CMU 15-112: Fundamentals of Programming and Computer Science
Day3 Practice (Due never)
- These problems will help you prepare for hw3, and quiz1.
- To start:
- Go to your folder named 'week1'
- Download day3_practice.py to that folder
- Edit day3_practice.py using pyzo
- Do not use lists or recursion.
- Do not hardcode the test cases in your solutions.
What will this code print? Figure it out by hand, then run the code to confirm. Then slightly edit the code and try again.
- Trace #1 of 3:
def ct1(s, t): result = "" for c in s: if (c.upper() not in "NO!!!"): i = t.find(c) if (result != ""): result += ":" result += "%d%s%s%s" % (i, c, s[i], t[i]) return result print(ct1("net", "two"))
- Trace #2 of 3:
def ct2(s): result = "" d = ord("a") for c in s.lower(): if (c.isalpha() and (ord(c) >= d)): result += str(ord(c) - d) + chr(d) d += 1 return result print(ct2("Be a CA?!?"))
- Trace #3 of 3:
def ct3(s): result = "" while (len(s) > 1): result += s[:1] + s[2:4] + "." s = s[1:-1:2] return result + s print(ct3("abcdefghi"))
Reasoning Over Code
Find parameter(s) to the following functions so that they return True. Figure it out by hand, then run the code to confirm. There may be more than one correct answer for each function, and you can provide any one of them.
- RC #1 of 2:
def rc1(s): if (not isinstance(s, str)): return False if ('0' in s): return False t,n = s[1:-1], int(s+s[-1]) return (t.isalpha() and (t == t*(n//2)))
- RC #2 of 2:
def rc2(s, t): assert((s != "") and (t != "") and (s in t) and (s != t)) result = "" for i in range(len(s)): if ((i % 2) == 0): result += t[i] else: result += t[-1-i] return (result == s)
Free Response (Problem-Solving)
Write the function vowelCount(s), that takes a string s, and returns the number of vowels in s, ignoring case, so "A" and "a" are both vowels. The vowels are "a", "e", "i", "o", and "u". So, for example, ("Abc def!!! a? yzyzyz!") returns 3 (two a's and one e).
- interleave(s1, s2)
Write the function interleave(s1, s2) that takes two strings, s1 and s2, and interleaves their characters starting with the first character in s1. For example, interleave('pto', 'yhn') would return the string "python". If one string is longer than the other, concatenate the rest of the remaining string onto the end of the new string. For example ('a#', 'cD!f2') would return the string "ac#D!f2". Assume that both s1 and s2 will always be strings.
Write the function hasBalancedParentheses, which takes a string and returns True if that code has balanced parentheses and False otherwise (ignoring all non-parentheses in the string). We say that parentheses are balanced if each right parenthesis closes (matches) an open (unmatched) left parenthesis, and no left parentheses are left unclosed (unmatched) at the end of the text. So, for example, "( ( ( ) ( ) ) ( ) )" is balanced, but "( ) )" is not balanced, and "( ) ) (" is also not balanced. Hint: keep track of how many right parentheses remain unmatched as you iterate over the string.
- rotateStringLeft(s, k)
Write the function rotateStringLeft that takes a string s and a non-negative integer k, and returns the string s rotated k places to the left.
- rotateStringRight(s, k)
Write the function rotateStringRight that takes a string s and a non-negative integer k, and returns the string s rotated k places to the right.
Write the function wordWrap(text, width) that takes a string of text (containing only lowercase letters or spaces) and a positive integer width, and returns a possibly-multiline string that matches the original string, only with line wrapping at the given width. So wordWrap("abc", 3) just returns "abc", but wordWrap("abc",2) returns a 2-line string, with "ab" on the first line and "c" on the second line. After you complete word wrapping in this way, only then: All spaces at the start and end of each resulting line should be removed, and then all remaining spaces should be converted to dashes ("-"), so they can be easily seen in the resulting string. Here are some test cases for you:
assert(wordWrap("abcdefghij", 4) == """\ abcd efgh ij""") assert(wordWrap("a b c de fg", 4) == """\ a-b c-de fg""")
largestNumber: Write the function largestNumber(text) that takes a string of text and returns the largest int value that occurs within that text, or None if no such value occurs. You may assume that the only numbers in the text are non-negative integers and that numbers are always composed of consecutive digits (without commas, for example). For example:
largestNumber("I saw 3 dogs, 17 cats, and 14 cows!")returns 17 (the int value 17, not the string "17"). And
largestNumber("One person ate two hot dogs!")returns None (the value None, not the string "None").
