May 07, 2020

#LeetCode: Find Numbers with Even Number of Digits

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
Explanation:
  • 12 contains 2 digits (even number of digits). 
  • 345 contains 3 digits (odd number of digits). 
  • 2 contains 1 digit (odd number of digits). 
  • 6 contains 1 digit (odd number of digits). 
  • 7896 contains 4 digits (even number of digits). 
  • Therefore only 12 and 7896 contain an even number of digits.
Example 2:
Input: nums = [555,901,482,1771]
Output: 1
Explanation:
  • Only 1771 contains an even number of digits.
Constraints:
1 <= nums.length <= 500
1 <= nums[i] <= 10^5

Algorithm
1). We need to check the length of each number, if the length is even, increment the counter.
2). To find the length we will convert each number into String by using Integer.toString() method.
3). Then we will divide the length of converted String by 2, if the remainder is 0, it is even.

Java Solution


ALSO CHECKOther LeetCode Solutions in Java

-K Himaanshu Shuklaa..

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