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This approach involves converting each number into a string to easily count the number of digits.
Time Complexity: O(n), where n is the number of elements in the array.
Space Complexity: O(1), as we're using a fixed amount of extra space.
1function findNumbers(nums) {
2 let count = 0;
3 for (let num of nums) {
4 if (num.toString().length % 2 === 0) {
5 count++;
6 }
7 }
8 return count;
9}
10
11const nums = [12, 345, 2, 6, 7896];
12console.log(findNumbers(nums));
This JavaScript solution iterates over the array of numbers, converts each number to a string, and checks whether the length of the string is even. It increments the count for numbers that have even digit lengths.
The logarithmic approach calculates the number of digits by using logarithms; specifically, the base-10 logarithm of a number is taken, and the result is incremented by 1 to get the total number of digits.
Time Complexity: O(n), the array is processed once.
Space Complexity: O(1), constant space utilized.
1#include <vector>
#include <cmath>
using namespace std;
int findNumbers(vector<int>& nums) {
int count = 0;
for (int num : nums) {
if (static_cast<int>(log10(num)) % 2 == 1) {
count++;
}
}
return count;
}
int main() {
vector<int> nums = {12, 345, 2, 6, 7896};
cout << findNumbers(nums) << endl;
return 0;
}
In this C++ code, log10()
is used to determine the number of digits. Adding 1 to static_cast
gives the digit count, which is checked for evenness.