# Count of unique digits in a given number N

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

Examples:

Input: N = 22342
Output:
Explanation:
The digits 3 and 4 occurs only once. Hence, the output is 2.

Input: N = 99677
Output: 1
Explanation:
The digit 6 occurs only once. Hence, the output is 1.

Naive Approach: By this approach, the problem can be solved using 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 no, then increase the required count by 1. In the end, print the calculated required count.
Time Complexity: O(L2
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 1. Follow the steps below to solve the problem:

1. Create a HashTable of size 10 for digits 0-9. Initially store each index as 0.
2. Now for each digit of number N, increment the count of that index in the hashtable.
3. Traverse the hashtable and count the indices that have value equal to 1.
4. At the end, print/return this count.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function that returns the count of ` `// unique digits of the given number ` `int` `countUniqueDigits(``int` `N) ` `{ ` `    ``// Initialize a variable to store ` `    ``// count of unique digits ` `    ``int` `res = 0; ` ` `  `    ``// Initialize cnt array to store ` `    ``// digit count ` `    ``int` `cnt = { 0 }; ` ` `  `    ``// Iterate through the digits of N ` `    ``while` `(N > 0) { ` ` `  `        ``// Retrieve the last digit of N ` `        ``int` `rem = N % 10; ` ` `  `        ``// Increase the count ` `        ``// of the last 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 1 ` `        ``if` `(cnt[i] == 1) { ` ` `  `            ``// Increment the count ` `            ``// of unique digits ` `            ``res++; ` `        ``} ` `    ``} ` ` `  `    ``// Return the count/ of unique digit ` `    ``return` `res; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// Given array arr[] ` `    ``int` `N = 2234262; ` ` `  `    ``// Function Call ` `    ``cout << countUniqueDigits(N); ` `    ``return` `0; ` `}`

## Java

 `// Java program for the above approach ` `import` `java.util.*; ` ` `  `class` `GFG { ` ` `  `    ``// Function that returns the count ` `    ``// of unique digits of number N ` `    ``public` `static` `void` `    ``countUniqueDigits(``int` `N) ` `    ``{ ` `        ``// Initialize a variable to ` `        ``// store count of unique digits ` `        ``int` `res = ``0``; ` ` `  `        ``// Initialize cnt array to ` `        ``// store digit count ` `        ``int` `cnt[] = { ``0``, ``0``, ``0``, ``0``, ``0``, ` `                      ``0``, ``0``, ``0``, ``0``, ``0` `}; ` ` `  `        ``// Iterate through digits of N ` `        ``while` `(N > ``0``) { ` ` `  `            ``// Retrieve the last ` `            ``// digit of N ` `            ``int` `rem = N % ``10``; ` ` `  `            ``// Increase the count ` `            ``// of the last digit ` `            ``cnt[rem]++; ` ` `  `            ``// Remove the last ` `            ``// digit of N ` `            ``N = N / ``10``; ` `        ``} ` ` `  `        ``// Iterate through the ` `        ``// cnt array ` `        ``for` `(``int` `i = ``0``; ` `             ``i < cnt.length; i++) { ` ` `  `            ``// If frequency of ` `            ``// digit is 1 ` `            ``if` `(cnt[i] == ``1``) { ` ` `  `                ``// Increment the count ` `                ``// of unique digits ` `                ``res++; ` `            ``} ` `        ``} ` ` `  `        ``// Return the count of unique digit ` `        ``System.out.println(res); ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``// Given Number N ` `        ``int` `N = ``2234262``; ` ` `  `        ``// Function Call ` `        ``countUniqueDigits(N); ` `    ``} ` `}`

## Python3

 `# Python3 program for the above approach  ` ` `  `# Function that returns the count of  ` `# unique digits of the given number  ` `def` `countUniqueDigits(N): ` ` `  `    ``# Initialize a variable to store ` `    ``# count of unique digits ` `    ``res ``=` `0` ` `  `    ``# Initialize cnt list 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 the last digit ` `        ``cnt[rem] ``+``=` `1` ` `  `        ``# Remove the last digit of N ` `        ``N ``=` `N ``/``/` `10` ` `  `    ``# Iterate through the cnt list ` `    ``for` `i ``in` `range``(``10``): ` ` `  `        ``# If frequency of ` `        ``# digit is 1 ` `        ``if` `(cnt[i] ``=``=` `1``): ` ` `  `            ``# Increment the count ` `            ``# of unique digits ` `            ``res ``+``=` `1` ` `  `    ``# Return the count of unique digit ` `    ``return` `res  ` ` `  `# Driver Code ` ` `  `# Given number N ` `N ``=` `2234262` ` `  `# Function call  ` `print``(countUniqueDigits(N))  ` ` `  `# This code is contributed by vishu2908 `

## C#

 `// C# program for the above approach ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Function that returns the count ` `// of unique digits of number N ` `public` `static` `void` `countUniqueDigits(``int` `N) ` `{ ` `     `  `    ``// Initialize a variable to ` `    ``// store count of unique digits ` `    ``int` `res = 0; ` ` `  `    ``// Initialize cnt array to ` `    ``// store digit count ` `    ``int``[] cnt = { 0, 0, 0, 0, 0, ` `                  ``0, 0, 0, 0, 0 }; ` ` `  `    ``// Iterate through digits of N ` `    ``while` `(N > 0)  ` `    ``{ ` `         `  `        ``// Retrieve the last ` `        ``// digit of N ` `        ``int` `rem = N % 10; ` ` `  `        ``// Increase the count ` `        ``// of the last digit ` `        ``cnt[rem]++; ` ` `  `        ``// Remove the last ` `        ``// digit of N ` `        ``N = N / 10; ` `    ``} ` ` `  `    ``// Iterate through the ` `    ``// cnt array ` `    ``for``(``int` `i = 0; i < cnt.Length; i++) ` `    ``{ ` `         `  `        ``// If frequency of ` `        ``// digit is 1 ` `        ``if` `(cnt[i] == 1)  ` `        ``{ ` `             `  `            ``// Increment the count ` `            ``// of unique digits ` `            ``res++; ` `        ``} ` `    ``} ` `     `  `    ``// Return the count of unique digit ` `    ``Console.WriteLine(res); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `     `  `    ``// Given Number N ` `    ``int` `N = 2234262; ` ` `  `    ``// Function Call ` `    ``countUniqueDigits(N); ` `} ` `} ` ` `  `// This code is contributed by jrishabh99 `

Output:

```3
```

Time Complexity: O(N), where N is the number of digits of the number.
Auxiliary Space: O(1)

Note: As the used hashtable is of size only 10, therefore its time and space complexity will be near to constant. Hence it is not counted in the above time and auxiliary space.

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Improved By : jrishabh99, vishu2908