Last Updated: 31 Mar, 2025

Count Repeating Digits

Easy

Problem statement

Let 'N' = 9397776.
The repeating digits in 'N' are 9 and 7.
Therefore, the answer is 2.

The first line contains an integer 'N'.

Return the count of repeating digits in 'N'.

You don’t need to print anything. Just implement the given function.

0 <= 'N' <= 10^9

Time Limit: 1 sec

Approach:

Create an array of size 10 to store the frequency of each digit (0-9).
Iterate through the digits of the number.
For each digit, increment its count in the frequency array.
Iterate through the frequency array and count the number of digits with a frequency greater than 1.

Algorithm:

Initialize an array 'freq' of size 10 with all elements as 0.
Convert the integer 'N' to a string 'S'.
Iterate using 'i' from 0 to length of 'S' - 1 :
- Let 'digit' be the integer value of 'S[i]'.
- Increment 'freq[digit]' by 1.
Initialize a variable 'count' to 0.
Iterate using 'i' from 0 to 9 :
- If ( freq[i] > 1 ) :
  - Increment 'count' by 1.
Return 'count'.