Create List_2.py

Author: TechnoBlogger14o3

List_2.py

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+# Medium python list problems -- 1 loop.. Use a[0], a[1], ... to access elements in a list, len(a) is the length.
+
+
+# ----------------------------------------------------------------------------------------------------------------------
+#
+# Return the number of even ints in the given array. Note: the % "mod" operator computes the remainder, e.g. 5 % 2 is 1.
+#
+#
+# count_evens([2, 1, 2, 3, 4]) → 3
+# count_evens([2, 2, 0]) → 3
+# count_evens([1, 3, 5]) → 0
+
+def count_evens(nums):
+    count = 0
+    for i in nums:
+        if i % 2 == 0:
+            count = count + 1
+
+    return count
+
+
+# ----------------------------------------------------------------------------------------------------------------------
+#
+# Given an array length 1 or more of ints, return the difference between the largest and smallest values in the array.
+# Note: the built-in min(v1, v2) and max(v1, v2) functions return the smaller or larger of two values.
+#
+#
+# big_diff([10, 3, 5, 6]) → 7
+# big_diff([7, 2, 10, 9]) → 8
+# big_diff([2, 10, 7, 2]) → 8
+
+def big_diff(nums):
+    return max(nums) - min(nums)
+
+
+# ----------------------------------------------------------------------------------------------------------------------
+#
+# Return the "centered" average of an array of ints, which we'll say is the mean average of the values,
+# except ignoring the largest and smallest values in the array. If there are multiple copies of the smallest value,
+# ignore just one copy, and likewise for the largest value. Use int division to produce the final average.
+# You may assume that the array is length 3 or more.
+
+#
+# centered_average([1, 2, 3, 4, 100]) → 3
+# centered_average([1, 1, 5, 5, 10, 8, 7]) → 5
+# centered_average([-10, -4, -2, -4, -2, 0]) → -3
+
+def centered_average(nums):
+    nums.sort()
+    return sum(nums[1:-1]) / (len(nums) - 2)
+
+
+# ----------------------------------------------------------------------------------------------------------------------
+#
+# Return the sum of the numbers in the array, returning 0 for an empty array.
+# Except the number 13 is very unlucky,
+# so it does not count and numbers that come immediately after a 13 also do not count.
+#
+#
+# sum13([1, 2, 2, 1]) → 6
+# sum13([1, 1]) → 2
+# sum13([1, 2, 2, 1, 13]) → 6
+
+def sum13(nums):
+    while 13 in nums:
+        if nums.index(13) < len(nums) - 1:
+            nums.pop(nums.index(13) + 1)
+        nums.pop(nums.index(13))
+
+    return sum(nums)
+
+
+# ----------------------------------------------------------------------------------------------------------------------
+#
+# Return the sum of the numbers in the array, except ignore sections of numbers starting with a 6 and extending to
+# the next 7 (every 6 will be followed by at least one 7). Return 0 for no numbers.
+#
+#
+# sum67([1, 2, 2]) → 5
+# sum67([1, 2, 2, 6, 99, 99, 7]) → 5
+# sum67([1, 1, 6, 7, 2]) → 4
+
+def sum67(nums):
+    count = 0
+    blocked = False
+
+    for n in nums:
+        if n == 6:
+            blocked = True
+            continue
+        if n == 7 and blocked:
+            blocked = False
+            continue
+        if not blocked:
+            count += n
+
+    return count
+
+
+# ----------------------------------------------------------------------------------------------------------------------
+#
+# Given an array of ints, return True if the array contains a 2 next to a 2 somewhere.
+#
+#
+# has22([1, 2, 2]) → True
+# has22([1, 2, 1, 2]) → False
+# has22([2, 1, 2]) → False
+
+def has22(nums):
+    for i, v in enumerate(nums[:-1]):
+        if v == 2 and nums[i + 1] == 2:
+            return True
+    return False
+
+# ----------------------------------------------------------------------------------------------------------------------