A common kind of puzzle question is \"how many numbers between 1 and 1000 contai

Question

A common kind of puzzle question is "how many numbers between 1 and 1000 contain the digit 7?". Write a function countNumsWithDigit upperNumber, digit) that returns the number of numbers from 1 through upperNumber that contain the given digit. For example countNumsWithDigit(10, 9) should return 1, countNumsWithDigit(10,1) should return 2, and countNumsWithDigit(30, 2) should return 12. Your function must use a list comprehension (but remember, it returns a number not a list!) and have only one line of code (in addition to the def line).

Explanation / Answer

import java.util.Scanner;

class GetInputFromUser

{

   public static void main(String args[])

   {

      int un,d;

     Scanner in = new Scanner(System.in);

//reading upper number AND DIGIT VALUES

System.out.println("Enter Upper number");

      un= in.nextInt();

System.out.println("Enter starting number");

      d= in.nextInt();

countNumsWithDigit(un,d);

/* 1 of every 10 numbers contains a 7 ("7")

10 of every 100 numbers contain an additional 7 ("70", "71",...)

100 of every 1000 numbers contain an additional 7 ("700", "701",...)

Number of 7s in 1000

= (1000/10)*1 + (1000/100)*10 + (1000/1000)*100 = 100*1 + 10*10 + 100*1 = 300

Using this logic we can implement the following function*/

int countNumsWithDigit(int uppernumber,int digit)

{

int count=0;

for(int i = 1; i <= uppernumber ; i++)

{

    for(char c : String.valueOf(i).toCharArray())

{

        if(c == 'digit') count++;

}

return count;

}

}

}//main

}//class

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