Python Program to print all distinct uncommon digits present in two given numbers

Given two positive integers A and B, the task is to print the distinct digits in descending order, which are not common in the two numbers.

Examples:

Input: A = 378212, B = 78124590
Output: 9 5 4 3 0
Explanation: All distinct digits present in the two numbers are {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. The digits {1, 2, 6, 7} are common in both numbers.

Input: A = 100, B = 273
Output: 7 3 2 1 0
Explanation: All distinct digits present in the two numbers are {0, 1, 2, 3, 7}. The digits {0, 1, 2, 3, 7} are common in both numbers.

Approach: The problem can be solved using sets, and lists in python. Follow the steps below to solve the problem:

• Convert both integers into strings.
• Now, store the equivalent integer values of the characters of the strings A and B in the lists lis1 and lis2, using map() function.
• Convert both the lists lis1 and lis2 to set lis1 and lis2 using the set() method.
• Now, find the uncommon digits of both sets lis1 and lis2 using the function symmetric_difference() and store it in a set, say lis.
• After completing the above steps, convert the set lis into list lis using list() method.
• Finally, Sort the list in descending order and print the list lis as the answer.

Below is the implementation of the above approach:

Python3

# Python implementation
# of the above approach

# Function to print uncommon digits
# of two numbers in descending order
def disticntUncommonDigits(A, B):

    # Stores digits of A as string
    A = str(A)

    # Stores digits of B as string
    B = str(B)

    # Stores the characters
    # of A in a integer list
    lis1 = list(map(int, A))

    # Stores the characters
    # of B in a integer list
    lis2 = list(map(int, B))

    # Converts lis1 to set
    lis1 = set(lis1)

    # Converts lis2 to set
    lis2 = set(lis2)

    # Stores the uncommon digits present
    # in the sets lis1 and lis2
    lis = lis1.symmetric_difference(lis2)

    # Converts lis to list
    lis = list(lis)

    # Sort the list in descending order
    lis.sort(reverse = True)

    # Print the list lis
    for i in lis:
        print(i, end =" ")

# Driver Code


# Input
A = 378212
B = 78124590

disticntUncommonDigits(A, B)

Output:

9 5 4 3 0

Time Complexity: O(log₁₀(A) + log₁₀(B))
Auxiliary Space: O(log₁₀(A) + log₁₀(B))

Method #2:Using Operator.countOf() method

Python3

# Python implementation
# of the above approach
import operator as op
# Function to print uncommon digits
# of two numbers in descending order


def disticntUncommonDigits(A, B):

    # Stores digits of A as string
    A = str(A)

    # Stores digits of B as string
    B = str(B)

    # Stores the characters
    # of A in a integer list
    lis1 = list(map(int, A))

    # Stores the characters
    # of B in a integer list
    lis2 = list(map(int, B))
    res = []
    for i in lis1:
        if op.countOf(lis2, i) == 0:
            res.append(i)
    for i in lis2:
        if op.countOf(lis1, i) == 0:
            res.append(i)
    # Sort the list in descending order
    res.sort(reverse=True)

    # Print the list lis
    for i in res:
        print(i, end=" ")

# Driver Code


# Input
A = 378212
B = 78124590

disticntUncommonDigits(A, B)

Output

9 5 4 3 0 

Time Complexity: O(log10(A) + log10(B))
Auxiliary Space: O(log10(A) + log10(B))

Approach#3: Using set() and difference() and sorted and union

Convert both numbers A and B to sets of digits. Find uncommon digits in both sets by taking the set difference and union of the sets. Convert the resulting set of uncommon digits to a list and sort it in descending order. Print the list of uncommon digits as a string.

Algorithm

1. Initialize A and B with given values.
2. Convert A and B to sets of digits.
3. Find uncommon digits using set difference and union.
4. Convert resulting set to a list and sort it in descending order.
5. Print the list of uncommon digits as a string.

Python3

A = 378212
B = 78124590

# find unique digits in each number
digits_A = set(str(A))
digits_B = set(str(B))

# find uncommon digits using set difference
uncommon_digits = digits_A.difference(digits_B).union(digits_B.difference(digits_A))

# convert set to list and sort in descending order
uncommon_digits = sorted(list(uncommon_digits), reverse=True)

# print the list of uncommon digits
print(' '.join(uncommon_digits))

Output

9 5 4 3 0

Time Complexity: O(m+n), where m and n are the number of digits in A and B, respectively. This is because the set() function and set difference and union operations take O(m) and O(n) time, respectively.

Space Complexity: O(m+n), where m and n are the number of digits in A and B, respectively. This is because we are storing the digits of A and B in sets, which take up O(m) and O(n) space, respectively. The resulting set of uncommon digits and the list of uncommon digits also take up O(m+n) space.