Print all repeating digits present in a given number in sorted order

Finding repeating digits in a number is a common programming task. Python provides several built-in functions like count(), Counter(), and operator.countOf() that can efficiently identify and sort duplicate digits.

Problem Statement

Given a number, we need to find all digits that appear more than once and display them in sorted order.

Input

inputNum = 5322789124199

Expected Output

1 2 9

In the input number 5322789124199, digits 1, 2, and 9 appear multiple times, so they are displayed in ascending order.

Method 1: Using count() Function

The count() method returns the number of occurrences of a substring in a string. We convert the number to a string and check each digit's frequency.

Algorithm

• Convert the input number to a string
• Initialize an empty list for storing duplicate digits
• Traverse each character and check if its count is greater than 1
• Add unique duplicates to the result list
• Sort the list and display the result

Example

# input number
inputNum = 5322789124199

# storing resultant repeating digits in a list
duplicatesList = []

# converting input number to a string
newString = str(inputNum)

# traversing through each character in the string
for c in newString:
    # checking if frequency > 1 and not already in duplicates list
    if newString.count(c) > 1 and c not in duplicatesList:
        # appending character to duplicates list
        duplicatesList.append(c)

# sorting the resultant duplicates list
duplicatesList.sort()

# converting list to string and printing
print(' '.join(duplicatesList))

1 2 9

Method 2: Using Hashing

This approach uses an array to count the frequency of each digit (0-9), then identifies digits with frequency greater than 1.

def getDuplicateNum(inputNum):
    # array to count frequency of digits 0-9
    count = [0] * 10
    
    # converting input number to string
    newString = str(inputNum)
    
    # counting frequency of each digit
    for c in newString:
        curr_digit = int(c)
        count[curr_digit] += 1
    
    # checking for digits with frequency > 1
    result = []
    for k in range(10):
        if count[k] > 1:
            result.append(str(k))
    
    return ' '.join(result)

# input number
inputNum = 5322789124199
print(getDuplicateNum(inputNum))

1 2 9

Method 3: Using Counter() Function

The Counter class from the collections module creates a frequency dictionary of characters automatically.

from collections import Counter

# input number
inputNum = 5322789124199

# converting input number to string
newString = str(inputNum)

# getting frequency of each character
charFrequency = Counter(newString)

# finding digits with frequency > 1
duplicatesList = []
for digit, frequency in charFrequency.items():
    if frequency > 1:
        duplicatesList.append(digit)

# sorting and displaying result
duplicatesList.sort()
print(' '.join(duplicatesList))

1 2 9

Method 4: Using operator.countOf() Function

The operator.countOf() function returns the number of occurrences of a value in a sequence.

import operator as op

# input number
inputNum = 5322789124199

# storing resultant repeating digits
duplicatesList = []

# converting input number to string
newString = str(inputNum)

# traversing each character
for c in newString:
    # checking if frequency > 1 and not already added
    if op.countOf(newString, c) > 1 and op.countOf(duplicatesList, c) == 0:
        duplicatesList.append(c)

# sorting and displaying result
duplicatesList.sort()
print(' '.join(duplicatesList))

1 2 9

Comparison

Conclusion

The Counter() method provides the most readable solution, while the hashing approach offers the best time complexity. Choose the method based on your specific requirements for readability and performance.