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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.
1function countDigitOne(n) {
2 let count = 0;
3 for (let i = 1; i <= n; i++) {
4 let num = i;
5 while (num > 0) {
6 if (num % 10 === 1) count++;
7 num = Math.floor(num / 10);
8 }
9 }
10 return count;
11}
12
13const n = 13;
14console.log(countDigitOne(n));This JavaScript function counts ones by iterating through each number and examining each digit.
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#
The C code iterates over each digit position and calculates how often a 1 would appear at that position using arithmetic based on the surrounding digits. The core idea is to count 1s in each digit position separately.