Given two integers L, R, and digit X. Find the number of occurrences of X in all the numbers in the range (L, R) excluding L and R.

Example 1:

Input:
L=10, R=19, X=1
Output:
9
Explanation:
There are 9 1s (11, 12, 13, 14, 15, 16, 17, 18) in the numbers in the range (10,19).

Example 2:

Input:
L=18, R=81, X=9
Output:
7
Explanation:
There are 7 occurrences of the digit 9 in the numbers in the range (18,81).

Your Task:
You don’t need to read input or print anything. Your task is to complete the function countX() which takes three integers L, R, and X as input parameters and returns the number of occurrences of X in the numbers in the range(L, R).

Constraints:
1 <= L< R <= 105
0 <= X <= 9

Explanation:

Here we want to find the number of X’s in between the L and R. [excluding L and R values]

Example: L = 10, R = 19, X = 1

The numbers contains 1 in between 10 to 19

11 12 13 14 15 16 17 18

11 – 2(1's)

12 - 1

13 – 1

count all the 1’s = 9

Approach:

Step 1:

Start a loop from L+1 to R (Here R is Excluded and it will iterate R-1)

Step 2:

Call the countValue(i,X)

Here the method take the number (L->R) and X.

Step 3:

In countValue(i,X) method

Take i as n and iterate until n!=0

take remainder and check the remainder equal to X then count.

after, do the n = n/10 (this will give u the new n value by dividing with 10)

return the count.

public static int countValue(int n,int X){
    int c=0;
    while(n!=0){
      int rem = n%10;
      if(rem==X){
        c++;
      }
      n/=10;
    }
    return c;
  }

Step 4:

Every return count is stored in the count variable and finally return the count as answer.

CODE:

public class XInRange {
  public static void main(String[] args) {
    int L=10, R=19, X=1;
    // 11(count=2(1's)) 12 13 14 15 16 17 18
    //excluding L and R
    System.out.println("How many X's: "+countX(L,R,X));
  }
  static int countX(int L, int R, int X) 
  {
    int count=0;
    for(int i=L+1;i<R;i++){
      count+=countValue(i,X);

    }
    return count;
  }

  public static int countValue(int n,int X){
    int c=0;
    while(n!=0){
      int rem = n%10;
      if(rem==X){
        c++;
      }
      n/=10;
    }
    return c;
  }
}

How many X's: 9

problem: https://www.geeksforgeeks.org/problems/how-many-xs4514/1

Thank you for reading.

Happy Coding :)

Please share any optimized code we can discuss and improve code.