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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.
1#include <iostream>
2using namespace std;
3
4int countDigitOne(int n) {
5 int count = 0;
6 for (int i = 1; i <= n; i++) {
7 int num = i;
8 while (num > 0) {
9 if (num % 10 == 1) count++;
10 num /= 10;
11 }
12 }
13 return count;
14}
15
16int main() {
17 int n = 13;
18 cout << countDigitOne(n) << endl;
19 return 0;
20}This C++ solution is functionally identical to the C solution. It iterates from 1 to n, extracts the digits of each number, and counts the 1s.
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.
1public
This Java solution counts 1s at each digit level using arithmetic and logical checks, efficiently counting occurrences of the digit 1.