Problem

Given an array nums of integers, return how many of them contain an even number of digits.

Examples

Example 1:

1
2
3
4
5
6
7
8
9

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:

1
2
3
4

Input: nums = [555,901,482,1771]
Output: 1 
Explanation: 
Only 1771 contains an even number of digits.

Solution

Method 1 - Convert numbers to string

  1. Convert each number to its string representation to count the number of digits.
  2. Check if the number of digits is even.
  3. Count how many numbers satisfy the condition.

Code

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17

public class Solution {
    public int countEvenDigitNumbers(int[] nums) {
        int count = 0;

        for (int num : nums) {
            if (numberOfDigits(num) % 2 == 0) {
                count += 1;
            }
        }

        return count;
    }

    private int numberOfDigits(int num) {
        return (int) Math.log10(num) + 1;
    }
}
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13

def find_floor(arr, x):
    start, end = 0, len(arr) - 1
    floor = -1

    while start <= end:
        mid = (start + end) // 2
        if arr[mid] == x:
            return arr[mid]
        elif arr[mid] < x:
            floor = arr[mid]
            start = mid + 1
        else:
            end = mid - 1

Complexity

  • Time: O(n), where n is the length of the array. Each number is processed individually.
  • Space: O(1), as we are not using additional space proportional to the input size, aside from the space required for the count and temporary variables.