What is the time complexity of Number of Digit One?

The optimal solution has a time complexity of O(log(n)).

What data structures are used to solve Number of Digit One?

This problem uses Math, Dynamic Programming, Recursion. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Number of Digit One?

Insight 1: We can iterate through each digit place (ones, tens, hundreds, etc.) and count the number of 1s appearing in that place. Insight 2: For each digit place, we can determine the number of 1s based on the digits to the left (higher digits) and the digit in the current place.

Which companies ask LeetCode 233. Number of Digit One in interviews?

This problem is frequently asked at Atlassian, Google.

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Number of Digit One - LeetCode 233 Solution

Q: How to solve Number of Digit One?

The solution iterates through each digit place (starting from the ones place) and calculates how many times the digit '1' appears in that specific place for all numbers from 0 to 'n'. It does this by considering the 'high' part (digits to the left of the current digit), the 'current' digit, and the 'low' part (digits to the right of the current digit). Based on the value of the 'current' digit, we use different formulas to determine the number of '1's in that place. For example, if the current digit is 0, the number of 1s is solely determined by the 'high' part. If the current digit is 1, the number of 1s is influenced by both the 'high' and 'low' parts. If the current digit is greater than 1, the number of 1s is determined by the 'high' part increased by 1.

Number of Digit One - Complete Solution Guide

Number of Digit One is LeetCode problem 233, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an integer n , count the total number of digit 1 appearing in all non-negative integers less than or equal to n . Example 1: Input: n = 13 Output: 6 Example 2: Input: n = 0 Output: 0 Constraints: 0 <= n <= 10 9

Detailed Explanation

The problem asks us to count the number of times the digit '1' appears in all non-negative integers from 0 up to a given integer 'n'. For example, if n = 13, we need to count the '1's in 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13. There are six 1s: one in '1', one in '10', two in '11', one in '12', and one in '13'. The input 'n' is constrained to be between 0 and 10^9.

Solution Approach

The solution iterates through each digit place (starting from the ones place) and calculates how many times the digit '1' appears in that specific place for all numbers from 0 to 'n'. It does this by considering the 'high' part (digits to the left of the current digit), the 'current' digit, and the 'low' part (digits to the right of the current digit). Based on the value of the 'current' digit, we use different formulas to determine the number of '1's in that place. For example, if the current digit is 0, the number of 1s is solely determined by the 'high' part. If the current digit is 1, the number of 1s is influenced by both the 'high' and 'low' parts. If the current digit is greater than 1, the number of 1s is determined by the 'high' part increased by 1.

Step-by-Step Algorithm

Step 1: Initialize a counter 'count' to 0. This variable will store the total number of 1s.
Step 2: Initialize a variable 'i' to 1. This variable represents the current digit place (1, 10, 100, 1000, etc.).
Step 3: Iterate while 'i' is less than or equal to 'n'. In each iteration, we process one digit place.
Step 4: Calculate 'high': The integer formed by digits to the left of the current digit. high = n / (i * 10)
Step 5: Calculate 'curr': The digit in the current place. curr = (n / i) % 10
Step 6: Calculate 'low': The integer formed by digits to the right of the current digit. low = n % i
Step 7: If 'curr' is 0, the number of 1s in this place is 'high * i'. Add this to 'count'.
Step 8: If 'curr' is 1, the number of 1s in this place is 'high * i + low + 1'. Add this to 'count'.
Step 9: If 'curr' is greater than 1, the number of 1s in this place is '(high + 1) * i'. Add this to 'count'.
Step 10: Multiply 'i' by 10 to move to the next digit place (ones -> tens -> hundreds, etc.).
Step 11: After the loop finishes, return the 'count'.

Key Insights

Insight 1: We can iterate through each digit place (ones, tens, hundreds, etc.) and count the number of 1s appearing in that place.
Insight 2: For each digit place, we can determine the number of 1s based on the digits to the left (higher digits) and the digit in the current place.
Insight 3: We can divide the problem into three parts: the higher digits, the current digit, and the lower digits.

Complexity Analysis

Time Complexity: O(log(n))

Space Complexity: O(1)

Topics

This problem involves: Math, Dynamic Programming, Recursion.

Companies

Asked at: Atlassian, Google.

