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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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

✓ Mathematical Approach

⏱️ Time: O(n × d) Space: O(1)

Instead of converting to string, use mathematical operations to count digits. Repeatedly divide by 10 until the number becomes 0, counting the divisions needed.

String Conversion Approach

⏱️ Time: O(n × d) Space: O(d)

For each number in the array, convert it to a string, count its characters (digits), and check if the count is even. This is straightforward but uses string operations.

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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Find Numbers with Even Number of Digits - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Mathematical Approach — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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