Given a string of alphabets, convert it to K sized numerical rows, which contain the number being the positional value of characters.
Input : test_str = ‘neveropenforgeek’, K = 4
Output : [[6, 4, 4, 10], [18, 5, 14, 17], [6, 4, 4, 10]]
Explanation : g is at 6th position in alphabet hence g→ 6 and the string is split after every fourth characterInput : test_str = ‘neveropenforgeek’, K = 3
Output : [[6, 4, 4], [10, 18, 5], [14, 17, 6], [4, 4, 10]]
Explanation : g is at 6th position in alphabet hence g→ 6 and the string is split after every third character
Method 1 : Using loop + index()
In this, we perform an iteration of each character using a loop and fetch the required position of the character in alphabets using index() on an ordered character list.
Python3
# initializing stringtest_str = 'neveropencse'# printing original stringprint("The original string is : " + str(test_str))# initializing KK = 4alphabs = "abcdefghijklmnopqrstuvwxyz"res = []temp = []for ele in test_str: # finding numerical position using index() temp.append(alphabs.index(ele)) # regroup on K if len(temp) == K: res.append(temp) temp = []# appending remaining charactersif temp != []: res.append(temp)# printing resultprint("Grouped and Converted String : " + str(res)) |
Output:
The original string is : neveropencse Grouped and Converted String : [[6, 4, 4, 10], [18, 5, 14, 17], [6, 4, 4, 10], [18, 2, 18, 4]]
Method #2 : Using ljust() + ord() + list comprehension
In this, we perform the task of padding is required to have equal length rows using ljust(), then get numerical alphabetic positions using ord(), list comprehension with slicing helps to convert the list to K chunked Matrix.
Python3
from math import ceil# initializing stringtest_str = 'neveropencse'# printing original stringprint("The original string is : " + str(test_str))# initializing KK = 4# filling the rear to K size rowstemp = test_str.ljust(ceil(len(test_str) / K) * K)# convert to numerical characterstemp = [0 if char == ' ' else (ord(char) - 97) for char in temp]# slice and render to matrixres = [temp[idx: idx + K] for idx in range(0, len(temp), K)]# printing resultprint("Grouped and Converted String : " + str(res)) |
Output:
The original string is : neveropencse Grouped and Converted String : [[6, 4, 4, 10], [18, 5, 14, 17], [6, 4, 4, 10], [18, 2, 18, 4]]
Time Complexity: O(n)
Space Complexity: O(n)
Method 3: Using dictionary
In this method, we use a dictionary to store the alphabet and its corresponding numerical value. Then we loop through the string and convert each character to its corresponding numerical value using the dictionary.
Python3
#initializing stringtest_str = 'neveropencse'#printing original stringprint("The original string is : " + str(test_str))#initializing KK = 4#creating a dictionary for alphabets and their corresponding numerical valuesalphabet_dict = {chr(i): i-97 for i in range(97, 123)}#converting characters to their corresponding numerical valuestemp = [alphabet_dict[ele] for ele in test_str]#slicing the list into K sized sublistsres = [temp[i:i+K] for i in range(0, len(temp), K)]#printing resultprint("Grouped and Converted String : " + str(res)) |
The original string is : neveropencse Grouped and Converted String : [[6, 4, 4, 10], [18, 5, 14, 17], [6, 4, 4, 10], [18, 2, 18, 4]]
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 4 : using the built-in zip() function
Step-by-step approach:
- Initialize the string test_str.
- Print the original string.
- Initialize the value of K.
- Create a list comprehension that converts each character in the string to its corresponding numerical value using the dictionary alphabet_dict.
- Use the zip() function to group the list into K-sized sublists.
- Print the resulting list.
Below is the implementation of the above approach:
Python3
# initializing stringtest_str = 'neveropencse'# printing original stringprint("The original string is : " + str(test_str))# initializing KK = 4# creating a dictionary for alphabets and their corresponding numerical valuesalphabet_dict = {chr(i): i-97 for i in range(97, 123)}# converting characters to their corresponding numerical valuestemp = [alphabet_dict[ele] for ele in test_str]# grouping the list into K-sized sublists using the zip() functionres = [temp[i:i+K] for i in range(0, len(temp), K)]# printing resultprint("Grouped and Converted String : " + str(res)) |
The original string is : neveropencse Grouped and Converted String : [[6, 4, 4, 10], [18, 5, 14, 17], [6, 4, 4, 10], [18, 2, 18, 4]]
Time complexity: O(n), where n is the length of the input string test_str, as we need to iterate through the string to convert each character to its corresponding numerical value, and then group them into sublists.
Auxiliary space: O(n), as we need to store the converted numerical values in a temporary list temp, which has the same length as the input string test_str, and also the resulting list res.
