I have two Strings Paula and Pole. If we check Paula with Pole then we will get three discrepancy a,u,a is present in Paula but not present in Pole so it should return a value 3.
Input:
enter string1: Paula
enter string2: Pole
Expected Output:
3
String 1 is always correct here for a row of names.
I have tried something like below so far
import itertools
def compare(string1, string2, no_match_c=' ', match_c='|'):
if len(string2) < len(string1):
string1, string2 = string2, string1
result = ''
n_diff = 0
for c1, c2 in itertools.izip(string1, string2):
if c1 == c2:
result += match_c
else:
result += no_match_c
n_diff += 1
delta = len(string2) - len(string1)
result += delta * no_match_c
n_diff += delta
return (result, n_diff)
def main():
string1 = 'paula'
string2 = 'pole'
result, n_diff = compare(string1, string2, no_match_c='_')
print(n_diff)
main()
Answer should be in a function
Example of other string
string1 = Michelle
string2 = Michele
Output : 1
This is a simple approach to do what you want, assuming you always want to find the number of chars in string 1 not in string 2.
def compare(str1, str2):
#Return count of chars in str1 not in str2
s1 = set([x for x in str1])
s2 = set([x for x in str2])
ms = s1 ^ s2 & s1 #Finds the chars in string 1 not in str2
rslt = 0
for v in ms:
rslt += str1.count(v)
return rslt
You may try Counter, it can use to count the number of each character in both strings and support subtraction between counts.
from collections import Counter
def diff(s1, s2):
c1 = Counter(s1)
c2 = Counter(s2)
return sum((c1 - c2).values())
print(diff("Paula", "Pole")) # Output: 3
print(diff("Pole", "Paula")) # Output: 2
print(diff("Michelle", "Michele")) # Output: 1
print(diff("Michele", "Michelle")) # Output: 0
You can try that, using a list of zeors in the size of 256 (number of ASCII characters) which represents a counter for all characters.
def compare(string1, string2):
chars_counter = [0]*256
for c1 in string1:
chars_counter[ord(c1)] += 1
for c2 in string2:
if chars_counter[ord(c2)] != 0:
chars_counter[ord(c2)] -= 1
return sum(chars_counter)
Let's say I have a string presented in the following fashion:
st = 'abbbccccaaaAAbccc'
The task is to encode it so that single characters are followed by a number of their occurences:
st = 'a1b3c4a3A2b1c3'
I know one possible solution but it's too bulky and primitive.
s = str(input())
l = len(s)-1
c = 1
t = ''
if len(s)==1:
t = t +s+str(c)
else:
for i in range(0,l):
if s[i]==s[i+1]:
c +=1
elif s[i]!=s[i+1]:
t = t + s[i]+str(c)
c = 1
for j in range(l,l+1):
if s[-1]==s[-2]:
t = t +s[j]+str(c)
elif s[-1]!=s[-2]:
t = t +s[j]+str(c)
c = 1
print(t)
Is there any way to solve this shortly and elegantly?
P.S: I'm an unexperienced Python user and a new StackOverflow member, so I beg my pardon if the question is asked incorrectly.
Take advantage of the standard library:
from itertools import groupby
st = "abbbccccaaaAAbccc"
print("".join("{}{}".format(key, len(list(group))) for key, group in groupby(st)))
Output:
a1b3c4a3A2b1c3
>>>
just loop through and count. There are more graceful snippets but this will get the job done and is clear:
count = 1
char = st[0]
new_st = []
for c in st[1:]:
if c == char:
count += 1
else:
new_st.append(char + str(count))
char = c
count = 1
new_st.append(char + str(count))
s2= "".join(new_st)
print(s2) # 'a1b3c4a3A2b1c3'
If you want a fancy recursive solution:
def compose(s):
if not s:
return ""
count = 1
for char in s[1:]:
if s[0] != char:
break
count += 1
return s[0] + str(count) + compose(s[count:])
This is a pretty standard interview question. Find the longest substring without repeating characters. Here are the test cases,
abcabcbb -> 3
bbbbb -> 1
pwwkew -> 3
bpfbhmipx -> 7
tmmzuxt -> 5
Here's my code which uses a pretty simple approach with 2 pointers.
def lengthOfLongestSubstring(s):
checklist = {}
starting_index_of_current_substring = 0
length_of_longest_substring = 0
for i, v in enumerate(s):
if v in checklist:
starting_index_of_current_substring = checklist[v] + 1
else:
length_of_current_substring = i - starting_index_of_current_substring + 1
length_of_longest_substring = max(length_of_current_substring, length_of_longest_substring)
checklist[v] = i
return length_of_longest_substring
My code passes all the test cases except the last one (actual 4, expected 5). Can someone help me modify the code to take care of the last test case. I don't wish to reinvent the algorithm.
