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almond37 [142]
3 years ago
10

Building a String Library

Computers and Technology
1 answer:
maw [93]3 years ago
6 0
Code:

def myAppend( str, ch ):
# Return a new string that is like str but with
# character ch added at the end
return str + ch

def myCount( str, ch ):
# Return the number of times character ch appears
# in str.

# initiaalizing count with 0
count = 0

# iterating over every characters present in str
for character in str:
# incrementing count by 1 if character == ch
if character == ch:
count += 1

# returning count
return count


def myExtend( str1, str2 ):
# Return a new string that contains the elements of
# str1 followed by the elements of str2, in the same
# order they appear in str2.

# concatenating both strings and returning its result
return str1 + str2

def myMin( str ):
# Return the character in str with the lowest ASCII code.

# If str is empty, print "Empty string: no min value"
# and return None.
if str == "":
print("Empty string: no min value")
return None

# storing first character from str in char
char = str[0]

# iterating over every characters present in str
for character in str:
# if current character is lower than char then
# assigning char with current character
if character < char:
char = character
# returning char
return char


def myInsert( str, i, ch ):
# Return a new string like str except that ch has been
# inserted at the ith position. I.e., the string is now
# one character longer than before.

# Print "Invalid index" if
# i is greater than the length of str and return None.

if i > len(str):
print("Invalid index")
return None

# str[:i] gives substring starting from 0 and upto ith position
# str[i:] gives substring starting from i and till last position
# returning the concatenated result of all three
return str[:i]+ch+str[i:]

def myPop( str, i ):
# Return two results:
# 1. a new string that is like str but with the ith
# element removed;
# 2. the value that was removed.
# Print "Invalid index" if i is greater than or
# equal to len(str), and return str unchanged and None
if i >= len(str):
print("Invalid index")
return str, None

# finding new string without ith character
new_str = str[:i] + str[i+1:]

# returning new_str and popped character
return new_str, str[i]

def myFind( str, ch ):
# Return the index of the first (leftmost) occurrence of
# ch in str, if any. Return -1 if ch does not occur in str.

# finding length of the string
length = len(str)

# iterating over every characters present in str
for i in range(length):
# returning position i at which character was found
if str[i]==ch:
return i
# returning -1 otherwise
return -1


def myRFind( str, ch ):
# Return the index of the last (rightmost) occurrence of
# ch in str, if any. Return -1 if ch does not occur in str.

# finding length of the string
length = len(str)

# iterating over every characters present in str from right side
for i in range(length-1, 0, -1):
# returning position i at which character was found
if str[i]==ch:
return i
# returning -1 otherwise
return -1

def myRemove( str, ch ):
# Return a new string with the first occurrence of ch
# removed. If there is none, return str.

# returning str if ch is not present in str
if ch not in str:
return str

# finding position of first occurence of ch in str
pos = 0

for char in str:
# stopping loop if both character matches
if char == ch:
break
# incrementing pos by 1
pos += 1

# returning strig excluding first occurence of ch
return str[:pos] + str[pos+1:]

def myRemoveAll( str, ch ):
# Return a new string with all occurrences of ch.
# removed. If there are none, return str.

# creating an empty string
string = ""

# iterating over each and every character of str
for char in str:
# if char is not matching with ch then adding it to string
if char!=ch:
string += char
# returning string
return string

def myReverse( str ):
# Return a new string like str but with the characters
# in the reverse order.

return str[::-1]
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3 years ago
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Assume you are given three variables, revenue, expenses, and profit, all of type Money (a structured type with two int fields, d
Andre45 [30]

Answer:

if(revenue.cents - expenses.cents < 0){

profit.dollars = revenue.dollars - expenses.dollars - 1;

profit.cents = 1 - revenue.cents - expenses.cents;

}

else{

profit.dollars = revenue.dollars - expenses.dollars;

profit.cents = revenue.cents - expenses.cents;

}

Explanation:

We know that profit is given as: revenue - expenses from the question.

From the given expression above;

if(revenue.cents - expenses.cents < 0)

then profit.dollar will be revenue.dollars - expenses.dollars - 1; the 1 is to be carry over to the cent part. And the profit.cent will be 1 - revenue.cents - expenses.cents;

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5. Testing:

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