Find Numbers with Even Number of Digits - Practice Coding Problems

Find Numbers with Even Number of Digits - Problem

Array Math Easy

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

A number has an even number of digits if the count of its digits is divisible by 2.

Example: The number 123 has 3 digits (odd), while 1234 has 4 digits (even).

Input & Output

Example 1 — Basic Mixed Array

$ Input: nums = [12,345,2,6,7896]

› Output: 2

💡 Note: 12 has 2 digits (even), 345 has 3 digits (odd), 2 has 1 digit (odd), 6 has 1 digit (odd), 7896 has 4 digits (even). Total numbers with even digits: 2

Example 2 — All Single Digits

$ Input: nums = [7,8]

› Output: 0

💡 Note: 7 has 1 digit (odd), 8 has 1 digit (odd). No numbers have even digit count.

Example 3 — Large Numbers

$ Input: nums = [555,901,482,1771]

› Output: 1

💡 Note: 555 has 3 digits (odd), 901 has 3 digits (odd), 482 has 3 digits (odd), 1771 has 4 digits (even). Only 1771 has even digits.

Constraints

1 ≤ nums.length ≤ 500
1 ≤ nums[i] ≤ 10⁵

Visualization

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Understanding the Visualization

Input Array

Array of integers to analyze

Count Digits

For each number, determine how many digits it has

Count Even

Count how many numbers have even digit counts

Key Takeaway

🎯 Key Insight: Count digits for each number and check if the count is divisible by 2

Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to count digits for each number and check if the count is even. Best approach uses either string conversion or mathematical division - both have same time complexity. Time: O(n × d), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Mathematical Approach	O(n × d)	O(1)	Use mathematical operations to count digits without string conversion
String Conversion Approach	O(n × d)	O(d)	Convert each number to string and check if length is even

Mathematical Approach — Algorithm Steps

For each number, start with digit count = 0
While number > 0, divide by 10 and increment count
Check if final count is even
Increment result if even

Visualization

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Step-by-Step Walkthrough

Start Division

Begin with number and digit count = 0

Divide by 10

Keep dividing by 10 and increment count

Check Result

When number becomes 0, check if count is even

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

int solution(int* nums, int numsSize) {
    int count = 0;
    for (int i = 0; i < numsSize; i++) {
        // Count digits mathematically
        int digitCount;
        if (nums[i] == 0) {
            digitCount = 1;
        } else {
            digitCount = 0;
            int temp = abs(nums[i]);
            while (temp > 0) {
                digitCount++;
                temp /= 10;
            }
        }
        
        if (digitCount % 2 == 0) {
            count++;
        }
    }
    return count;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array manually
    int nums[1000];
    int size = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != '\n') {
            nums[size++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    int result = solution(nums, size);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × d)

n numbers, each taking O(d) time to count digits where d is average digits

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables for counting

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Mathematical Approach — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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