Given a String, the following program converts the alphabetic character around any digit to its uppercase.
Input : test_str = ‘neveropen4neveropen is best1 f6or ge8eks’
Output : geekS4Geeks is besT1 F6Or gE8Ek
Explanation : S and G are uppercased as surrounded by 4.
Input : test_str = ‘neveropen4neveropen best1 f6or ge8eks’
Output : geekS4Geeks besT1 F6Or gE8Ek
Explanation : S and G are uppercased as surrounded by 4.
Method 1 : Using upper(), loop and isdigit()
In this, we iterate for each character, check whether it’s a digit and if it is then, transform the surrounding alphabets(next and previous) to uppercase using upper().
Python3
# initializing stringtest_str = 'neveropen4neveropen is best1 for ge8eks'# printing original stringprint("The original string is : " + str(test_str))res = ''for idx in range(len(test_str) - 1): if test_str[idx + 1].isdigit() or test_str[idx - 1].isdigit(): res += test_str[idx].upper() else: res += test_str[idx]#Adding last index characterelse: res += test_str[idx+1] # printing result print("Transformed String : " + str(res)) |
The original string is :neveropen4neveropen is best1 for ge8eks Transformed String : geekS4Geeks is besT1 for gE8Eks
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 2: Using list comprehension
This is similar to the above method, the only difference being that the following tackles the same problem in a single line.
Python3
# initializing stringtest_str = 'neveropen4neveropen is best1 for ge8eks'# printing original stringprint("The original string is : " + str(test_str))# list comprehension offering 1 liner solutionres = [test_str[idx].upper() if test_str[idx + 1].isdigit() or test_str[idx - 1].isdigit() else test_str[idx] for idx in range(len(test_str) - 1)]res = res + list(test_str[-1]) # printing result print("Transformed String : " + ''.join(res)) |
The original string is :neveropen4neveropen is best1 for ge8eks Transformed String : geekS4Geeks is besT1 for gE8Eks
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #3 : By using array
Python3
# Python Program to Converts Characters To# Uppercase Around Numbers initializing# stringtest_str = 'neveropen4neveropen is best1 for ge8eks'# printing original stringprint("The original string is : " + str(test_str))res = ''digits = "0123456789"for idx in range(len(test_str) - 1): if test_str[idx + 1] in digits or test_str[idx - 1] in digits: res += test_str[idx].upper() else: res += test_str[idx]# Adding last index characterelse: res += test_str[idx+1]# printing resultprint("Transformed String : " + str(res)) |
The original string is :neveropen4neveropen is best1 for ge8eks Transformed String : geekS4Geeks is besT1 for gE8Eks
Time complexity: O(n), where n is the length of the input string.
Auxiliary space: O(n), where n is the length of the input string. A new string is created, and its size is proportional to the size of the input string.
Method #4 : Using operator.countOf() method
Python3
# Python Program to Converts Characters To# Uppercase Around Numbers initializing# stringimport operator as optest_str = 'neveropen4neveropen is best1 for ge8eks'# printing original stringprint("The original string is : " + str(test_str))res = ''digits = "0123456789"for idx in range(len(test_str) - 1): if op.countOf(digits, test_str[idx + 1]) > 0 or op.countOf(digits, test_str[idx - 1]) > 0: res += test_str[idx].upper() else: res += test_str[idx]# Adding last index characterelse: res += test_str[idx+1]# printing resultprint("Transformed String : " + str(res)) |
The original string is :neveropen4neveropen is best1 for ge8eks Transformed String : geekS4Geeks is besT1 for gE8Eks
Time Complexity: O(n)
Auxiliary Space: O(n)
Method#5 : Using a for loop and string slicing
Python3
test_str = 'neveropen4neveropen is best1 for ge8eks'result = ''# printing original stringprint("The original string is : " + str(test_str))for i in range(len(test_str)): if test_str[i-1:i+2].isdigit(): result += test_str[i].upper() else: result += test_str[i]print(result)#This code is contributed by Vinay pinjala. |
The original string is :neveropen4neveropen is best1 for ge8eks neveropen4neveropen is best1 for ge8eks
Time Complexity: O(n)
Auxiliary Space: O(n)
Method#6: Using re.finditer() and re.compile() functions:
1. Import the re module.
2.Define the input string.