Write the function longestSubpalindrome(s), that takes a string s and returns the longest palindrome that occurs as consecutive characters (not just letters, but any characters) in s. So:
longestSubpalindrome("ab-4-be!!!")returns "b-4-b". If there is a tie, return the lexicographically larger value -- in Python, a string s1 is lexicographically greater than a string s2 if (s1 > s2). So:
longestSubpalindrome("abcbce")returns "cbc", since ("cbc" > "bcb"). Note that unlike the previous functions, this function is case-sensitive (so "A" is not treated the same as "a" here). Also, from the explanation above, we see that longestSubpalindrome("aba") is "aba", and longestSubpalindrome("a") is "a".
Write the function leastFrequentLetters(s), that takes a string s, and ignoring case (so "A" and "a" are treated the same), returns a lowercase string containing the least-frequent alphabetic letters that occur in s, each included only once in the result and then in alphabetic order. So:
leastFrequentLetters("aDq efQ? FB'daf!!!")returns "be". Note that digits, punctuation, and whitespace are not letters! Also note that seeing as we have not yet covered lists, sets, maps, or efficiency, you are not expected to write the most efficient solution. Finally, if s does not contain any alphabetic characters, the result should be the empty string ("").
- sameChars(s1, s2)
Write the function sameChars(s1, s2) that takes two strings and returns True if the two strings are composed of the same characters (though perhaps in different numbers and in different orders) -- that is, if every character that is in the first string, is in the second, and vice versa -- and False otherwise. This test is case-sensitive, so "ABC" and "abc" do not contain the same characters. The function returns False if either parameter is not a string, but returns True if both strings are empty (why?).
- areAnagrams(s1, s2)
Write the function areAnagrams(s1, s2) that takes two strings, s1 and s2, that you may assume contain only upper and/or lower case letters, and returns True if the strings are anagrams, and False otherwise. Two strings are anagrams if each can be reordered into the other. Treat "a" and "A" as the same letters (so "Aba" and "BAA" are anagrams). You may not use sort() or sorted() or any other list-based functions or approaches. Hint: you may use s.count(), which could be quite handy here.
Without using the s.replace() method, write the function collapseWhitespace(s), that takes a string s and returns an equivalent string except that each occurrence of whitespace in the string is replaced by a single space. So, for example, collapseWhitespace("a\t\t\tb\n\nc") replaces the three tabs with a single space, and the two newlines with another single space , returning "a b c". Here are a few more test cases for you:
assert(cw("a\nb") == "a b") assert(cw("a\n \t b") == "a b") assert(cw("a\n \t b \n\n \t\t\t c ") == "a b c ")Once again, do not use s.replace() in your solution.
- replace(s1, s2, s3)
Without using the builtin method s.replace(), write its equivalent. Specifically, write the function replace(s1, s2, s3) that returns a string equal to s1.replace(s2, s3), but again without calling s.replace().
- encodeOffset(s, d)
Write the function encodeOffset(s, d) that takes a string and a possibly-negative int offset d (for "delta"), and returns the string formed by replacing each letter in s with the letter d steps away in the alphabet. So: encodeOffset("ACB", 1) return "BDC" encodeOffset("ACB", 2) return "CED" This works with wraparound, so: encodeOffset("XYZ", 1) returns "YZA" And with negative offsets, so: encodeOffset("ABC", -1) returns "ZAB" And the wraparound repeats with d>26, so: encodeOffset("ABC", -27) returns "ZAB" And it is case-preserving, so: encodeOffset("Abc", -27) returns "Zab" And it does not affect non-alphabetic characters (non-letters), so: encodeOffset("A2b#c", -27) returns "Z2a#b"
- decodeOffset(s, d)
Write the function decodeOffset(s, d) that takes a string that was encoded by encodeOffset using the given offset d, and returns the original string.
- encrypt and decrypt
Write the encrypt and decrypt functions described in part 6 (Simple Encryption) here.
Write the Mastermind functions described here. (Note: this exercise is not in the starter file.)