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Create merge_tools.py #152

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Create merge_tools.py #152

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21 changes: 21 additions & 0 deletions merge_tools.py
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Original file line number Diff line number Diff line change
@@ -0,0 +1,21 @@
def merge_the_tools(string, k):
#divide string into k length substrings
x = 0
y = k
subStrs = []
while (y <= len(string)):
subStrs.append(string[x:y])
x += k
y += k

#remove duplicates in each and print each
for subStr in subStrs:
print remove_dups(subStr)

def remove_dups(string):
newstr = []
for i in range(len(string)):
if string[i] not in string[:i]:
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@madhuri7112 madhuri7112 Jun 25, 2021 •
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You are iterating through the entire string until ith character for every character so the time complexity of this loop is O(n^2) which is not ideal.

Instead, you can use a set to do this

def remove_dups(string):
   seen_characters = {}
   result = ""
   for i in range(len(string)):
      if  string[i] not in seen_characters:
           result = result + string[i]
      seen_characters.add(string[i])

The time complexity of this loop is only O(n) because it iterates through the string only once and look up in set is O(1) time operation.

newstr.append(string[i])
newstr1 = ""
return newstr1.join(newstr)