Given an array nums of integers, return how many of them contain an even number of digits.
Example 1:
Input: nums = [12,345,2,6,7896]
Output: 2
Example 2:
Input: nums = [555,901,482,1771]
Output: 1
Solution
- Loop through Array
- Find number of digits in each number
- Increase counter by 1 if number of digits in the number is even
/**
* FindNumbersWithEvenNumberOfDigits.java
* Purpose: Find Numbers with Even Number of Digits
*
* @author Megha Rastogi
* @version 1.0 08/09/2020
*/
class FindNumbersWithEvenNumberOfDigits {
/**
* Find Numbers with Even Number of Digits
*
* @param Array, nums
* @return integer, numberOfEvenDigitNumbers
*/
public int findNumbers(int[] nums) {
// initialize the variable to store the count of
// even digit numbers
int numberOfEvenDigitNumbers = 0;
// loop through the array and increase the counter
// if number of digits are even
for(int num: nums){
if(numberOfDigits(num) % 2 == 0)
numberOfEvenDigitNumbers++;
}
//return the result
return numberOfEvenDigitNumbers;
}
/**
* Find number of digits in an integer
*
* @param integer, num
* @return integer, numberOfDigits
*/
public int numberOfDigits(int num){
// initialize the counter to store number of digits
int numberOfDigits = 0;
// loop until number is greater than 0
// Continuously divide num by 10
// increase the numberOfDigits by 1
while(num > 0){
num /= 10;
numberOfDigits++;
}
// return the result
return numberOfDigits;
}
}
Source – Leetcode