Python program to convert String to K sized Numerical Rows

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 = ‘geeksforgeek’, 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 character

Input : test_str = ‘geeksforgeek’, 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 string
test_str = 'geeksforgeekscse'

# printing original string
print("The original string is : " + str(test_str))

# initializing K
K = 4

alphabs = "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 characters
if temp != []:
    res.append(temp)

# printing result
print("Grouped and Converted String : " + str(res))

Output:

The original string is : geeksforgeekscse 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 string
test_str = 'geeksforgeekscse'

# printing original string
print("The original string is : " + str(test_str))

# initializing K
K = 4

# filling the rear to K size rows
temp = test_str.ljust(ceil(len(test_str) / K) * K)

# convert to numerical characters
temp = [0 if char == ' ' else (ord(char) - 97) for char in temp]

# slice and render to matrix
res = [temp[idx: idx + K] for idx in range(0, len(temp), K)]

# printing result
print("Grouped and Converted String : " + str(res))

Output:

The original string is : geeksforgeekscse 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 string
test_str = 'geeksforgeekscse'

#printing original string
print("The original string is : " + str(test_str))

#initializing K
K = 4

#creating a dictionary for alphabets and their corresponding numerical values
alphabet_dict = {chr(i): i-97 for i in range(97, 123)}

#converting characters to their corresponding numerical values
temp = [alphabet_dict[ele] for ele in test_str]

#slicing the list into K sized sublists
res = [temp[i:i+K] for i in range(0, len(temp), K)]

#printing result
print("Grouped and Converted String : " + str(res))

Output

The original string is : geeksforgeekscse
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 string
test_str = 'geeksforgeekscse'

# printing original string
print("The original string is : " + str(test_str))

# initializing K
K = 4

# creating a dictionary for alphabets and their corresponding numerical values
alphabet_dict = {chr(i): i-97 for i in range(97, 123)}

# converting characters to their corresponding numerical values
temp = [alphabet_dict[ele] for ele in test_str]

# grouping the list into K-sized sublists using the zip() function
res = [temp[i:i+K] for i in range(0, len(temp), K)]

# printing result
print("Grouped and Converted String : " + str(res))

Output

The original string is : geeksforgeekscse
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.