Question

Must be written in python 3:

☆ Dictionary Ordered Array Given two arrays of numbers, we can use "dictionary ordering" to identify which array is "smaller" (similar to which array would come first in a dictionary). Just like the order of words in a dictionary, we will determine which array is "smaller" by finding the left-most place where the arrays differ and compare the two at that point. This is just like how in a dictionary you would compare "hamper" and "hamburger" by finding the first position where the words differ, here it's "p" and "b", and then taking the smaller letter "b". Thus "hamburger" comes first. Similarly, if you have (-3,9,4, 8, 15] and [-3,9,4, -19,7] we will find the first place they differ, here it's 8 and -19, and take the smaller of the two. Thus, (-3,9,4, -19, 7] is the "smaller" array. Also, just like dictionary ordering, if either array is a prefix of the other, then it is smaller. For example [4,0,-5) is smaller than all of the following: [5], [9, 8], [4, 1], [4, 0, -4], [4, 0, -5, -1000], [4,0,-5, 0], [4, 0, -5, 9], [4, 0, -5, 1, 7, 6). Note that the empty array, [], is smaller than all arrays (except itself, for which it is equal). You need to implement the function SmallestDictionaryOrderedArray to return the array that is smaller, by the definition above.

#
# Complete the 'SmallestDictionaryOrderedArray' function below.
#
# The function is expected to return an INTEGER_ARRAY.
# The function accepts following parameters:
# 1. INTEGER_ARRAY a
# 2. INTEGER_ARRAY b
#

def SmallestDictionaryOrderedArray(a, b):
# Write your code here

Answer 1

[-3, 9, 4, -19, 7] £ GG GGgg main.py saved def SmallestDictionaryOrderedArray(a, b): #if any of the array has elements it is the smaller if len(a) == 0: return a if len(b) == : return b for i in range (len(a): #if array a has more elements than array if i>=len(b): return b #check if element in a is greater than b if a[i] > b[i]: return b elif b[i] > a[i]: return a #if no results found then array b has more elements than array a #array a is smaller return a print(SmallestDictionaryOrderedArray ([-3,9,4,8,15], [-3,9,4, -19,7])) print(SmallestDictionaryOrderedArray([4,0,-5],[5])) print(SmallestDictionaryOrderedArray([4,0,-5],[9,8])) print(SmallestDictionaryOrderedArray ( [4,1],[4,0, -5])) print(SmallestDictionaryOrderedArray ( [4,0,-5],[4,,4])) print(SmallestDictionaryOrderedArray ( [4,0,-5],[4,0,-5,0])) print(SmallestDictionaryOrderedArray ([4,,-5,0],[4,0,-5])) 24 25 26

code:

def SmallestDictionaryOrderedArray(a, b):

#if any of the array has 0 elements it is the smaller

if len(a) == 0:

return a

if len(b) == 0:

return b

for i in range(len(a)):

#if array a has more elements than array b

if i>=len(b):

return b

#check if element in a is greater than b

if a[i] > b[i]:

return b

elif b[i] > a[i]:

return a

#if no results found then array b has more elements than array a

#array a is Smaller

return a

print(SmallestDictionaryOrderedArray([-3,9,4,8,15],[-3,9,4,-19,7]))

print(SmallestDictionaryOrderedArray([4,0,-5],[5]))

print(SmallestDictionaryOrderedArray([4,0,-5],[9,8]))

print(SmallestDictionaryOrderedArray([4,1],[4,0,-5]))

print(SmallestDictionaryOrderedArray([4,0,-5],[4,0,4]))

print(SmallestDictionaryOrderedArray([4,0,-5],[4,0,-5,0]))

print(SmallestDictionaryOrderedArray([4,0,-5,0],[4,0,-5]))