Elso hazi

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
+    count = 0
+    for word in words:
+        if len(word) >= 2 and word[0] == word[-1]:
+            count += 1
+    return count
 
 
 # B. front_x
@@ -33,8 +36,14 @@ 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
+    list = []
+    xList = []
+    for n in words:
+        if n.startswith('x'):
+            xList.append(n)
+        else:
+            list.append(n)
+    return sorted(xList) + sorted(list)
 
 
 # C. sort_last
@@ -44,8 +53,9 @@ 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
+    def last(a):
+        return a[-1]
+    return sorted(tuples, key=last)
 
 
 # Simple provided test() function used in main() to print

basic/list2.py

@@ -13,17 +13,27 @@
 # 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
+    result = []
+    for num in nums:
+        if len(result) == 0 or num != result[-1]:
+            result.append(num)
+    return result
 
 
 # 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
+    result = []
+    while len(list1) and len(list2):
+        if list1[0] < list2[0]:
+            result.append(list1.pop(0))
+        else:
+            result.append(list2.pop(0))
+    result.extend(list1)
+    result.extend(list2)
+    return result
 
 
 # Note: the solution above is kind of cute, but unforunately list.pop(0)

basic/string1.py

@@ -24,8 +24,10 @@
 # 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:
+        return 'Number of donuts: ' + str(count)
+    else:
+        return 'Number of donuts: many'
 
 
 # B. both_ends
@@ -34,8 +36,11 @@ 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 ''
+    firstTwo = s[0:2]
+    lastTwo = s[-2:]
+    return firstTwo + lastTwo
 
 
 # C. fix_start
@@ -48,8 +53,10 @@ 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
+    front = s[0]
+    back = s[1:]
+    back_f = back.replace(front, '*')
+    return front + back_f
 
 
 # D. MixUp
@@ -60,8 +67,9 @@ 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
+    a_changed = b[:2] + a[2:]
+    b_changed = a[:2] + b[2:]
+    return a_changed + ' ' + b_changed
 
 
 # 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:
+        if s[-3:] != 'ing':
+            s += 'ing'
+        else:
+            s += 'ly'
+    return s
 
 
 # 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
+    a = s.find('not')
+    b = s.find('bad')
+    if a != -1 and b != -1 and b > a:
+        s = s[:a] + 'good' + s[b+3:]
+    return s
 
 
 # F. front_back
@@ -41,8 +48,13 @@ def not_bad(s):
 # Given 2 strings, a and b, return a string of the form
 #  a-front + b-front + a-back + b-back
 def front_back(a, b):
-    # +++your code here+++
-    return
+    a_kozep = len(a) / 2
+    b_kozep = len(b) / 2
+    if len(a) % 2 == 1:
+        a_kozep += 1
+    if len(b) % 2 == 1:
+        b_kozep += 1
+    return a[:a_kozep] + b[:b_kozep] + a[a_kozep:] + b[b_kozep:]
 
 
 # Simple provided test() function used in main() to print