# 06. How Many X's?

The problem can be found at the following link: [Question Link](https://www.geeksforgeeks.org/problems/how-many-xs4514/1)

![](https://badgen.net/badge/Level/Medium/yellow)

## My Approach

Simply go with the flow of question.

* I iterate through the numbers in the given range (`L` to `R`). For each number,
* I count the occurrences of the digit `X` by using the `cntX` function.
  * The function divides the number by 10 in each iteration and checks if the last digit is equal to `X`. If it is, I increment the count.

## Time and Auxiliary Space Complexity

* **Time Complexity** : `O((R-L) * log(R))`, where `log(R)` is the number of digits in R.
* **Auxiliary Space Complexity**: `O(1)`, as no extra space is used.

## Code (C++)

```cpp
class Solution {
public: 
    void cntX(int n, int& X, int& cnt){
        while(n){
            if(n % 10 == X)
                ++cnt;
            n /= 10;
        }
    }

    int countX(int L, int R, int X) {
        int cnt = 0;
        
        for(int i = L + 1; i < R; i++)
            cntX(i, X, cnt);

        return cnt;
    }
};
```

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