How do you solve Number of Digit One on LeetCode?

Number of Digit One (LeetCode #233) can be solved using Brute Force. The optimal approach is Brute Force with O(n log n) time complexity and O(1) space complexity. Key insight: The brute force approach involves checking each number from 0 to n and counting how many times the digit '1' appears in each number. It's straightforward but inefficient for large values of n.

What is the time complexity of Number of Digit One?

The optimal solution for Number of Digit One runs in O(n log n) time and uses O(1) space. The time complexity is O(n log n) because converting each number to a string takes log n time, and we do this for n numbers.

What algorithm or data structure is used to solve Number of Digit One?

Number of Digit One uses the following concepts: Math, Dynamic Programming, Recursion. The recommended patterns to study are: Mathematical Counting, Digit Manipulation.

What are the interview tips for Number of Digit One?

Always clarify the constraints and edge cases with the interviewer. Explain your thought process while coding to show your understanding. If you get stuck, discuss your thought process and ask for hints.

What are common mistakes when solving Number of Digit One?

Not considering edge cases like n = 0. Forgetting to handle the current digit correctly in the optimal solution.

Number of Digit One — LeetCode #233 (Hard)

Tags: Math, Dynamic Programming, Recursion

Related patterns: Mathematical Counting, Digit Manipulation

Brute Force approach

Time complexity: O(n log n). Space complexity: O(1).

The brute force approach involves checking each number from 0 to n and counting how many times the digit '1' appears in each number. It's straightforward but inefficient for large values of n.

The time complexity is O(n log n) because converting each number to a string takes log n time, and we do this for n numbers.

Step 1: Initialize a counter to zero.
Step 2: Loop through each number from 0 to n.
Step 3: For each number, convert it to a string and count the occurrences of '1'.
Step 4: Add the count to the counter.
Step 5: Return the counter.

For n = 13: 1. i = 0, count = 0 (0 has no '1') 2. i = 1, count = 1 (1 has one '1') 3. i = 2, count = 1 (2 has no '1') 4. i = 3, count = 1 (3 has no '1') 5. i = 4, count = 1 (4 has no '1') 6. i = 5, count = 1 (5 has no '1') 7. i = 6, count = 1 (6 has no '1') 8. i = 7, count = 1 (7 has no '1') 9. i = 8, count = 1 (8 has no '1') 10. i = 9, count = 1 (9 has no '1') 11. i = 10, count = 2 (10 has one '1') 12. i = 11, count = 4 (11 has two '1's) 13. i = 12, count = 5 (12 has one '1') 14. i = 13, count = 6 (13 has one '1') Final count = 6.

Common Mistakes

Not considering edge cases like n = 0.
Forgetting to handle the current digit correctly in the optimal solution.

Interview Tips

Always clarify the constraints and edge cases with the interviewer.
Explain your thought process while coding to show your understanding.
If you get stuck, discuss your thought process and ask for hints.

#233

Number of Digit One

Hard

Math↗Dynamic Programming↗Recursion↗Mathematical Counting↗Digit Manipulation↗

LeetCode ↗

Approaches

💡

Intuition

Time O(n log n)Space O(1)

The brute force approach involves checking each number from 0 to n and counting how many times the digit '1' appears in each number. It's straightforward but inefficient for large values of n.

⚙️

Algorithm

5 steps

1Step 1: Initialize a counter to zero.
2Step 2: Loop through each number from 0 to n.
3Step 3: For each number, convert it to a string and count the occurrences of '1'.
4Step 4: Add the count to the counter.
5Step 5: Return the counter.

solution.py9 lines

1# Full working Python code
2
3def countDigitOne(n):
4    count = 0
5    for i in range(n + 1):
6        count += str(i).count('1')
7    return count
8
9print(countDigitOne(13))  # Output: 6

ℹ

Complexity note: The time complexity is O(n log n) because converting each number to a string takes log n time, and we do this for n numbers.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.