Find Numbers with Even Number of Digits

Leetcode Problem:

• Find Numbers with Even Number of Digits

Summary

• Count numbers with even number of digits in an array

Approach

• The approach used is to iterate through each number in the array and calculate the number of digits in each number using a helper function
• If the number of digits is even, it increments the counter
• Finally, it returns the count of numbers with even number of digits.

Complexity

• O(n * log(max_num)) where n is the size of the array and max_num is the maximum number in the array

Explanation

• The solution iterates through each number in the array
• For each number, it calculates the number of digits using a helper function
• If the number of digits is even, it increments the counter
• The time complexity of the helper function is O(log(max_num)) and it is called for each number in the array
• Therefore, the overall time complexity is O(n * log(max_num)) where n is the size of the array and max_num is the maximum number in the array.

Solution Code:

class Solution {
public:
    int getDigitNum(int num){
        int cnt = 0;
        while(num){
            num/=10;
            cnt++;
        }
        return cnt;
    }
    int findNumbers(vector& nums) {
        int ans = 0;
        for(int num : nums){
            if(getDigitNum(num) % 2){
                continue;
            }
            ans++;
        }
        return ans;
    }
};