LeetCode #1295 — EASY

Find Numbers with Even Number of Digits

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

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 
Explanation: 
Only 1771 contains an even number of digits.

Constraints:

1 <= nums.length <= 500
1 <= nums[i] <= 10⁵

Step 01

Brute Force Baseline

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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math

Example 1

[12,345,2,6,7896]

Example 2

[555,901,482,1771]

Core Insight

What unlocks the optimal approach

How to compute the number of digits of a number ?
Divide the number by 10 again and again to get the number of digits.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1295: Find Numbers with Even Number of Digits
class Solution {
    public int findNumbers(int[] nums) {
        int ans = 0;
        for (int x : nums) {
            if (String.valueOf(x).length() % 2 == 0) {
                ++ans;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1295: Find Numbers with Even Number of Digits
func findNumbers(nums []int) (ans int) {
	for _, x := range nums {
		if len(strconv.Itoa(x))%2 == 0 {
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #1295: Find Numbers with Even Number of Digits
class Solution:
    def findNumbers(self, nums: List[int]) -> int:
        return sum(len(str(x)) % 2 == 0 for x in nums)

// Accepted solution for LeetCode #1295: Find Numbers with Even Number of Digits
impl Solution {
    pub fn find_numbers(nums: Vec<i32>) -> i32 {
        nums.iter()
            .filter(|&x| x.to_string().len() % 2 == 0)
            .count() as i32
    }
}

// Accepted solution for LeetCode #1295: Find Numbers with Even Number of Digits
function findNumbers(nums: number[]): number {
    return nums.filter(x => x.toString().length % 2 === 0).length;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log M)

Space

O(log M)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.