06. How Many X's?

The problem can be found at the following link: Question Link

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++)

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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