Develop Reference

Python Inserting a string

I need to insert a string (character by character) into another string at every 3rd position
For example:- string_1:-wwwaabkccgkll
String_2:- toadhp
Now I need to insert string2 char by char into string1 at every third position
So the output must be wwtaaobkaccdgkhllp
Need in Python.. even Java is ok
So i tried this
Test_str="hiimdumbiknow"
challenge="toadh"
new_st=challenge [k]
Last=list(test_str)
K=0
For i in range(Len(test_str)):
if(i%3==0):
last.insert(i,new_st)
K+=1
and the output i get
thitimtdutmbtiknow

You can split test_str into sub-strings to length 2, and then iterate merging them with challenge:
def concat3(test_str, challenge):
chunks = [test_str[i:i+2] for i in range(0,len(test_str),2)]
result = []
i = j = 0
while i<len(chunks) or j<len(challenge):
if i<len(chunks):
result.append(chunks[i])
i += 1
if j<len(challenge):
result.append(challenge[j])
j += 1
return ''.join(result)
test_str = "hiimdumbiknow"
challenge = "toadh"
print(concat3(test_str, challenge))
# hitimoduambdikhnow
This method works even if the lengths of test_str and challenge are mismatching. (The remaining characters in the longest string will be appended at the end.)

You can split Test_str in to groups of two letters and then re-join with each letter from challenge in between as follows;
import itertools
print(''.join(f'{two}{letter}' for two, letter in itertools.zip_longest([Test_str[i:i+2] for i in range(0,len(Test_str),2)], challenge, fillvalue='')))
Output:
hitimoduambdikhnow
*edited to split in to groups of two rather than three as originally posted

you can try this, make an iter above the second string and iterate over the first one and select which character should be part of the final string according the position
def add3(s1, s2):
def n():
try:
k = iter(s2)
for i,j in enumerate(s1):
yield (j if (i==0 or (i+1)%3) else next(k))
except:
try:
yield s1[i+1:]
except:
pass
return ''.join(n())

def insertstring(test_str,challenge):
result = ''
x = [x for x in test_str]
y = [y for y in challenge]
j = 0
for i in range(len(x)):
if i % 2 != 0 or i == 0:
result += x[i]
else:
if j < 5:
result += y[j]
result += x[i]
j += 1
get_last_element = x[-1]
return result + get_last_element
print(insertstring(test_str,challenge))
#output: hitimoduambdikhnow

How to find the longest repeating sequence using python

I went through an interview, where they asked me to print the longest repeated character sequence.
I got stuck is there any way to get it?
But my code prints only the count of characters present in a string is there any approach to get the expected output
import pandas as pd
import collections
a = 'abcxyzaaaabbbbbbb'
lst = collections.Counter(a)
df = pd.Series(lst)
df
Expected output :
bbbbbbb
How to add logic to in above code?

A regex solution:
max(re.split(r'((.)\2*)', a), key=len)
Or without library help (but less efficient):
s = ''
max((s := s * (c in s) + c for c in a), key=len)
Both compute the string 'bbbbbbb'.

Without any modules, you could use a comprehension to go backward through possible sizes and get the first character multiplication that is present in the string:
next(c*s for s in range(len(a),0,-1) for c in a if c*s in a)
That's quite bad in terms of efficiency though
another approach would be to detect the positions of letter changes and take the longest subrange from those
chg = [i for i,(x,y) in enumerate(zip(a,a[1:]),1) if x!=y]
s,e = max(zip([0]+chg,chg+[len(a)]),key=lambda se:se[1]-se[0])
longest = a[s:e]
Of course a basic for-loop solution will also work:
si,sc = 0,"" # current streak (start, character)
ls,le = 0,0 # longest streak (start, end)
for i,c in enumerate(a+" "): # extra space to force out last char.
if i-si > le-ls: ls,le = si,i # new longest
if sc != c: si,sc = i,c # new streak
longest = a[ls:le]
print(longest) # bbbbbbb

A more long winded solution, picked wholesale from:
maximum-consecutive-repeating-character-string
def maxRepeating(str):
len_s = len(str)
count = 0
# Find the maximum repeating
# character starting from str[i]
res = str[0]
for i in range(len_s):
cur_count = 1
for j in range(i + 1, len_s):
if (str[i] != str[j]):
break
cur_count += 1
# Update result if required
if cur_count > count :
count = cur_count
res = str[i]
return res, count
# Driver code
if __name__ == "__main__":
str = "abcxyzaaaabbbbbbb"
print(maxRepeating(str))
Solution:
('b', 7)

Longest Prefix That is Also a Substring in Second String

This code (adapted from a Prefix-Suffix code) is quite slow for larger corpora:
s1 = 'gafdggeg'
s2 = 'adagafrd'
Output: gaf
def pref_also_substr(s):
n = len(s)
for res in range(n, 0, -1):
prefix = s[0: res]
if (prefix in s1):
return res
# if no prefix and string2 match occurs
return 0
Any option for an efficient alternative?

I have another approach to solve this question. First you can find all substrings of s2 and replace the key in dictionary d with highest size.
s2 = "'adagafrd'"
# Get all substrings of string
# Using list comprehension + string slicing
substrings = [test_str[i: j] for i in range(len(test_str))
for j in range(i + 1, len(test_str) + 1)]
Now you can use startswith() function to check longest prefix from this list of substring and compare the size of substring.
s1 = 'gafdggeg'
d={}
for substring in substrings:
if s1.startswith(substring):
if not d:
d[substring]=len(substring)
else:
if len(substring)>list(d.values())[0]:
d={}
d[substring]=len(substring)
print(d)
Output:
{'gaf': 3}

def f(s1, s2):
for i in range(len(s1)):
i += 1
p = s1[:i]
if p in s2:
s2 = s2[s2.index(p):]
else:
return i - 1
Check the prefixes starting from length 1.
If find a prefix, discard the chars behind the prefix founded and continue searching.

