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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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 

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Output: 

0

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

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