Niven number.
  
niven: Nat -> Bool
  
Examples:
niven(0) => False
niven(1) => True
niven(132) => True
niven(143) => False

- Write the body of a function find_max(L) using loops that returns the maximum of a non-empty list of integers L. Do not use recursion, abstract list functions, or the command max.

Returns the maximum in a non-empty list of integers L
  
find_max: (listof Int) -> Int
  
Examples:
find_max([1]) => 1
find_max([-10,-1,-5]) => -1

Answer 1

program plan:

second_instance():

• set count = 0
• loop for i = 0 to range(len(s))
  • if s[i] == c
    • increment count
    • if count == 2
      • return i

make_list():

• set L = ['']*(n+1)
• loop for i = 1 to n
  • set L[i] str(i)*i

nivel(n):

• set total = 0 and n1 = n
• while n1>0
  • add n1%10 to total
  • set n = n/10
• if n%total = 0 return true, else return false

find_max()

• set max = L[0]
• loop through list
  • if current element is greater than max, then set max = current element

program:

def second_instance(s, c):
count = 0
for i in range(len(s)):
if c == s[i]:
count+=1
if count == 2:
return i

def make_list(n):
l = ['']*(n+1)
for i in range(1,n+1):
l[i] = str(i)*i

return l

def niven(n):
total = 0
n1 = n
while n1>0:
total += n1%10
n1 = n1//10
return n%total == 0

def find_max(L):
m = L[0]
for x in L:
if m < x:
m = x
return m

print(second_instance("banana","a"))
print(make_list(3))
print(niven(18))
print(find_max([-10,-1,-5]))

output: