Python | Check possible bijection between sequence of characters and digits

Given a string ‘char_seq'(sequence of characters) and a positive integer ‘dig_seq'(sequence of digits), Write a Python program to find possible bijection or one-one onto relationship between ‘char_seq’ and ‘dig_seq’ such that each character matches to one and only one digit. 

Examples:

Input : char_seq = 'bxdyxb'
        dig_seq = 123421
Output : True

Input : char_seq = 'bxdyxb'
        dig_seq = 123321
Output : False

Method #1: Using zip method This method simply zips the ‘char_seq’ and ‘dig_seq’ and checks if corresponding digits and characters matches or not. 

Python3

# Python3 program to Check possible bijection
# between sequence of characters and digits

def is_bijection(char_seq, dig_seq):
    z = zip(str(char_seq), str(dig_seq))
    res = all(
    (z1[0] == z2[0]) == (z1[1] == z2[1]) for z1 in z for z2 in z)

    return res

# Driver code
char_seq = 'bxdyxb'
dig_seq = 123421
print(is_bijection(char_seq, dig_seq))

Output

True

Method #2: Using itertools.groupby method This method uses the same approach with a slight difference, it uses itertools.groupby to match characters with digits. 

Python3

# Python3 program to Check possible bijection
# between sequence of characters and digits
import itertools

def is_bijection(char_seq, dig_seq):
    z = sorted(zip(str(char_seq), str(dig_seq)))
    res = all(gx == gy
          for k, g in itertools.groupby(z, key = lambda res: res[0])
          for gx in g for gy in g)

    return res

# Driver code
char_seq = 'bxdyxb'
dig_seq = 123421
print(is_bijection(char_seq, dig_seq))

Output

True

Method #3: It works by iterating through the input sequences and creating a dictionary that maps characters to digits. For each character and digit pair, if the character is already in the dictionary and the digit does not match the one already mapped to the character, then the bijection is not valid and the function returns False. Otherwise, the character and digit are added to the dictionary. If the loop completes, then the bijection is valid and the function returns True.

Python3

def is_bijection(char_seq, dig_seq):
    # Create a dictionary to map characters to digits
    dig_seq=str(dig_seq)
    char_dict = {}
    # Iterate through the character and digit pairs
    for c, d in zip(char_seq, dig_seq):
        # If the character is already in the dictionary
        if c in char_dict:
            # If the digit does not match the one already mapped to the character,
            # then the bijection is not valid
            if char_dict != d:
                return False
        else:
            # Add the character and digit to the dictionary
            char_dict = d
    # If the loop completes, then the bijection is valid
    return True

char_seq = 'bxdyxb'
dig_seq = 123421
print(is_bijection(char_seq, dig_seq))
#This code is contributed by Edula Vinay Kumar Reddy

Output

True

 Time complexity: O(n) and 
Auxiliary space: O(n), where n is the length of the input sequences.

Method #4: By repeatedly going through the input sequences, it builds a dictionary that converts characters to numbers. The bijection is invalid and the function returns False for any character and digit pair where the character is already present in the dictionary and the digit does not match the one previously mapped to the character. The character and digit are added to the vocabulary in all other cases. The bijection is true and the function returns True if the loop is finished.

Python

# Python program to Check possible bijection
# between sequence of characters and digits
def is_bijection(char_seq, dig_seq):
    dig_seq_str = str(dig_seq)
    if len(char_seq) != len(dig_seq_str):
        return False

    char_to_dig = {}
    dig_to_char = {}

    for char, dig in zip(char_seq, dig_seq_str):
        if char in char_to_dig and char_to_dig[char] != dig:
            return False
        if dig in dig_to_char and dig_to_char[dig] != char:
            return False
        char_to_dig[char] = dig
        dig_to_char[dig] = char

    return True

# drive code
char_seq = 'bxdyxb'
dig_seq = 123421
print(is_bijection(char_seq, dig_seq))

Output

True

Time Complexity : O(n)
Auxiliary Space: O(n)