How to change uppercase & lowercase alternatively in a string?

I want to create a new string from a given string with alternate uppercase and lowercase.
I have tried iterating over the string and changing first to uppercase into a new string and then to lower case into another new string again.
def myfunc(x):
even = x.upper()
lst = list(even)
for itemno in lst:
if (itemno % 2) !=0:
even1=lst[1::2].lowercase()
itemno=itemno+1
even2=str(even1)
print(even2)
Since I cant change the given string I need a good way of creating a new string alternate caps.

Here's a onliner
"".join([x.upper() if i%2 else x.lower() for i,x in enumerate(mystring)])

You can simply randomly choose for each letter in the old string if you should lowercase or uppercase it, like this:
import random
def myfunc2(old):
new = ''
for c in old:
lower = random.randint(0, 1)
if lower:
new += c.lower()
else:
new += c.upper()
return new

Here's one that returns a new string using with alternate caps:
def myfunc(x):
seq = []
for i, v in enumerate(x):
seq.append(v.upper() if i % 2 == 0 else v.lower())
return ''.join(seq)

This does the job also
def foo(input_message):
c = 0
output_message = ""
for m in input_message:
if (c%2==0):
output_message = output_message + m.lower()
else:
output_message = output_message + m.upper()
c = c + 1
return output_message

Here's a solution using itertools which utilizes string slicing:
from itertools import chain, zip_longest
x = 'inputstring'
zipper = zip_longest(x[::2].lower(), x[1::2].upper(), fillvalue='')
res = ''.join(chain.from_iterable(zipper))
# 'iNpUtStRiNg'

Using a string slicing:
from itertools import zip_longest
s = 'example'
new_s = ''.join(x.upper() + y.lower()
for x, y in zip_longest(s[::2], s[1::2], fillvalue=''))
# ExAmPlE
Using an iterator:
s_iter = iter(s)
new_s = ''.join(x.upper() + y.lower()
for x, y in zip_longest(s_iter, s_iter, fillvalue=''))
# ExAmPlE
Using the function reduce():
def func(x, y):
if x[-1].islower():
return x + y.upper()
else:
return x + y.lower()
new_s = reduce(func, s) # eXaMpLe

This code also returns alternative caps string:-
def alternative_strings(strings):
for i,x in enumerate(strings):
if i % 2 == 0:
print(x.upper(), end="")
else:
print(x.lower(), end= "")
return ''
print(alternative_strings("Testing String"))

def myfunc(string):
# Un-hash print statements to watch python build out the string.
# Script is an elementary example of using an enumerate function.
# An enumerate function tracks an index integer and its associated value as it moves along the string.
# In this example we use arithmetic to determine odd and even index counts, then modify the associated variable.
# After modifying the upper/lower case of the character, it starts adding the string back together.
# The end of the function then returns back with the new modified string.
#print(string)
retval = ''
for space, letter in enumerate(string):
if space %2==0:
retval = retval + letter.upper()
#print(retval)
else:
retval = retval + letter.lower()
#print(retval)
print(retval)
return retval
myfunc('Thisisanamazingscript')

Check the most frequent letter(s) in a word. Python

My task is:
To write a function that gets a string as an argument and returns the letter(s) with the maximum appearance in it.
Example 1:
s = 'Astana'
Output:
a
Example 2:
s = 'Kaskelen'
Output:
ke
So far, I've got this code(click to run):
a = input()
def most_used(w):
a = list(w)
indexes = []
g_count_max = a.count(a[0])
for letter in a:
count = 0
i = int()
for index in range(len(a)):
if letter == a[index] or letter == a[index].upper():
count += 1
i = index
if g_count_max <= count: //here is the problem.
g_count_max = count
if i not in indexes:
indexes.append(i)
letters = str()
for i in indexes:
letters = letters + a[i].lower()
return letters
print(most_used(a))
The problem is that it automatically adds first letter to the array because the sum of appearance of the first element is actually equal to the starter point of appearance(which is basically the first element).
Example 1:
s = 'hheee'
Output:
he
Example 2:
s = 'malaysia'
Output:
ma

I think what you're trying to can be much simplified by using the standard library's Counter object
from collections import Counter
def most_used(word):
# this has the form [(letter, count), ...] ordered from most to least common
most_common = Counter(word.lower()).most_common()
result = []
for letter, count in most_common:
if count == most_common[0][1]:
result.append(letter) # if equal largest -- add to result
else:
break # otherwise don't bother looping over the whole thing
return result # or ''.join(result) to return a string

You can use a dictionary comprehension with a list comprehension and max():
s = 'Kaskelen'
s_lower = s.lower() #convert string to lowercase
counts = {i: s_lower.count(i) for i in s_lower}
max_counts = max(counts.values()) #maximum count
most_common = ''.join(k for k,v in counts.items() if v == max_counts)
Yields:
'ke'

try this code using list comprehensions:
word = input('word=').lower()
letters = set(list(word))
max_w = max([word.count(item) for item in letters])
out = ''.join([item for item in letters if word.count(item)==max_w])
print(out)

Also you can import Counter lib:
from collections import Counter
a = "dagsdvwdsbd"
print(Counter(a).most_common(3)[0][0])
Then it returns:
d