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In this approach, we iterate through each number from 0 to n. For each number, we count the number of times digit 1 appears. Although this approach is simple, it's not optimal for large values of n due to its high time complexity.
Time Complexity: O(n * log10(n)), as it checks each digit of each number from 1 to n.
Space Complexity: O(1), as it uses a fixed amount of space.
1def countDigitOne(n: int) -> int:
2 count = 0
3 for i in range(1, n + 1):
4 count += str(i).count('1')
5 return count
6
7n = 13
8print(countDigitOne(n))This Python code converts each number to a string and uses the count method to find occurrences of '1'.
This optimized approach examines digit positions and calculates how many times 1 appears at each position up to n. It leverages the structure of numbers and is more efficient for large values of n.
Time Complexity: O(log10(n)), as it iterates digit by digit.
Space Complexity: O(1), as it uses a fixed amount of space.
1#include <iostream>
using namespace std;
int countDigitOne(int n) {
int count = 0;
for (long long i = 1; i <= n; i *= 10) {
long long divider = i * 10;
count += (n / divider) * i + ((n % divider >= i) ? (n % divider - i + 1) : 0);
}
return count;
}
int main() {
int n = 13;
cout << countDigitOne(n) << endl;
return 0;
}Similar to the C solution, the C++ solution uses mathematical operations to determine the count of digit 1s by handling each digit position.