Here is a simple tweak in your code with 2 pointers to find the longest sub-string without repeating characters.
Change in your code is instead of calculating the length of longest substring when v is not present in checklist, I am calculating length of longest substring for all cases.
def lengthOfLongestSubstring(s):
checklist = {}
starting_index_of_current_substring = 0
length_of_longest_substring = 0
for i, v in enumerate(s):
if v in checklist:
starting_index_of_current_substring = max(starting_index_of_current_substring, checklist[v] + 1)
checklist[v] = i
length_of_longest_substring = max(length_of_longest_substring, i - starting_index_of_current_substring + 1)
return length_of_longest_substring
## Main
result = {}
for string in ['abcabcbb', 'bbbbb', 'ppwwkew', 'wcabcdeghi', 'bpfbhmipx', 'tmmzuxt', 'AABGAKGIMN', 'stackoverflow']:
result[string] = lengthOfLongestSubstring(string)
print(result)
Sample run:
{'abcabcbb': 3, 'bbbbb': 1, 'ppwwkew': 3, 'wcabcdeghi': 8, 'bpfbhmipx': 7, 'tmmzuxt': 5, 'AABGAKGIMN': 6, 'stackoverflow': 11}
This post is pretty old, but I think my solution fixes the bug in the original code.
def lengthOfLongestSubstring(s):
checklist = {}
starting_index_of_current_substring = 0
length_of_longest_substring = 0
for i, v in enumerate(s):
if v in checklist:
if checklist[v] >= starting_index_of_current_substring:
starting_index_of_current_substring = checklist[v] + 1
length_of_current_substring = i - starting_index_of_current_substring + 1
length_of_longest_substring = max(length_of_current_substring, length_of_longest_substring)
checklist[v] = i
return length_of_longest_substring
This doesnt really iterate upon your solution, but it's a bit simpler approach, just to give you an idea how it could be also solved.
def longest_substr(s):
longest = 0
for start_index in range(len(s)):
contains = set()
for letter in s[start_index:]:
if letter in contains:
break
contains.add(letter)
longest = max(longest, len(contains))
return longest
0
I would prefer this solution>>
Time and Space Management Optimised:
def lengthOfLongestSubstring(self, s: str) -> int:
curlen = maxlen = 0 # curlen saves the max len of substrings ending with current num
for i, num in enumerate(s):
curlen -= s[i-curlen:i].find(num)
maxlen = max(maxlen, curlen)
return maxlen
Find Longest Substring in the string without repeating characters.
def find_non_repeating_substring(input_str):
output_length = 0
longest_sub_str = ''
len_str = len(input_str)
index = 0
while len_str != 1:
l_str = ''
for i in range(index, len(input_str)):
if input_str[i] not in l_str:
l_str = l_str + input_str[i]
else:
break
sub_str_length = len(l_str)
if sub_str_length > output_length:
output_length = sub_str_length
longest_sub_str = l_str
len_str = len_str -1
index = index + 1
return output_length,longest_sub_str
if __name__ == '__main__':
input_str = raw_input("Please enter the string: ")
sub_str_length, sub_str = find_non_repeating_substring(input_str)
print ('Longest Substing lenght is "{0}" and the sub string is "{1}"'.format(sub_str_length, sub_str))```
I need to create a function called compress that compresses a string by replacing any repeated letters with a letter and number. My function should return the shortened version of the string. I've been able to count the first character but not any others.
