I have been working on this problem for quite some time now and I can\'t seem to

Question

I have been working on this problem for quite some time now and I can't seem to figure it out, any help would be greatly appreciated.

Write a function named matchUp(strArray, word) that takes 2 parameters, an array of strings, and another lone string.

Return a dictionary such that for every string in strArray, it contains the index at which it is a valid substring in the second string parameter. If it does not exist, set the value to -1.

E.g. matchUp(["a", "b, "c", "d"], "abc") would return:

{"a":0, "b":1, "c":2, "d":-1}

Note there exists a find function in Python of the form some_sample_string.find(some_other_string) which will return the index at which it funds the substring. You may use this at first, but I would like you to also write your own code to find the index of a substring as this will help you further hone your Pythonic skillset.

My Code:

def matchUp(strArray, word):
d={}
for i in strArray:
for j in range(len(word)):
if (i == word[j]):
d=i
if (i != word[j]):
d=-1   
  
return d

Test Cases:

print cmp({"a":0, "b":1, "c":2, "d":-1}, matchUp(["a", "b", "c", "d"], "abc")) Expected: 0 Got: -1

print cmp({"a":0, "b":1, "c":2, "d":-1, "abc":0}, matchUp(["a", "b", "c", "abc", "d"], "abc"))  Expected: 0 Got: -1

print cmp({"a":0, "b":1, "c":2, "d":-1, "abc":0, "ac":-1}, matchUp(["a", "ac", "b", "c", "abc", "d"], "abc"))  Expected: 0 Got: -1

Explanation / Answer

strArray = input().split()
word = input()

def matchup(a,b):
   l1=[]
   for i in range(len(b)):
       for j in range(i,len(b)):
           l1.append(b[i:j+1])
   dict1 = {}
   for i in range(len(a)):
       if len(a[i]) == 1:
           if a[i] not in dict1 and a[i] in l1:
               dict1[a[i]] = i
           else:
               dict1[a[i]] = -1
       else:
           if a[i] not in dict1 and a[i] in l1:
               s = a[i][0]
               dict1[a[i]] = dict1[s]
           else:
               dict1[a[i]] = -1
   print(dict1)

matchup(strArray,word)

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