What is the time complexity of Find Numbers with Even Number of Digits?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Find Numbers with Even Number of Digits?

This problem uses Array, Math. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Find Numbers with Even Number of Digits?

The most direct and often idiomatic way to count digits in many programming languages is through string conversion. `len(str(number))` instantly provides the digit count, obviating the need for iterative division by 10. The problem simplifies the "even/odd" digit count check to a basic modulo 2 operation (`length % 2 == 0`), which is extremely efficient and immediately reveals the parity of the digit count.

Which companies ask LeetCode 1295. Find Numbers with Even Number of Digits in interviews?

This problem is frequently asked at Quora.

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Find Numbers with Even Number of Digits - LeetCode 1295 Solution

Find Numbers with Even Number of Digits - Complete Solution Guide

Find Numbers with Even Number of Digits is LeetCode problem 1295, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 Ex

Detailed Explanation

The core task in "Find Numbers with Even Number of Digits" is straightforward: iterate through a given list of integers, `nums`, and for each number, determine if it possesses an even count of digits. The ultimate goal is to return a total tally of such numbers. For instance, if `nums` contains `[12, 345, 2, 6, 7896]`, we'd be looking to identify `12` (two digits) and `7896` (four digits) as contributors to our final count of 2.

Solution Approach

The provided solution embraces a wonderfully pragmatic and highly readable approach: leveraging string conversion. For each integer `num` in the `nums` array, the algorithm first casts `num` into its string representation using `str(num)`. Once it's a string, determining the number of digits becomes trivial—it's simply the `len()` of that string. The elegance here is that built-in string functions handle all the heavy lifting of digit counting, including potential edge cases like zero (which would be '0' with length 1). After obtaining the length, a simple modulo operation (`% 2 == 0`) efficiently checks for an even digit count, incrementing a `count` variable if true. This method sidesteps manual arithmetic loops, resulting in concise and idiomatic code.

Step-by-Step Algorithm

Step 1: Initialize a counter variable `count` to 0.
Step 2: Iterate through each number `num` in the input array `nums`.
Step 3: Determine the number of digits in `num` (using either string conversion or repeated division).
Step 4: Check if the number of digits is even using the modulo operator (`% 2 == 0`).
Step 5: If the number of digits is even, increment the `count`.
Step 6: After iterating through all numbers, return the final value of `count`.

Key Insights

The most direct and often idiomatic way to count digits in many programming languages is through string conversion. `len(str(number))` instantly provides the digit count, obviating the need for iterative division by 10.
The problem simplifies the "even/odd" digit count check to a basic modulo 2 operation (`length % 2 == 0`), which is extremely efficient and immediately reveals the parity of the digit count.
The solution's cleanliness stems from its clear functional decomposition: one language feature (`str()`) for digit representation, another (`len()`) for counting, and an operator (`%`) for parity checking. This, combined with direct iteration, makes the code highly readable and maintainable.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Math.

Companies

Asked at: Quora.

Find Numbers with Even Number of Digits | LeetCode 1295

Easy

Array Math

Time: O(n)Space: O(1)

Solution Code

Available in:

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

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Problem Info

Description

Problem 1295: Find Numbers with Even Number of Digits - Detailed explanation and solution approaches.

Topics

Array Math

Companies

Quora

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1295

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize a counter variable `count` to 0.
2
Step 2: Iterate through each number `num` in the input array `nums`.
3
Step 3: Determine the number of digits in `num` (using either string conversion or repeated division).
4
Step 4: Check if the number of digits is even using the modulo operator (`% 2 == 0`).
5
Step 5: If the number of digits is even, increment the `count`.
6
Step 6: After iterating through all numbers, return the final value of `count`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The dominant operation is the iteration through the array. The digit counting operation for each number is a constant-time operation (or at most logarithmic in the magnitude of the number, which is still dominated by the linear iteration of n elements). The space used is independent of the array size, hence the constant space complexity.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using Logarithm

Instead of converting to string or iterative division, you could use the base-10 logarithm to estimate the number of digits. `floor(log10(num)) + 1` gives the number of digits. However, this approach might introduce slight inaccuracies due to floating-point precision and adds complexity for minimal performance gain.

Time: O(n)Space: O(1)

Trade-offs: Logarithm approach is slightly less readable and can be less accurate, but it might offer marginal performance improvement for extremely large numbers. The standard iterative or string conversion approach is generally preferred for its clarity and simplicity.

✨

Key Insights

1The most direct and often idiomatic way to count digits in many programming languages is through string conversion. `len(str(number))` instantly provides the digit count, obviating the need for iterative division by 10.
2The problem simplifies the "even/odd" digit count check to a basic modulo 2 operation (`length % 2 == 0`), which is extremely efficient and immediately reveals the parity of the digit count.
3The solution's cleanliness stems from its clear functional decomposition: one language feature (`str()`) for digit representation, another (`len()`) for counting, and an operator (`%`) for parity checking. This, combined with direct iteration, makes the code highly readable and maintainable.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: An array with only one element. The solution correctly handles this by iterating once and checking the digit count of that single element.
Edge case 2: An array with all numbers having an odd number of digits. The solution correctly returns 0 in this case.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the number of digits (e.g., off-by-one errors in the loop or logic).

✗Common mistake 2: Forgetting to handle the case where a number has only one digit (which is an odd number of digits).

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data validation: Checking if input data (e.g., account numbers, identification numbers) conforms to a specific length requirement (even number of digits).

🏦Real-world scenario 2: Data analysis: Analyzing numerical data sets and classifying entries based on the number of digits they have. For example, identifying numbers with a certain level of precision.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 6, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Find Numbers with Even Number of Digits"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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