Find Numbers with Even Number of Digits

EASY

Description

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] <= 10⁵

Approaches

Checkout 4 different approaches to solve Find Numbers with Even Number of Digits. Click on different approaches to view the approach and algorithm in detail.

Brute Force using String Conversion

This approach iterates through each number in the input array. For each number, it converts it into a string and then checks the length of the string. If the length is an even number, a counter is incremented.

Algorithm

Initialize a counter evenDigitCount to 0.
Loop through each integer num in the nums array.
Convert the integer num to its string representation using String.valueOf(num).
Find the length of the resulting string.
Check if the length is divisible by 2 (i.e., length % 2 == 0).
If the length is even, increment evenDigitCount.
After iterating through all the numbers, return evenDigitCount.

This method provides a straightforward way to solve the problem by leveraging built-in string functionalities. The core idea is that the number of digits in an integer is equivalent to the length of its string representation.

We initialize a counter evenDigitCount to 0.
We then loop through each integer num in the input nums array.
Inside the loop, we convert the current number num to a string. For example, the integer 123 becomes the string "123".
We then get the length of this string. The length of "123" is 3.
We check if this length is an even number using the modulo operator (length % 2 == 0).
If the condition is true, we increment our evenDigitCount.
After the loop has processed all numbers in the array, the evenDigitCount holds the total count of numbers with an even number of digits, which we then return.

class Solution {
    public int findNumbers(int[] nums) {
        int evenDigitCount = 0;
        for (int num : nums) {
            // Convert the number to a string
            String s = String.valueOf(num);
            // Check if the length of the string is even
            if (s.length() % 2 == 0) {
                evenDigitCount++;
            }
        }
        return evenDigitCount;
    }
}

Complexity Analysis

Time Complexity: O(N * D), where N is the number of elements in the array and D is the maximum number of digits in a number. Converting a number with D digits to a string takes O(D) time. Given the problem constraints, D is small, so the complexity is often simplified to O(N).Space Complexity: O(D), where D is the maximum number of digits in a number. This space is required to store the string representation of a number. Since the constraints limit numbers to 10^5, D is at most 6, making the space complexity effectively O(1).

Pros and Cons

Pros:

Simple to understand and implement.
The logic is very direct and easy to reason about.

Cons:

Incurs the overhead of string conversion and memory allocation for each number, which is generally less efficient than purely mathematical approaches.

Code Solutions

Checking out 4 solutions in different languages for Find Numbers with Even Number of Digits. Click on different languages to view the code.

class Solution {
public
  int findNumbers(int[] nums) {
    int ans = 0;
    for (int v : nums) {
      if (String.valueOf(v).length() % 2 == 0) {
        ++ans;
      }
    }
    return ans;
  }
}

Video Solution

Watch the video walkthrough for Find Numbers with Even Number of Digits

Patterns:

Math

Data Structures:

Array

Companies:

Quora |