first week homework

basic/list1.py

@@ -21,8 +21,11 @@
 # and last chars of the string are the same.
 # Note: python does not have a ++ operator, but += works.
 def match_ends(words):
-    # +++your code here+++
-    return
+    ends = 0
+    for word in words:
+        if len(word)>1 and word[0] == word[-1:]:
+            ends += 1
+    return ends
 
 
 # B. front_x
@@ -33,8 +36,15 @@ def match_ends(words):
 # Hint: this can be done by making 2 lists and sorting each of them
 # before combining them.
 def front_x(words):
-    # +++your code here+++
-    return
+    words.sort()
+    l = []
+    for word in words:
+        if word[0] == 'x':
+            l.append(word)
+    for word in words:
+        if word[0] != 'x':
+            l.append(word)
+    return l
 
 
 # C. sort_last
@@ -44,8 +54,9 @@ def front_x(words):
 # [(2, 2), (1, 3), (3, 4, 5), (1, 7)]
 # Hint: use a custom key= function to extract the last element form each tuple.
 def sort_last(tuples):
-    # +++your code here+++
-    return
+    tuples = sorted(tuples, key=lambda tuple: tuple[-1])
+    #source: https://docs.python.org/3/howto/sorting.html
+    return tuples
 
 
 # Simple provided test() function used in main() to print

basic/list2.py

@@ -13,17 +13,29 @@
 # so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or
 # modify the passed in list.
 def remove_adjacent(nums):
-    # +++your code here+++
-    return
+    i = 0
+    while True:
+        if i+1 >= len(nums):
+            break
+        elif nums[i] == nums[i+1]:
+            nums.pop(i+1)
+        else:
+            i+=1
+    return nums
 
 
 # E. Given two lists sorted in increasing order, create and return a merged
 # list of all the elements in sorted order. You may modify the passed in lists.
 # Ideally, the solution should work in "linear" time, making a single
 # pass of both lists.
 def linear_merge(list1, list2):
-    # +++your code here+++
-    return
+    ans = []
+    for a in list1:
+        ans.append(a)
+    for a in list2:
+        ans.append(a)
+    ans.sort()
+    return ans
 
 
 # Note: the solution above is kind of cute, but unforunately list.pop(0)

basic/string1.py

@@ -24,8 +24,9 @@
 # So donuts(5) returns 'Number of donuts: 5'
 # and donuts(23) returns 'Number of donuts: many'
 def donuts(count):
-    # +++your code here+++
-    return
+    if count >=10:
+        count = 'many'
+    return ('Number of donuts: ' + str(count))
 
 
 # B. both_ends
@@ -34,8 +35,9 @@ def donuts(count):
 # so 'spring' yields 'spng'. However, if the string length
 # is less than 2, return instead the empty string.
 def both_ends(s):
-    # +++your code here+++
-    return
+    if len(s) < 2:
+        return ''
+    return s[0] + s[1] + s[-2] + s[-1]
 
 
 # C. fix_start
@@ -48,8 +50,7 @@ def both_ends(s):
 # Hint: s.replace(stra, strb) returns a version of string s
 # where all instances of stra have been replaced by strb.
 def fix_start(s):
-    # +++your code here+++
-    return
+    return s[0] + s[1:].replace(s[0], '*')
 
 
 # D. MixUp
@@ -60,8 +61,7 @@ def fix_start(s):
 #   'dog', 'dinner' -> 'dig donner'
 # Assume a and b are length 2 or more.
 def mix_up(a, b):
-    # +++your code here+++
-    return
+    return b[0:2] + a[2:] + ' ' + a[0:2] + b[2:]
 
 
 # Provided simple test() function used in main() to print

basic/string2.py

@@ -16,8 +16,12 @@
 # If the string length is less than 3, leave it unchanged.
 # Return the resulting string.
 def verbing(s):
-    # +++your code here+++
-    return
+    if len(s) < 3:
+        return s
+    elif s[-3:] == 'ing':
+        return s + 'ly'
+    else:
+        return s + 'ing'
 
 
 # E. not_bad
@@ -29,8 +33,11 @@ def verbing(s):
 # So 'This dinner is not that bad!' yields:
 # This dinner is good!
 def not_bad(s):
-    # +++your code here+++
-    return
+    n = s.find('not')
+    b = s.find('bad')
+    if n < b:
+        s = s[:n] + 'good' + s[b+3:]
+    return s
 
 
 # F. front_back
@@ -39,10 +46,19 @@ def not_bad(s):
 # If the length is odd, we'll say that the extra char goes in the front half.
 # e.g. 'abcde', the front half is 'abc', the back half 'de'.
 # Given 2 strings, a and b, return a string of the form
-#  a-front + b-front + a-back + b-back
+# a-front + b-front + a-back + b-back
+def halve(n):
+    if len(n) % 2 == 0:
+        return len(n)/2
+    else:
+        return len(n)/2+1
+
 def front_back(a, b):
-    # +++your code here+++
-    return
+    a_front = a[:halve(a)]
+    a_back = a[halve(a):]
+    b_front = b[:halve(b)]
+    b_back = b[halve(b):]
+    return a_front + b_front + a_back + b_back
 
 
 # Simple provided test() function used in main() to print