Ex:
>>> compress("ddaaaff")
'd2a3f2'
def compress(s):
count=0
for i in range(0,len(s)):
if s[i] == s[i-1]:
count += 1
c = s.count(s[i])
return str(s[i]) + str(c)
Here is a short python implementation of a compression function:
def compress(string):
res = ""
count = 1
#Add in first character
res += string[0]
#Iterate through loop, skipping last one
for i in range(len(string)-1):
if(string[i] == string[i+1]):
count+=1
else:
if(count > 1):
#Ignore if no repeats
res += str(count)
res += string[i+1]
count = 1
#print last one
if(count > 1):
res += str(count)
return res
Here are a few examples:
>>> compress("ddaaaff")
'd2a3f2'
>>> compress("daaaafffyy")
'da4f3y2'
>>> compress("mississippi")
'mis2is2ip2i'
Short version with generators:
from itertools import groupby
import re
def compress(string):
return re.sub(r'(?<![0-9])[1](?![0-9])', '', ''.join('%s%s' % (char, sum(1 for _ in group)) for char, group in groupby(string)))
(1) Grouping by chars with groupby(string)
(2) Counting length of group with sum(1 for _ in group) (because no len on group is possible)
(3) Joining into proper format
(4) Removing 1 chars for single items when there is a no digit before and after 1
There are several reasons why this doesn't work. You really need to try debugging this yourself first. Put in a few print statements to trace the execution. For instance:
def compress(s):
count=0
for i in range(0, len(s)):
print "Checking character", i, s[i]
if s[i] == s[i-1]:
count += 1
c = s.count(s[i])
print "Found", s[i], c, "times"
return str(s[i]) + str(c)
print compress("ddaaaff")
Here's the output:
Checking character 0 d
Found d 2 times
Checking character 1 d
Found d 2 times
Checking character 2 a
Found a 3 times
Checking character 3 a
Found a 3 times
Checking character 4 a
Found a 3 times
Checking character 5 f
Found f 2 times
Checking character 6 f
Found f 2 times
f2
Process finished with exit code 0
(1) You throw away the results of all but the last letter's search.
(2) You count all occurrences, not merely the consecutive ones.
(3) You cast a string to a string -- redundant.
Try working through this example with pencil and paper. Write down the steps you use, as a human being, to parse the string. Work on translating those to Python.
x="mississippi"
res = ""
count = 0
while (len(x) > 0):
count = 1
res= ""
for j in range(1, len(x)):
if x[0]==x[j]:
count= count + 1
else:
res = res + x[j]
print(x[0], count, end=" ")
x=res
Just another simplest way to perform this:
def compress(str1):
output = ''
initial = str1[0]
output = output + initial
count = 1
for item in str1[1:]:
if item == initial:
count = count + 1
else:
if count == 1:
count = ''
output = output + str(count)
count = 1
initial = item
output = output + item
print (output)
Which gives the output as required, examples:
>> compress("aaaaaaaccddddeehhyiiiuuo")
a7c2d4e2h2yi3u2o
>> compress("lllhhjuuuirrdtt")
l3h2ju3ir2dt
>> compress("mississippi")
mis2is2ip2i
from collections import Counter
def string_compression(string):
counter = Counter(string)
result = ''
for k, v in counter.items():
result = result + k + str(v)
print(result)
input = "mississippi"
count = 1
for i in range(1, len(input) + 1):
if i == len(input):
print(input[i - 1] + str(count), end="")
break
else:
if input[i - 1] == input[i]:
count += 1
else:
print(input[i - 1] + str(count), end="")
count = 1
Output : m1i1s2i1s2i1p2i1
s=input("Enter the string:")
temp={}
result=" "
for x in s:
if x in temp:
temp[x]=temp[x]+1
else:
temp[x]=1
for key,value in temp.items():
result+=str(key)+str(value)
print(result)
Here is something I wrote.
def stringCompression(str1):
counter=0
prevChar = str1[0]
str2=""
charChanged = False
loopCounter = 0
for char in str1:
if(char==prevChar):
counter+=1
charChanged = False
else:
str2 += prevChar + str(counter)
counter=1
prevChar = char
if(loopCounter == len(str1) - 1):
str2 += prevChar + str(counter)
charChanged = True
loopCounter+=1
if(not charChanged):
str2+= prevChar + str(counter)
return str2
Not the best code I guess. But works well.
a -> a1
aaabbbccc -> a3b3c3
This is a solution to the problem. But keep in mind that this method only effectively works if there's a lot of repetition, specifically if consecutive characters are repetitive. Otherwise, it will only worsen the situation.
