# 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:

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.

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