Count of repeating digits in a given Number

Given a number N, the task is to count the total number of repeating digits in the given number.

Examples:

Input: N = 99677  
Output: 2
Explanation:
In the given number only 9 and 7 are repeating, hence the answer is 2.

Input: N = 12
Output: 0
Explanation:
In the given number no digits are repeating, hence the answer is 0.

Naive Approach: The idea is to use two nested loops. In the first loop, traverse from the first digit of the number to the last, one by one. Then for each digit in the first loop, run a second loop and search if this digit is present anywhere else as well in the number. If yes, then increase the required count by 1. In the end, print the calculated count.

Time Complexity: O(N²)
Auxiliary Space: O(1)

Efficient Approach: The idea is to use Hashing to store the frequency of the digits and then count the digits with a frequency equal to more than 1. Follow the steps below to solve the problem:

• Create an array of size 10 to store the count of digits 0 – 9. Initially store each index as 0.
• Now for each digit of number N, increment the count of that index in the array.
• Traverse the array and count the indices that have value more than 1.
• In the end, print this count.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function that returns the count of
// repeating digits of the given number
int countRepeatingDigits(int N)
{
    // Initialize a variable to store
    // count of Repeating digits
    int res = 0;

    // Initialize cnt array to
    // store digit count

    int cnt[10] = { 0 };

    // Iterate through the digits of N
    while (N > 0) {

        // Retrieve the last digit of N
        int rem = N % 10;

        // Increase the count of digit
        cnt[rem]++;

        // Remove the last digit of N
        N = N / 10;
    }

    // Iterate through the cnt array
    for (int i = 0; i < 10; i++) {

        // If frequency of digit
        // is greater than 1
        if (cnt[i] > 1) {

            // Increment the count
            // of Repeating digits
            res++;
        }
    }

    // Return count of repeating digit
    return res;
}

// Driver Code
int main()
{
    // Given array arr[]
    int N = 12;

    // Function Call
    cout << countRepeatingDigits(N);
    return 0;
}

Java

// Java program for the above approach
class GFG{

// Function that returns the count of
// repeating digits of the given number
static int countRepeatingDigits(int N)
{
    // Initialize a variable to store
    // count of Repeating digits
    int res = 0;

    // Initialize cnt array to
    // store digit count

    int cnt[] = new int[10];

    // Iterate through the digits of N
    while (N > 0)
    {

        // Retrieve the last digit of N
        int rem = N % 10;

        // Increase the count of digit
        cnt[rem]++;

        // Remove the last digit of N
        N = N / 10;
    }

    // Iterate through the cnt array
    for (int i = 0; i < 10; i++)
    {

        // If frequency of digit
        // is greater than 1
        if (cnt[i] > 1)
        {

            // Increment the count
            // of Repeating digits
            res++;
        }
    }

    // Return count of repeating digit
    return res;
}

// Driver Code
public static void main(String[] args)
{
    // Given array arr[]
    int N = 12;

    // Function Call
    System.out.println(countRepeatingDigits(N));
}
}

// This code is contributed by Ritik Bansal

Python3

# Python3 program for the above approach

# Function that returns the count of
# repeating digits of the given number
def countRepeatingDigits(N):

    # Initialize a variable to store
    # count of Repeating digits
    res = 0

    # Initialize cnt array to
    # store digit count
    cnt = [0] * 10

    # Iterate through the digits of N
    while (N > 0):

        # Retrieve the last digit of N
        rem = N % 10

        # Increase the count of digit
        cnt[rem] += 1

        # Remove the last digit of N
        N = N // 10

    # Iterate through the cnt array
    for i in range(10):

        # If frequency of digit
        # is greater than 1
        if (cnt[i] > 1):

            # Increment the count
            # of Repeating digits
            res += 1

    # Return count of repeating digit
    return res

# Driver Code

# Given array arr[]
N = 12

# Function call
print(countRepeatingDigits(N))

# This code is contributed by sanjoy_62

C#

// C# program for the above approach
using System;
class GFG{

// Function that returns the count of
// repeating digits of the given number
static int countRepeatingDigits(int N)
{
    // Initialize a variable to store
    // count of Repeating digits
    int res = 0;

    // Initialize cnt array to
    // store digit count
    int []cnt = new int[10];

    // Iterate through the digits of N
    while (N > 0)
    {

        // Retrieve the last digit of N
        int rem = N % 10;

