CoreTask#3: CountDigits - Submit Your Code in Comment Section

[ErSKS]

CoreTask#3: CountDigits - Write a function named countDigit that returns the number of times that a given digit appears in a positive number. For example countDigit(32121, 1) would return 2 because there are two 1s in 32121. Other examples: countDigit(33331, 3) returns 4 countDigit(33331, 6) returns 0 countDigit(3, 3) returns 1 The function should return -1 if either argument is negative, so countDigit(-543, 3) returns -1. The function signature is int ountDigit(int n, int digit) Hint: Use modulo base 10 and integer arithmetic to isolate the digits of the number.

[Niranjan2054]

package countdigit;

/**

• @author: Niranjan Bekoju

• @address: Liwali Bhaktapur

• @College: Khwopa College of Engineering
  */
  public class CountDigit {

  public static void main(String[] args) {
  System.out.println("Count Digit 2 in 2343241 :" + countDigit(2343241, 2));
  System.out.println("Count Digit 6 in 9833475 :" + countDigit(9833475, 6));
  System.out.println("Count Digit 7 in 903458 :" + countDigit(903458, 7));
  System.out.println("Count Digit 4 in 39485 :" + countDigit(39485, 4));
  System.out.println("Count Digit 6 in 32535 :" + countDigit(32535, 6));
  System.out.println("Count Digit 9 in 0 :" + countDigit(0, 9));
  System.out.println("Count Digit 2 in -3525 :" + countDigit(-3525, 2));
  System.out.println("Count Digit 4 in -3525235 :" + countDigit(-3525235, 4));
  }

  private static int countDigit(int n, int digit) {
  int count = 0;
  if (n < 0 || digit < 0) {
  return -1;
  }
  while (n > 0) {
  if (n % 10 == digit) {
  count++;
  }
  n /= 10;
  }
  return count;
  }
  }