3.Define a function called “transform_string” that takes the input string as a parameter.
4.Inside the function, compile a regular expression pattern that matches characters that are either preceded by a digit or followed by a digit.
5.Initialize an empty string called “res” to store the result.
6.Initialize a variable called “last_index” to keep track of the last index.
7.Iterate over all the matches found in the input string using the “re.finditer()” method.
8.For each match, add the characters from the last index to the start of the current match to the result string.
9.Add the uppercased character to the result string.
10.Update the last index to the end of the current match.
11.Add the remaining characters from the input string to the result string.
12.Return the result string.
13.Call the function with the input string as a parameter.
14.Print the transformed string.
Python3
# import the re moduleimport retest_str='neveropen4neveropen is best1 for ge8eks'# printing original stringprint("The original string is : " + str(test_str))# define a function to transform the stringdef transform_string(test_str): # compile a pattern to match characters that are either # preceded by a digit or followed by a digit pattern = re.compile(r'(?<=\d)[a-zA-Z]|[a-zA-Z](?=\d)') # initialize an empty string to store the result res = '' # initialize a variable to keep track of the last index last_index = 0 # iterate over all the matches found in the input string for match in re.finditer(pattern, test_str): # add the characters from the last index to the start of the current match # to the result string res += test_str[last_index:match.start()] # add the uppercased character to the result string res += match.group().upper() # update the last index to the end of the current match last_index = match.end() # add the remaining characters from the input string to the result string res += test_str[last_index:] # return the result string return res# call the function to transform the stringresult = transform_string(test_str)# print the resultprint("Transformed String : " + result)#This code is contributed by Jyothi pinjala. |
The original string is :neveropen4neveropen is best1 for ge8eks Transformed String : geekS4Geeks is besT1 for gE8Eks
Time Complexity:
Compiling the regular expression pattern takes O(m) time where m is the length of the pattern.
The iteration over all the matches in the input string takes O(n) time where n is the length of the input string.
The time complexity of adding characters to a string is O(1) time per character.
Therefore, the overall time complexity of the algorithm is O(m + n), where m is the length of the pattern and n is the length of the input string.
Space Complexity:
The space complexity of the algorithm is O(n), where n is the length of the input string.
This is because we are creating a new string to store the result, which has a maximum size of n.
Method 7: Using a lambda function and map() method
Iterating over each character in the input string, checking if the previous two characters and the current character form a digit using the isdigit() method, and then appending the modified character to an empty result string. The modified string is then printed as output.
Step-by-step approach:
- Define a lambda function that takes a character as input and returns the uppercase version if the previous character is a digit, otherwise returns the same character.
- Use the map() method to apply the lambda function to each character in the input string.
- Join the resulting list of characters back into a single string.
Below is the implementation of the above approach:
Python3
# Define the input stringtest_str = 'neveropen4neveropen is best1 for ge8eks'# Initialize an empty string to store the modified string.result = ''# Iterate over each character in the input string.for i in range(len(test_str)): # Check if the current character and the previous two characters form a digit. if test_str[i-1:i+2].isdigit(): # If the previous two characters and the current character form a digit, append the uppercase version of the current character to the result string. result += test_str[i].upper() else: # Otherwise, append the current character to the result string. result += test_str[i]# Print the resultprint("The modified string is : " + str(result)) |
The modified string is :neveropen4neveropen is best1 for ge8eks
Time complexity: O(n), where n is the length of the input string. The lambda function and map() method iterate over each character in the string once.
Auxiliary space: O(n), where n is the length of the input string. A list is created to hold the modified characters before they are joined back into a single string.