        // Increase the count of digit
        cnt[rem]++;

        // Remove the last digit of N
        N = N / 10;
    }

    // Iterate through the cnt array
    for (int i = 0; i < 10; i++)
    {

        // If frequency of digit
        // is greater than 1
        if (cnt[i] > 1)
        {

            // Increment the count
            // of Repeating digits
            res++;
        }
    }

    // Return count of repeating digit
    return res;
}

// Driver Code
public static void Main(String[] args)
{
    // Given array []arr
    int N = 12;

    // Function Call
    Console.WriteLine(countRepeatingDigits(N));
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// Javascript program for the above approach

// Function that returns the count of
// repeating digits of the given number
function countRepeatingDigits(N)
{

    // Initialize a variable to store
    // count of Repeating digits
    var res = 0;

    // Initialize cnt array to
    // store digit count

    var cnt = Array(10).fill(0);

    // Iterate through the digits of N
    while (N > 0) {

        // Retrieve the last digit of N
        var rem = N % 10;

        // Increase the count of digit
        cnt[rem]++;

        // Remove the last digit of N
        N = N / 10;
    }

    // Iterate through the cnt array
    for (var i = 0; i < 10; i++) {

        // If frequency of digit
        // is greater than 1
        if (cnt[i] > 1) {

            // Increment the count
            // of Repeating digits
            res++;
        }
    }

    // Return count of repeating digit
    return res;
}

// Driver Code
// Given array arr[]
var N = 12;

// Function Call
document.write( countRepeatingDigits(N));

// This code is contributed by rrrtnx.
</script>

Output

0

Time Complexity: O(N)
Auxiliary Space: O(1)

Method #2:Using built in python functions:

• Convert integer to string.
• Use the Counter function to count the frequency of characters.
• If the frequency is greater than 1 increment the count

Below is the implementation:

C++

#include <iostream>
#include <unordered_map>
#include <string>

using namespace std;

// Function that returns the count of
// repeating digits of the given number
int countRepeatingDigits(int N)
{

  // Converting int to string
  string number = to_string(N);

  // Initializing count
  int count = 0;
  unordered_map<char, int> frequency;

  // Finding frequency of each character
  for (char c : number) {
    frequency++;
  }
  for (auto& it : frequency) {
    if (it.second > 1) {
      count++;
    }
  }
  // return the count
  return count;
}

int main()
{
  int N = 1232145;

  // Function call
  cout << countRepeatingDigits(N) << endl;
  return 0;
}

Java

import java.util.HashMap;

public class Main {
    //Function that returns the count of
    //repeating digits of the given number
    static int countRepeatingDigits(int N)
    {
        // Converting int to string
        String number = Integer.toString(N);
          // Initializing count
        int count = 0;
        HashMap<Character, Integer> frequency
            = new HashMap<>();
        // Finding frequency of each character
        for (char c : number.toCharArray()) {
            frequency.put(c,
                          frequency.getOrDefault(c, 0) + 1);
        }
        for (char key : frequency.keySet()) {
            if (frequency.get(key) > 1) {
                count++;
            }
        }
        // return the count
        return count;
    }

    public static void main(String[] args)
    {
        int N = 1232145;
          // Function call
        System.out.println(countRepeatingDigits(N));
    }
}

Python3

# Python3 program for the above approach
from collections import Counter

# Function that returns the count of
# repeating digits of the given number
def countRepeatingDigits(N):

    # converting integer to string
    number = str(N)

    # initializing count = 0
    count = 0
    frequency = Counter(number)

    # Traversing frequency
    for i in frequency:
        if(frequency[i] > 1):

            # increase the count
            count = count+1
    return count

# Driver Code


# Given array arr[]
N = 1232145

# Function call
print(countRepeatingDigits(N))

# This code is contributed by vikkycirus

Javascript

// js equivalent

// Function that returns the count of
// repeating digits of the given number
function countRepeatingDigits(N)
{

    // Converting int to string
    let number = N.toString();

    // Initializing count
    let count = 0;
    let frequency = {};

    // Finding frequency of each character
    for (let c of number) {
        frequency = (frequency || 0) + 1;
    }
    for (let key in frequency) {
        if (frequency[key] > 1) {
            count++;
        }
    }

    // return the count
    return count;
}

// Function call
let N = 1232145;
console.log(countRepeatingDigits(N));

Output

2