- Given an input string with lowercase letters, the task is to write a python program to identify the repeated characters in the string and capitalize them.
Examples:
Input: programming language
Output: pRoGRAMMiNG lANGuAGe
Explanation: r,m,n,a,g are repeated elementsInput: geeks for geeks
Output: GEEKS for GEEKS
Explanation: g,e,k,s are repeated elements
Approach 1:
- We have to keep the character of a string as a key and the frequency of each character of the string as a value in the dictionary.
- Traverse the string and check the frequency of each character using a dictionary if the frequency of the character is greater than one then change the character to the uppercase using the upper() function.
Implementation:
# function for changing the # repeated characters to uppercase def RepeatedUpper(s):
# declaring dictionary
dic = {}
# Traversing the string
for i in s:
# If the character is already
# present in dictionary then increment
# the frequency of the character
if i in dic:
dic[i] = dic[i] + 1
# If the character is not present in
# the dictionary then inserting
# the character in the dictionary
else :
dic[i] = 1
ans = ''
# traversing the string
for i in s:
# if the frequency of the character is
# greater than one
if dic[i] > 1 :
# change into uppercase
i = i.upper()
# appending each character to the ans
ans = ans + i
return ans
# Driver code s = 'geeks for geeks'
# function call print (RepeatedUpper(s))
|
GEEKS for GEEKS
Time Complexity: O(n)
Auxiliary Space: O(n)
Approach 2:
Using count() function.If count is greater than 1 then the character is repeated.Later on used upper() to convert to uppercase
s = "programming language"
new_str = ""
for i in s:
if (i! = "" and s.count(i)> 1 ):
new_str + = i.upper()
else :
new_str + = i
print (new_str)
|
pRoGRAMMiNG lANGuAGe
Time Complexity: O(n2) -> (count function + loop)
Auxiliary Space: O(n)
Approach 3: Using replace() and len() methods
#Capitalize repeated letters s = "programming language"
length = len (s)
new_str = ""
for i in s:
x = s.replace(i,"")
if ( len (x)! = length - 1 ):
new_str + = i.upper()
else :
new_str + = i
print (new_str)
|
pRoGRAMMiNG lANGuAGe
Time Complexity: O(n2) -> (replace function + loop)
Auxiliary Space: O(n)
Approach #4 : Using Counter() function
from collections import Counter
s = "programming language"
new_str = ""
freq = Counter(s)
for i in s:
if (i ! = "" and freq[i] > 1 ):
new_str + = i.upper()
else :
new_str + = i
print (new_str)
|
pRoGRAMMiNG lANGuAGe
Time Complexity: O(n)
Auxiliary Space: O(n)
Approach #5 : Using List Comprehension
In this approach, we first create a Counter object to store the frequency of each character in the string. Then, we use a list comprehension to iterate over the characters in the string, and for each character, we check if its frequency is greater than 1. If it is, we convert it to uppercase using upper(). Finally, we use join() to concatenate the characters and return the result as a single string.
from collections import Counter
def capitalize_repeated_chars(s):
freq = Counter(s)
return "".join([char.upper() if freq[char] > 1 else char for char in s])
s = "programming language"
print (capitalize_repeated_chars(s))
|
pRoGRAMMiNG lANGuAGe
Time complexity: O(n), where n is the length of the input string.
Auxiliary space: O(n), for storing the frequency of each character in the Counter object.
Approach #6 : Using operator.countOf() method
- In this approach we iterate over the characters in the string.
- For each character, we check if its frequency(operator.countOf()) is greater than 1.
- Initialize a new_str with empty string.
- If it is, we convert it to uppercase using upper() and concatenate to new_str.
- If not concatenate the character as it is to new_str.
- Finally print the new_str.
s = "programming language"
new_str = ""
import operator
for i in s:
if (i! = "" and operator.countOf(s,i)> 1 ):
new_str + = i.upper()
else :
new_str + = i
print (new_str)
|
pRoGRAMMiNG lANGuAGe
Time complexity: O(n), where n is the length of the input string.
Auxiliary space: O(n), for storing the frequency of each character.