Hello Everyone,
Given a number N , the task is to count the number of unique digits in the given number.
Examples:
Input: N = 22342
Output: 2
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:
- Create a HashTable of size 10 for digits 0-9. Initially store each index as 0.
- Now for each digit of number N, increment the count of that index in the hashtable.
- Traverse the hashtable and count the indices that have value equal to 1.
- At the end, print/return this count.
Below is the implementation of the above approach:
// C++ program for the above approach
#include <iostream>
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[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 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;
}
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.