e.g.,
AABCD --> A2B1C1D1
BcDG ---> B1c1D1G1
def compress_string(s):
result = [""] * len(s)
visited = None
index = 0
count = 1
for c in s:
if c == visited:
count += 1
result[index] = f"{c}{count}"
else:
count = 1
index += 1
result[index] = f"{c}{count}"
visited = c
return "".join(result)
You can simply achieve that by:
gstr="aaabbccccdddee"
last=gstr[0]
count=0
rstr=""
for i in gstr:
if i==last:
count=count+1
elif i!=last:
rstr=rstr+last+str(count)
count=1
last=i
rstr=rstr+last+str(count)
print ("Required string for given string {} after conversion is {}.".format(gstr,rstr))
Here is a short python implementation of a compression function:
#d=compress('xxcccdex')
#print(d)
def compress(word):
list1=[]
for i in range(len(word)):
list1.append(word[i].lower())
num=0
dict1={}
for i in range(len(list1)):
if(list1[i] in list(dict1.keys())):
dict1[list1[i]]=dict1[list1[i]]+1
else:
dict1[list1[i]]=1
s=list(dict1.keys())
v=list(dict1.values())
word=''
for i in range(len(s)):
word=word+s[i]+str(v[i])
return word
Below logic will work irrespective of
Data structure
Group By OR Set or any sort of compression logic
Capital or non-capital characters
Character repeat if not sequential
def fstrComp_1(stng):
sRes = ""
cont = 1
for i in range(len(stng)):
if not stng[i] in sRes:
stng = stng.lower()
n = stng.count(stng[i])
if n > 1:
cont = n
sRes += stng[i] + str(cont)
else:
sRes += stng[i]
print(sRes)
fstrComp_1("aB*b?cC&")
I wanted to do it by partitioning the string.
So aabbcc would become: ['aa', 'bb', 'cc']
This is how I did it:
def compression(string):
# Creating a partitioned list
alist = list(string)
master = []
n = len(alist)
for i in range(n):
if alist[i] == alist[i-1]:
master[-1] += alist[i]
else:
master += alist[i]
# Adding the partitions together in a new string
newString = ""
for i in master:
newString += i[0] + str(len(i))
# If the newString is longer than the old string, return old string (you've not
# compressed it in length)
if len(newString) > n:
return string
return newString
string = 'aabbcc'
print(compression(string))
string = 'aabccccd'
output = '2a3b4c4d'
new_string = " "
count = 1
for i in range(len(string)-1):
if string[i] == string[i+1]:
count = count + 1
else:
new_string = new_string + str(count) + string[i]
count = 1
new_string = new_string + str(count) + string[i+1]
print(new_string)
For a coding interview, where it was about the algorithm, and not about my knowledge of Python, its internal representation of data structures, or the time complexity of operations such as string concatenation:
def compress(message: str) -> str:
output = ""
length = 0
previous: str = None
for char in message:
if previous is None or char == previous:
length += 1
else:
output += previous
if length > 1:
output += str(length)
length = 1
previous = char
if previous is not None:
output += previous
if length > 1:
output += str(length)
return output
For code I'd actually use in production, not reinventing any wheels, being more testable, using iterators until the last step for space efficiency, and using join() instead of string concatenation for time efficiency:
from itertools import groupby
from typing import Iterator
def compressed_groups(message: str) -> Iterator[str]:
for char, group in groupby(message):
length = sum(1 for _ in group)
yield char + (str(length) if length > 1 else "")
def compress(message: str) -> str:
return "".join(compressed_groups(message))
Taking things a step further, for even more testability:
from itertools import groupby
from typing import Iterator
from collections import namedtuple
class Segment(namedtuple('Segment', ['char', 'length'])):
def __str__(self) -> str:
return self.char + (str(self.length) if self.length > 1 else "")
def segments(message: str) -> Iterator[Segment]:
for char, group in groupby(message):
yield Segment(char, sum(1 for _ in group))
def compress(message: str) -> str:
return "".join(str(s) for s in segments(message))
Going all-out and providing a Value Object CompressedString:
from itertools import groupby
from typing import Iterator
from collections import namedtuple
class Segment(namedtuple('Segment', ['char', 'length'])):
def __str__(self) -> str:
return self.char + (str(self.length) if self.length > 1 else "")
class CompressedString(str):
#classmethod
def compress(cls, message: str) -> "CompressedString":
return cls("".join(str(s) for s in cls._segments(message)))
#staticmethod
def _segments(message: str) -> Iterator[Segment]:
for char, group in groupby(message):
yield Segment(char, sum(1 for _ in group))
def compress(message: str) -> str:
return CompressedString.compress(message)
def compress(val):
print(len(val))
end=0
count=1
result=""
for i in range(0,len(val)-1):
#print(val[i],val[i+1])
if val[i]==val[i+1]:
count=count+1
#print(count,val[i])
elif val[i]!=val[i+1]:
#print(end,i)
result=result+val[end]+str(count)
end=i+1
count=1
result=result+val[-1]+str(count)
return result
res=compress("I need to create a function called compress that compresses a string by replacing any repeated letters with a letter and number. My function should return the shortened version of the string. I've been able to count the first character but not any others.")
