This problem requires calculating how many times the digit '1' appears in all numbers from 0 to n. Using a state transition dynamic programming approach, we track contributions of each digit position efficiently. The solution leverages mathematical counting logic and careful handling of digit boundaries to avoid overflow and redundant computations.
Problem Statement
Given a non-negative integer n, determine the total number of times the digit 1 appears in all numbers from 0 to n inclusive. You must handle large n efficiently without iterating through every number.
For example, if n = 13, the numbers from 0 to 13 contain six occurrences of the digit 1. You need to compute this count using an algorithm that combines math reasoning with state transition dynamic programming to account for each digit position.
Examples
Example 1
Input: n = 13
Output: 6
Example details omitted.
Example 2
Input: n = 0
Output: 0
Example details omitted.
Constraints
- 0 <= n <= 109
Solution Approach
Digit Position Analysis
Break down n by each digit and count the contribution of 1s at each position. For every position, consider higher digits, current digit, and lower digits to compute how many times 1 appears in that place.
Recursive or Iterative DP
Implement a state transition DP where each state represents the current digit index and tightness constraint. Either recursive memoization or iterative bottom-up DP works, tracking counts while respecting number boundaries.
Overflow Prevention and Aggregation
Accumulate counts carefully to avoid integer overflow. Use long or appropriate integer types when summing contributions from each digit position, ensuring correctness for large n.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | O(log_{10}(n)) |
| Space | O(1) |
The algorithm processes each digit of n once, leading to O(log_{10}(n)) time complexity. Space complexity is O(1) if iterative aggregation is used, or O(log_{10}(n)) for recursive memoization.
What Interviewers Usually Probe
- Checks if candidate recognizes that brute-force counting is too slow and will fail for large n.
- Wants a solution that properly models each digit's contribution using DP or mathematical counting.
- Looks for handling of boundary cases like n = 0 or digits with zeros in higher places.
Common Pitfalls or Variants
Common pitfalls
- Forgetting to handle the current digit correctly when it is 1, leading to undercounting.
- Using naive iteration through all numbers, which exceeds time limits for large n.
- Overflowing integer types when summing counts without considering large n.
Follow-up variants
- Count digit 'd' instead of 1 across numbers up to n.
- Compute number of times a substring appears in decimal representation of numbers up to n.
- Adapt the solution for base-k numbers rather than decimal.
How GhostInterview Helps
- Breaks down the digit position contributions automatically to guide correct DP states.
- Highlights overflow risks and ensures candidate tracks sums safely for each digit position.
- Provides step-by-step analysis showing how state transition DP accumulates the total count efficiently.
Topic Pages
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FAQ
What is the main pattern used in Number of Digit One?
The primary pattern is state transition dynamic programming applied to digit positions.
Why is naive iteration over all numbers inefficient for this problem?
Because n can be up to 10^9, iterating each number would exceed time limits, making DP necessary.
How do I handle large n without integer overflow?
Use larger integer types or carefully sum counts from each digit position to prevent exceeding limits.
Can this solution be adapted for other digits besides 1?
Yes, the same state transition DP can be applied to count occurrences of any digit across numbers up to n.
Is recursion necessary for this problem?
No, iterative DP can also be used; recursion is optional but must respect state transitions and digit boundaries.
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