LeetCode 233: Number of Digit One Solution
Master LeetCode problem 233 (Number of Digit One), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
233. Number of Digit One
Problem Explanation
Explanation
To solve this problem, we can iterate through each digit of the numbers from 1 to n and count the occurrence of digit 1 in each position (ones place, tens place, etc.). We can then sum up these counts to get the total number of digit 1's in all numbers from 1 to n.
The algorithm involves analyzing the current digit being considered, the higher digits, and the lower digits. By considering each digit individually and calculating the contribution of that digit to the total count of digit 1's, we can determine the final result.
Time Complexity
The time complexity of this algorithm is O(logn) because we are iterating through the digits of the number n.
Space Complexity
The space complexity of this algorithm is O(1) as we are using a constant amount of extra space.
Solution Code
class Solution {
public int countDigitOne(int n) {
int count = 0;
for (long i = 1; i <= n; i *= 10) {
long divider = i * 10;
count += (n / divider) * i + Math.min(Math.max(n % divider - i + 1, 0), i);
}
return count;
}
}Try It Yourself
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