Number of Digit One | LeetCode 233

Hard

Math Dynamic Programming Recursion

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

Solution Code

Available in:

class Solution {
    public int countDigitOne(int n) {
        if (n <= 0) {
            return 0;
        }

        int count = 0;
        int i = 1;

        while (i <= n) {
            int high = n / (i * 10);
            int curr = (n / i) % 10;
            int low = n % i;

            if (curr == 0) {
                count += high * i;
            } else if (curr == 1) {
                count += high * i + low + 1;
            } else {
                count += (high + 1) * i;
            }
            
            i *= 10;
            
        }

        return count;
    }
}

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Problem Info

Description

Problem 233: Number of Digit One - Detailed explanation and solution approaches.

Topics

Math Dynamic Programming Recursion

Companies

AtlassianGoogle

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:233

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

11 steps

1
Step 1: Initialize a counter 'count' to 0. This variable will store the total number of 1s.
2
Step 2: Initialize a variable 'i' to 1. This variable represents the current digit place (1, 10, 100, 1000, etc.).
3
Step 3: Iterate while 'i' is less than or equal to 'n'. In each iteration, we process one digit place.
4
Step 4: Calculate 'high': The integer formed by digits to the left of the current digit. high = n / (i * 10)
5
Step 5: Calculate 'curr': The digit in the current place. curr = (n / i) % 10
6
Step 6: Calculate 'low': The integer formed by digits to the right of the current digit. low = n % i
7
Step 7: If 'curr' is 0, the number of 1s in this place is 'high * i'. Add this to 'count'.
8
Step 8: If 'curr' is 1, the number of 1s in this place is 'high * i + low + 1'. Add this to 'count'.
9
Step 9: If 'curr' is greater than 1, the number of 1s in this place is '(high + 1) * i'. Add this to 'count'.
10
Step 10: Multiply 'i' by 10 to move to the next digit place (ones -> tens -> hundreds, etc.).
11
Step 11: After the loop finishes, return the 'count'.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(log(n))

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity depends on the number of digits in n, which is logarithmic in n. The space complexity is constant as we only use a few integer variables.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Iterate through all numbers from 0 to 'n' and, for each number, count the number of '1's in its digits. This approach is simple but inefficient.

Time: O(n log(n)). For each of the 'n' numbers, we potentially iterate through all its digits, which is O(log n).Space: O(1)

Trade-offs: The brute-force approach is easy to understand but much slower for larger values of 'n'. It is a good approach for very small values of n, but the provided solution is far more efficient in general.

✨

Key Insights

1Insight 1: We can iterate through each digit place (ones, tens, hundreds, etc.) and count the number of 1s appearing in that place.
2Insight 2: For each digit place, we can determine the number of 1s based on the digits to the left (higher digits) and the digit in the current place.
3Insight 3: We can divide the problem into three parts: the higher digits, the current digit, and the lower digits.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Recursion

A function that calls itself to solve smaller versions of the same problem.

Edge Cases

Edge case 1: n = 0. The solution correctly returns 0 as the number of 1s.
Edge case 2: n = a small number like 1. The solution returns 1, which is also correct
Edge case 3: n = a number with multiple 1s in different positions, like 111. The solution accurately counts each occurrence of 1.
Edge case 4: A number containing zero digits, like 203. The solution handles these cases correctly in calculating the digits

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the number of 1s for each digit place. For example, not considering the 'low' digits when the 'curr' digit is 1.

✗Common mistake 2: Integer overflow. For large values of n, intermediate calculations (particularly when multiplying `high` and `i`) might exceed the maximum value of an `int`, leading to incorrect results (fixed in C and C++ examples by using `long long`).

✗Common mistake 3: Not handling the edge case when n is 0 correctly.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Analyzing website traffic. You might need to count the occurrences of a specific digit in user IDs or timestamps to identify patterns or anomalies.

🏦Real-world scenario 2: Financial analysis. When processing large datasets, one might need to count specific digits in transaction IDs or account numbers for data validation or statistical analysis.

🎮Real-world scenario 3: Counting specific events in log files. Say you want to count number of times error code 1XX appears in log files, this can be counted using same principles.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 24, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Number of Digit One"?

For this hard-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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