print(len(res))
Use python's standard library re.
def compress(string):
import re
p=r'(\w+?)\1+' # non greedy, group1 1
sub_str=string
for m in re.finditer(p,string):
num=m[0].count(m[1])
sub_str=re.sub(m[0],f'{m[1]}{num}',sub_str)
return sub_str
string='aaaaaaaabbbbbbbbbcccccccckkkkkkkkkkkppp'
string2='ababcdcd'
string3='abcdabcd'
string4='ababcdabcdefabcdcd'
print(compress(string))
print(compress(string2))
print(compress(string3))
print(compress(string4))
Resut:
a8b9c8k11p3
ab2cd2
abcd2
ab2cdabcdefabcd2
Using generators:
input = "aaaddddffwwqqaattttttteeeeeee"
from itertools import groupby
print(''.join(([char+str(len(list(group))) for char, group in groupby(input)])))
def compress(string):
# taking out unique characters from the string
unique_chars = []
for c in string:
if not c in unique_chars:
unique_chars.append(c)
# Now count the characters
res = ""
for i in range(len(unique_chars)):
count = string.count(unique_chars[i])
res += unique_chars[i]+str(count)
return res
string = 'aabccccd'
compress(string)
from collections import Counter
def char_count(input_str):
my_dict = Counter(input_str)
print(my_dict)
output_str = ""
for i in input_str:
if i not in output_str:
output_str += i
output_str += str(my_dict[i])
return output_str
result = char_count("zddaaaffccc")
print(result)
This is the modification of Patrick Yu's code. It code fails for the below test cases.
SAMPLE INPUT:
c
aaaaaaaaaabcdefgh
EXPECTED OUTPUT:
c1
a10b1c1d1e1f1g1h1
OUPUT OF Patrick's Code:
c
a10bcdefgh
Below is the modified code:
def Compress(S):
Ans = S[0]
count = 1
for i in range(len(S)-1):
if S[i] == S[i+1]:
count += 1
else:
if count >= 1:
Ans += str(count)
Ans += S[i+1]
count = 1
if count>=1:
Ans += str(count)
return Ans
Just the condition must be changed from greater(">") to greater than equal to(">=") when comparing the count with 1.
I try to create a modified LZW which will find patterns of words inside a string. My problem is that 1st element is '' and last is not checked if it is in the list. I saw the pseudo-code from here : https://www.cs.duke.edu/csed/curious/compression/lzw.html . Here is my script for compression:
string = 'this is a test a test this is pokemon'
diction = []
x = ""
count = 0
for c in string.split():
print (c)
print (x)
#x = x + " " + c
if x in diction:
x += " " + c
#print("debug")
else:
#print(x)
diction.append(x)
x = c
count +=1
#print(count)
print (diction)
I tried to fix the 2nd problem by 'appending' a random word to the end of the string but I don't think that's the best solution.
For the 1st problem I tried just to define the variable "x" as str or None but I get this < class 'str' > inside the list.
The link deals with character and splitting a string will give an array of words.
In order to get not an empty string in the dictionary and parsing the last element.
string = 'this is a test a test this is pokemon'
diction = []
x = ""
count = 0
for c in string.split():
print (c)
if x+" "+c in diction:
x += " " + c
else:
diction.append(x+" "+c)
x = c
count +=1
print (diction)
But perhaps you would like something like :
string = 'this is a test a test this is pokemon'
diction = []
x = ""
count = 0
for c in string:
print (c)
if x+c in diction:
x += c
else:
diction.append(x+c)
x = c
count +=1
print (diction)
I'm not sure what the code pretends, but to fix the issues that you mentioned I think you could do this:
string = 'this is a test a test this is pokemon'
diction = []
x = None
count = 0
for c in string.split():
if x in diction:
x += " " + c
else:
if x: diction.append(x)
x = c
count += 1
if not x in diction: diction.append(x)
print (diction)
The output for that code would be:
['this', 'is', 'a', 'test', 'a test', 'this is', 'pokemon']